User:YohanN7/sandbox

First Mondave summer school

Color

 * $$\color{blue}{a} + \color{black}{b}$$

History
Vacuum polarization was first discussed in papers by P. A. M. Dirac and W. Heisenberg in 1934. Effects of vacuum polarization were calculated to first order in the coupling constant by R. Serber and E. A. Uehling in 1935.

Measurement of the Electromagnetic Coupling at Large Momentum Transfer

Linear Modifications in the Maxwell Field Equations

Polarization Effects in the Positron Theory

Representation theory of the Lorentz group
The Lorentz group is a symmetry group of the unbound Kepler problem. This is manifested by the Laplace–Runge–Lenz vector $D$ together with the angular momentum vector $L$. The components of these vectors form a Lie algebra under the Poisson bracket. In the bound problem, this Lie algebra is $so(4)$, but in the unbound problem, the Lie algebra is $so(3, 1)$. Thus the connected component $SO(3, 1)^{+}$ is isomorphic to the group of canonical transformations for this problem.

Symmetry groups


The mathematical notion of a group and the notion of symmetry in both mathematics and physics are intimately related. A group has the simple property that if one element of a group is multiplied by another, the result is another element of the group. The same can, mutatis mutandis, be said of symmetries. Apply one symmetry operation (physically or by changing coordinate system), and then another one. The result is that of applying a single symmetry operation. Group theory is thus the mathematical language in which symmetries of nature are expressed. Symmetries may relate to very concrete symmetries of physical objects, like the symmetries of a square. One then speaks of the symmetry group associated with the object, in this case the finite group Dihedral group $D_{2}$.

In the case of a rectangle, the symmetry group is a subgroup of $D_{2}$ called the Klein four-group $K_{4}$. Only one rotation, and two reflections in the plane, will make the transformed rectangle look exactly like it did before the symmetry operation.

Other objects possess higher symmetry. The sphere is an extreme example. It possesses full rotational symmetry and reflection symmetry. Rotate or reflect a ball with any kind of rotation or reflection about any plane through the origin, and it will look exactly the same as before the symmetry operation. In this case of the symmetries of a sphere, the matrix group is the orthogonal group of three dimensions. These are $3 × 3$ matrices.

Multiplication table
The basic feature of every finite group is its multiplication table, also called Cayley table, that records the result of multiplying any two elements. A representation of a group can be thought of new set of elements, finite-dimensional or infinite-dimensional matrices, giving the same multiplication table after mapping the old elements to the new elements in a one-to-one fashion.

The multiplication table of the Klein four-group is displayed on the right. The elements are the identity $V$, vertical reflection $V$, the horizontal reflection $H$, and a 180 degree rotation $E$.

The same holds true in the case of an infinite group like the rotation group SO(3) or the Lorentz group. The multiplication table is just harder to visualize in the case of a group of uncountable size (same size as the set of reals). One way to do this is to wellorder the elements of the group with an ordinal number $P$ being the order type. The "infinite Cayley table" is then indexed by two ordinals $Π(gh) = Π(g)Π(h)$ written on Cantor normal form.

Representations
A central fact is that the symmetry groups can be represented by matrices. In the case of $GL(V)$ for the rectangle, one matrix representation is composed of the four $H$ matrices:


 * $$E = \left(\begin{matrix}1&0\\0&1\end{matrix}\right), \qquad P = \left(\begin{matrix}1&0\\0&-1\end{matrix}\right), \qquad T = \left(\begin{matrix}-1&0\\0&1\end{matrix}\right), \qquad R = \left(\begin{matrix}-1&0\\0&-1\end{matrix}\right).$$

The matrices satisfy by direct calculation the multiplication table in the Cayley table above, and hence represent $B(H)$. The Klein four-group emerges as the zeroth homotopy group $GL(V)$ or component group of the Lorentz group.

Symmetry of space and time
Space itself possesses symmetry. It looks the same no matter how one rotates it, and the resulting rotational symmetry is referred to as isotropy of space. In the present case it is common to use passive rotations, meaning that the observer rotates himself. Mathematically, the active operation of a rotation is performed by multiplying position vectors by a rotation matrix. A passive rotation is accomplished by rotating only the basis vectors of the coordinate system. (The coordinate system can be thought of as being fixed in the rotated observer. The observer is physically rotated.) In this way, every point in space obtains new coordinates as if it was somehow physically rigidly rotated.

The Lorentz group contains all rotation matrices, extended to four dimensions with zeros in the first row and the first column except for the upper left element which is one, as elements.

There are, in addition, matrices that effect Lorentz boosts. These can be thought of, in the passive view, as (instantly!) giving the coordinate system (and with it the observer) a velocity in a chosen direction. A representative rotation and a representative boost are given by


 * $$R = \left(\begin{smallmatrix}1&0&0&0\\0&\cos \theta&-\sin \theta&0\\0&\sin \theta&\cos \theta&0\\0&0&0&1\end{smallmatrix}\right), \qquad B = \left(\begin{smallmatrix}\cosh \zeta&\sinh \zeta&0&0\\\sinh \zeta&\cosh \zeta&0&0\\0&0&1&0\\0&0&0&1\end{smallmatrix}\right),$$

where the rotation is $T$ radians about the z-axis and the boost is with rapidity $R$ in the x-direction.

Finally, two special transformations are used to invert the coordinate system in space, space inversion, and in time, time reversal. In the first case, the space coordinate axes are reversed. The latter is reversal of the time direction. This is thought of, in the passive view, as having the observer set his clock at minus what it shows and then have the clock's hands move counterclockwise. Physical time progresses forward. The identity $ρ$ and inversions $θ$ and $ς$, together with their product $0 ≤ α, β ≤ ρ$ have matrices


 * $$E = \left(\begin{smallmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{smallmatrix}\right), \qquad P = \left(\begin{smallmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{smallmatrix}\right), \qquad T = \left(\begin{smallmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{smallmatrix}\right), \qquad PT = \left(\begin{smallmatrix}-1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{smallmatrix}\right).$$

Direct calculation confirms that the new $K_{4}$ and $2 × 2$ represent the Klein four-group.

Mathematically the Lorentz group is defined as the set of transformations preserving the bilinear form


 * $$(ct_1, x_1, y_1, z_1) \cdot (ct_2, x_2, y_2, z_2) = -c^2t_1t_2 + x_1x_2 + y_1y_2 + z_1z_2,$$

where the left-hand side is the Minkowski inner product of two events in spacetime, and the right-hand side is the spacetime interval, see classical group for mathematical detail.

Infinite-dimensional unitary representations
There are several features of the general infinite-dimensional theory of representations of non-compact semisimple Lie groups that differs profoundly from that of compact or finite groups. There are agreeable features: These results surfaced already in the earliest investigations, referenced below.
 * 1) There are irreducible unitary representations that do not occur in the regular representation. The complementary series is of this sort.
 * 2) The regular representation may have (with respect to parameters) both discrete and continuous parts.
 * 1) There is an intimate connection between representations of the Lie group and the Lie algebra. This enables the study of general representations, not necessarily unitary, from a purely algebraic point of viw.
 * 2) The notion of characters to the non-compact case, making it possible to express a Plancherel formula.

Equivalence of representations
A G-map or an intertwining map of representations $$(\Pi, V), (\Sigma, W)$$ of a Lie group $$G$$ and vector spaces $$V, W$$ is a linear map $$A:V \rightarrow W$$ satisfying $$\Sigma \circ A = A \circ \Pi.$$. Two representations are isomorphic or equivalent if there exists bijective G-map (an isomorphism or equivalence). If $$V = W$$ and $$S$$ is an equivalence, then $$S$$ is a change-of basis matrix.

Lie group representations
The Lie correspondence may be employed for obtaining group representations of the connected component of the $E$. This is effected by taking the matrix exponential of the matrices of the Lie algebra representation. A subtlety arises if $P$ is not simply connected. This results in projective representations or $T$-valued representations that are actually representations of the universal covering group.

The Lie correspondence gives results only for the connected component of the groups, and thus the other components of the full group are treated separately representatives for matrices representing these components, one for each component. These form the zeroth homotopy group.

The Lie correspondence


The Lie correspondence for linear groups and Lie algebras is stated for reference. If $G$ denotes a linear Lie group (i.e. a group of matrices) and $$\mathfrak{g}$$ a linear Lie algebra (again a set of matrices), let $$\Gamma(\mathfrak{g})$$ denote the group generated by $$\exp(\mathfrak{g}),$$ the image of the Lie algebra under the exponential mapping (which is the matrix exponential in this case), and let $K_{2}$ denote the Lie algebra of $G$ (interpreted as the set of matrices $n$ such that $π_{0}$ for all $$t \in \R$$). The Lie correspondence reads in modern language, here specialized to linear Lie groups, as follows:


 * There is a one-to-one correspondence between connected linear Lie groups and linear Lie algebras given by $$G \leftrightarrow \mathfrak{g}$$ with $$\mathfrak{g} = \text{Lie}(G)$$ or, equivalently $$ G = \Gamma(\mathfrak{g}),$$ expressed as $$\Gamma(\text{Lie}(G)) = G,$$ respectively $$\text{Lie}(\Gamma (\mathfrak{g})) = \mathfrak{g}.$$ $G$

Lie algebra representations from group representations
Using the above theorem it is always possible to pass from a representation of a Lie group $G$ to a representation of its Lie algebra $$\mathfrak{g}.$$ If $PT$ is a group representation for some vector space $G$, then its pushforward (differential) at the identity, or Lie map, $$\pi : \mathfrak{g} \to \text{End} V$$ is a Lie algebra representation. It is explicitly computed using

Not all Lie algebra representations arise this way because their corresponding group representations may not exist as proper representations, i.e. they are projective, see below.

Group representations from Lie algebra representations


If $$\pi : \mathfrak{g} \to \mathfrak{gl}(V)$$ for some vector space $X$ is a representation, a representation $E, P, T$ of the connected component of $$ is tentatively defined by setting

It can be shown that simple connectedness of $G$ is a sufficient condition for $V$ to yield a representation, but it is not a necessary condition.
 * The simply connected case is the statement of the theorem If $$\pi : \mathfrak{g} \to \mathfrak{h}$$ is a Lie algebra homomorphism and $$ is simply connected, then there is a unique Lie group homomorphism $R = PT$ ''satisfying the first line in $V$.
 * If $V$ is not simply connected, then there is a unique representation $Lie(G)$ of the universal covering group $π$ of $V$ satisfying the same equation as the first equation in $G$. It is a consequence of the above theorem.
 * If the kernel of the covering map is included in the kernel of $e^{itX} ∈ G$, then the representation of $Π : G → GL(V)$ descends to a unique representation of $$. This is essentially a consequence of a variant of the first isomorphism theorem.
 * If the kernel of the covering map is not included in the kernel of $GL(V)$, then a (non-unique) projective representation of $G$ results.

All representations have the following properties:
 * Near the identity i.e. for $$ in a small enough open neighborhood, $G$ yields, by the Baker–Campbell–Hausdorff formula and that $Π$ is one-to-one on that neighborhood, a unique local homomorphism.
 * Representatives of elements $$ far off the identity are defined by selecting a path from the identity to $G$, partitioning it finely enough so that the above property can be used. The result using $G_{c}$ can then only depend on the homotopy class (in the standard representation of $G$) of the path used in the (attempted) definition of $exp$. In turn, this depends only on which $$ in the Lie algebra is used to represent an element $G$ in the standard representation (and is used in $G$.
 * In the simply connected case, there is only one homotopy class, and $X$ is unambiguous even far off the identity.
 * In the non-simply connected case, there are $Π$ homotopy classes, and $$ is to a certain extent ambiguous far off the identity.

A pictorial view of how the universal covering group contains all such homotopy classes, and a technical definition of it (as a set and as a group) is given in geometric view.

For example, when this is specialized to the doubly connected $Π$, the universal covering group is $$\text{SL}(2,\Complex)$$, and whether its corresponding representation $Π : G → H$ is faithful decides whether $Π_{c}$ is projective.

Talk:Electromagnetic tensor
This is a statement most would say is wrong. There can, on the face of things, be at most four since the entries are derived from the four-potential. Then there are the Maxwell equations, and gauge freedom, reducing the degrees of freedom to only two. It is true that there are field configurations satisfying any given value (respecting anti-symmetry) at an event. But the above needs to be clarified in the article.

Physics stack exchange
Irreducibility is not an absolute demand in the classification. For instance, photons of helicity $$\pm 1$$ are both considered to be the same particle. These helicities correspond each to a different one-dimensional representation of the two-dimensional Euclidean group, denoted $$\mathrm E(2)$$ or $$\mathrm{ISO}(2)$$, which is the "little group" that leaves a light-like vector invariant. These induce, in turn, different irreducible unitary representations of the Poincaré group.

The one-dimensionality of these representations is also one way to explain why helicity invariant under the connected component of the Lorentz group, and does not mix like spin $$z$$-component of ordinary spin does. Moreover, a would-be zero mass helicity $$0$$ particle does not exist since the corresponding representation of $$\mathrm E(2)$$ is the trivial representation.

However, helicity changes sign under space inversion, and since space inversion is a symmetry of the theory (QED), photons of helicity $$\pm 1$$ are both considered to be the same particle.

The representation theory of $$\mathrm E(2)$$ is detailed in Wu-Ki Tung, Group Theory in Physics (1985). It can be pursued by recycling the same method that leads to the unitary representation of the Poincare group, the method of induced representations (the Mackey machine).

Answer
According to [Wigner's classification](https://en.wikipedia.org/wiki/Wigner%27s_classification), the task in the case of a lightlike four-vector is to find all irreducible unitary representations of the "little group" that leaves this vector invariant, $\mathrm{ISO}(2)$. As it happens in this case, $\mathrm{ISO}(2)$ itself is amenable to the method of induced representations (i. e. the little group approach). The invariant subgroup is in this case isomorphic to the group of translations in two dimensions. (These translations have nothing to do with translations in spacetime, nor do the corresponding momentum operators correspond to those of the Poincare algebra.)

Now, given a vector in the plane, which is the little group (subgroup of the factor group $\mathrm{SO}(2)$) preserving standard vectors in the plane? It is, in the case of the null-vector all of $\mathrm{SO}(2)$. The unitary representations of $\mathrm{SO}(2)$ are all one-dimensional, labeled by an integer $m$ and act by multiplication by a phase factor. The null-vector is an eigenvector corresponding to eigenvalue zero of the momentum operators of $\mathrm{ISO}(2)$.

For a non-zero vector in the plane the short little group is the trivial group. The corresponding irreducible unitary representation is infinite-dimensional. This follows, as is outlined in the cited passage, from the commutation relations. Again, this vector is an eigenvector corresponding to eigenvalue zero of the momentum operators of $\mathrm{ISO}(2)$. These vectors do get "rotated" by the action of $\mathrm{SO}(2)$.

Another way to see that these are infinite-dimensional is to use another basis in which the $\mathrm{SO}(2)$ rotation operator is diagonal and to construct "raising" and "lowering" operators from the momentum operators that shift the eigenvalue of the rotation operator. For a given eigenvalue of $P^2$ (the Casimir operator of $\mathrm{ISO}(2)$), $m$ assumes all values $0, \pm1, \pm2,\ldots$.

This exhausts the irreducible unitary representations. The infinite-dimensional ones do not correspond to known particles, since there is no known quantum number (like spin z-component) that could vary in these representation spaces. Only the one-dimensional cases remains. (When including space-inversion, these become two-dimensional in the case of photons, but that is another matter.)

Affine combinations and barycenter
Collections $Π_{c}$ and $$\{\lambda_1, \ldots, \lambda_n\}$$, being $Π_{c}$ points in an affine space $$(A, \overrightarrow A)$$ and $Π_{c}$ elements  of the ground field of the associated vector space $$\overrightarrow A$$ respectively, are considered below.

If $$\lambda_1 + \cdots + \lambda_n = 0,$$ then for any two points $exp$ and $Π$,


 * $$\begin{align}\lambda_1 \overrightarrow{oa_1} + \cdots + \lambda_n \overrightarrow{oa_n}

&= \lambda_1(\overrightarrow{oo'} + \overrightarrow{o'a_1}) + \cdots \lambda_n (\overrightarrow{oo'} + \overrightarrow{o'a_n})\\ &= \lambda_1 \overrightarrow{o'a_1} + \cdots + \lambda_n \overrightarrow{o'a_n}.\end{align}$$

Thus this sum is independent of the choice of the origin, and for the resulting unambiguous vector in $$\overrightarrow A$$ the notation is $$v \equiv \lambda_1 a_1 + \cdots + \lambda_n a_n.$$ In particular, when $$n = 2, \lambda_1 = 1, \lambda_2 = -1,$$ the $$v$$ occuring in the definition of the subtraction of points is retrieved.

For $$\lambda_1 + \cdots + \lambda_n = 1, $$ $$g \in A$$ is defined, given $$o$$, by


 * $$\overrightarrow{og} \equiv \lambda_1 \overrightarrow{oa_1} + \cdots + \lambda_n \overrightarrow{oa_n}.$$

Then
 * $$\overrightarrow{o'g} = \lambda_1 \overrightarrow{o'a_1} + \cdots + \lambda_n \overrightarrow{o'a_n} = \lambda_1(\overrightarrow{o'o} + \overrightarrow{oa_1}) + \cdots \lambda_n (\overrightarrow{o'o} + \overrightarrow{oa_n}) = \overrightarrow{o'o} + \overrightarrow{og} = \overrightarrow{og},$$

where the second equality is by the first defining equality. Hence this quantity is also independent from the choice of the origin. Therefore, when the condition holds, the notation is $$g = \lambda_1 a_1 + \cdots + \lambda_n a_n. $$

The point $$g$$ is called the barycenter (or center of mass) of the $$a_i$$ for the weights $$\lambda_i.$$ The quantity $$g$$ is also called an affine combination of the $$a_i$$ with coefficients $$\lambda_i.$$

When letting the $$\lambda_i$$ (hence $$g$$) vary, subject only to the condition, an affine subspace results. With the additional condition $$0 \le \lambda_i \le 1$$, a simplex is obtained.

Representation theory of the Poincaré group
The Poincaré group is the full symmetry group of Albert Einstein's special relativity underlying all of foundational physics.

Induced representations
In order to expose the idea behind representations induced by representations of subgroups, it is advantageous to examine the finite-dimensional framework developed by Frobenius.

If a Lie group has an abelian invariant subgroup, and if this subgroup is unitarily represented on a space, then there is a subgroup of the factor group of the semidirect product that leaves this subspace invariant.

Lightlike four-vectors: Euclidean group E(2)
The little group leaving a timelike vector invariant is the two-dimensional Euclidean group. This group is a semidirect product


 * $$\mathrm E(2) = \mathrm{SO}(2) \rtimes \mathrm T(2),$$

where $$\mathrm T(2)$$ is the translation group in two dimensions. Since $$\mathrm T(2)$$ is abelian, the same analysis as pertain to the Poincaré group can be applied here. $$\mathrm E(2)$$ acts on the Euclidean plane $$\mathbb R^2$$.

Representation theory from the method of induced representations

 * For each $$m \in \mathbb Z$$ there is a one-dimensional unitary representation of the full $$\mathrm E(2).$$ It is labeled by $$(0, m)$$, where the first coordinate refers to the eigenvalue of the Casimir operator $$P^2$$, and the second coordinate is a further label referring to the eigenvalue of the Casimir operator $$J$$ of the little group $$\mathrm{SO}(2)$$. The action of the Lie algebra is given by

$g$
 * At the group level,

$g$
 * is obtained.


 * For each $$p > 0 \in \mathbb Z$$ there is an infinite-dimensional unitary representation of the full $$\mathrm E(2).$$ It is labeled by $$p$$, the square root of eigenvalue of the Casimir operator. The action of the Lie algebra is given by

$$


 * At the group level,

$G$

To derive these results, the standard representation on $$\mathbb R^2$$ is examined for subgroups leaving invariant a vector $$\mathbf p \in \mathbb R^2$$. There are only two cases. Either $$(p_1, p_2) = (0, 0)$$ in which case the little group is $$\mathrm{SO}(2)$$, or $$(p_1, p_2) \ne (0, 0)$$ in which case the little group is the trivial group $$1.$$

= Nonzero vector: The one-element group
= Here the labeling of states $$\left|p_1, p_2, \ldots\right\rangle$$ is introduced. By definition, the operators $$P_i$$ act by


 * $$\begin{align} P_1\left|p_1, p_2, \ldots\right\rangle &= \left|p_1, p_2, \ldots\right\rangle p_1,\\ P_2\left|p_1, p_2, \ldots\right\rangle &= \left|p_1, p_2, \ldots\right\rangle p_2.\end{align}$$

At the group level this is


 * $$e^{-i\mathbf a \cdot \mathbf P}\left|p_1, p_2, \ldots\right\rangle = \left|p_1, p_2, \ldots\right\rangle e^{-i\mathbf a \cdot \mathbf p},$$

where $$\mathbf p = (p_1, p_2).$$

The Lie algebra $$\mathfrak 1$$ of the little group $$1$$ is trivial, $$\mathfrak 1 = \{\emptyset\},$$ and $$1$$ has only one irreducible unitary representation, the trivial one.

To deduce the action of the full group $$\mathrm E(2)$$, the action of $$\mathrm {SO}(2)$$ is examined by examining the effect of $$P_k, k = 1, 2$$ on rotated states. To facilitate notation, write $$(p_1, p_2)$$ as $$\mathbf p.$$


 * $$\begin{align}

P_kR(\theta)\left|\mathbf p, \ldots\right\rangle &= R(\theta)R(\theta)^{-1}P_kR(\theta)\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[R(\theta)^{-1}P_kR(\theta)\right]\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[\sum_{l=1}^2R(-\theta)_{kl}P_l\right]\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[R(-\theta)_{k1}P_1 + R(-\theta)_{k2}P_2\right]\left|p_1, p_2, \ldots\right\rangle,\\ \end{align}$$

or


 * $$\begin{align}

P_1R(\theta)\left|\mathbf p, \ldots\right\rangle &= R(\theta)\left[R(-\theta)_{11}P_1 + R(-\theta)_{12}P_2\right]\left|p_1, p_2, \ldots\right\rangle = R(\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 \cos \theta + p_2 \sin \theta \right],\\ P_2R(\theta)\left|\mathbf p, \ldots\right\rangle &= R(\theta)\left[R(-\theta)_{21}P_1 + R(-\theta)_{22}P_2\right]\left|p_1, p_2, \ldots\right\rangle = R(\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 (-\sin \theta) + p_2 \cos \theta \right]. \end{align}$$

On infinitesimal form, this is


 * $$\begin{align}

P_1R(\delta\theta)\left|\mathbf p, \ldots\right\rangle &= R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 \cos \delta\theta + p_2 \sin \delta\theta \right] \approx R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 + p_2 \delta\theta \right],\\ P_2R(\delta\theta)\left|\mathbf p, \ldots\right\rangle &= R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 (-\sin \delta\theta) + p_2 \cos \delta\theta \right] \approx R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 (-\delta\theta) + p_2 \right]. \end{align}$$

Since this is deduced from the postulated behavior of $$P_1$$ and $$P_2$$ on a single vector, and the result are eigenvalues different from the postulated ones for the single vector, the only reasonable conclusion is that $$R(\theta)\left|\mathbf p\right\rangle$$ is a new eigenvector of $$P_1$$ and $$P_2$$ orthogonal to $$\mathbf p$$. Evidently,


 * $$\begin{align}R(\theta)\left|\mathbf p\right\rangle &= \left|\mathbf pR(\theta)\right\rangle\\J\left|\mathbf p\right\rangle &= \left|\mathbf pJ\right\rangle.\end{align}$$

Since elements are orthogonal matrices, the norm of $$\left|\mathbf pR(\theta)\right\rangle$$ is the same as the norm of $$\mathbf p.$$ Thus the eigenvalue of the Casimir operator remains the same, and an infinite-dimensional unitary representation of $$\mathrm E(2)$$ is characterized by this eigenvalue.

= Zero vector: Rotation group SO(2)
= Here the labeling is $$\left|0, 0\, \ldots\right\rangle$$ is introduced. The first two zeros refer to the eigenvalues of the $$P_i$$. They act by definition according to


 * $$\begin{align}P_1\left|0, 0, \ldots\right\rangle &= \left|0, 0, \ldots\right\rangle 0 = \emptyset,\\ P_2\left|0, 0, \ldots\right\rangle &= \left|0, 0, \ldots\right\rangle 0 = \emptyset.\end{align}$$

It remains to work out how the little group acts. If the representation is to be irreducible, it must be one-dimensional, since only one-dimensional irreducible representations of $$\mathbf{so}(2)$$ exist. In these representations, labeled by $$m \in \mathbb Z$$ the generator $$J$$ of $$\mathbf{so}(2)$$ acts by


 * $$J|0, 0, ...\rangle = |0, 0, \cdots\rangle m.$$

This suggests the labeling $$\left|0, 0, m\right\rangle, m \in \mathbb Z$$ for the basis vector.

At the group level, one obtains


 * $$\begin{align}e^{-i\mathbf a \cdot \mathbf P}|0, 0, m\rangle &= |0, 0, m\rangle,\\ e^{-i\theta J}|0, 0, m\rangle &= e^{-im\theta}|0, 0, m\rangle,\end{align}$$

As it happens, the actions of the abelian subgroup $$\mathrm T(2)$$ and of the little group $$\mathrm{SO}(2)$$ described so far exhausts the action of all of $$\mathrm E(2)$$.

Euclidean group E(2)
This group is a semidirect product


 * $$\mathrm E(2) = \mathrm{SO}(2) \rtimes \mathrm T(2),$$

where $$\mathrm T(2)$$ is the translation group in two dimensions. Since $$\mathrm T(2)$$ is abelian, the same analysis as pertain to the Poincaré group can be applied here. $$\mathrm E(2)$$ acts on the Euclidean plane $$\mathbb R^2$$.

Representation theory of the Euclidean group E(2)
The Euclidean group $card π_{1}$ in two dimensions is the group of isometries of the Euclidean plane. It is also denoted $SO(3, 1)^{+}$ provided reflections are excluded. The $Π_{c}$ stands for inhomogeneous, referring to the translational part, and $Π$ stands for special orthogonal, referring to the rotational part. Its elements are called rigid Euclidean motions. When reflections are included, the group is sometimes denoted ${a_{1}, ..., a_{n}}|undefined$ (but rarely $n$). The elements are then rigid motions.

Group multiplication rule
A rigid motion can be written as

$$x' = gx = Rx + b,\quad x', x, b \in \mathbb R^2, R \in \mathrm{SO}(2), g \in \mathrm{E}(2),$$

where the vector is first rotated in the plane and then a translation is added. The group has a faithful three-dimensional representation. The idea is to embed $n$ as the affine plane $o$ in $o'$. Then $E(2)$ is represented by $ISO(2)$, and

$$\left(\begin{matrix}x'\\1\end{matrix}\right) = \left(\begin{matrix}R&b\\0&1\end{matrix}\right)\left(\begin{matrix}x\\1\end{matrix}\right) = \left(\begin{matrix}x'_1\\x'_2\\1\end{matrix}\right) = \left(\begin{matrix}\cos \theta&-\sin\theta&b_1\\ \sin \theta& \cos \theta&b_2\\0&0&1\end{matrix}\right)\left(\begin{matrix}x_1\\x_2\\1\end{matrix}\right) = \left(\begin{matrix}Rx+b\\1\end{matrix}\right) = \left(\begin{matrix}Rx+b\\1\end{matrix}\right).$$

The group multiplication rule

$X$

and the inverse operation

$g$

are read off directly from the multiplication law.

Representation theory from the method of induced representations

 * For each $$m \in \mathbb Z$$ there is a one-dimensional unitary representation of the full $$\mathrm E(2).$$ It is labeled by $$(0, m)$$, where the first coordinate refers to the eigenvalue of the Casimir operator $$P^2$$, and the second coordinate is a further label referring to the eigenvalue of the Casimir operator $$J$$ of the little group $$\mathrm{SO}(2)$$. The action of the Lie algebra is given by

$$
 * At the group level,

$$
 * is obtained.


 * For each $$p > 0 \in \mathbb Z$$ there is an infinite-dimensional unitary representation of the full $$\mathrm E(2).$$ It is labeled by $$p$$, the square root of eigenvalue of the Casimir operator. The action of the Lie algebra is given by

$$


 * At the group level,

$$

To derive these results, the standard representation on $$\mathbb R^2$$ is examined for subgroups leaving invariant a vector $$\mathbf p \in \mathbb R^2$$. There are only two cases. Either $$(p_1, p_2) = (0, 0)$$ in which case the little group is $$\mathrm{SO}(2)$$, or $$(p_1, p_2) \ne (0, 0)$$ in which case the little group is the trivial group $$1.$$

Nonzero vector: The one-element group
Here the labeling of states $$\left|p_1, p_2, \ldots\right\rangle$$ is introduced. By definition, the operators $$P_i$$ act by


 * $$\begin{align} P_1\left|p_1, p_2, \ldots\right\rangle &= \left|p_1, p_2, \ldots\right\rangle p_1,\\ P_2\left|p_1, p_2, \ldots\right\rangle &= \left|p_1, p_2, \ldots\right\rangle p_2.\end{align}$$

At the group level this is


 * $$e^{-i\mathbf a \cdot \mathbf P}\left|p_1, p_2, \ldots\right\rangle = \left|p_1, p_2, \ldots\right\rangle e^{-i\mathbf a \cdot \mathbf p},$$

where $$\mathbf p = (p_1, p_2).$$

The Lie algebra $$\mathfrak 1$$ of the little group $$1$$ is trivial, $$\mathfrak 1 = \{\emptyset\},$$ and $$1$$ has only one irreducible unitary representation, the trivial one.

To deduce the action of the full group $$\mathrm E(2)$$, the action of $$\mathrm {SO}(2)$$ is examined by examining the effect of $$P_k, k = 1, 2$$ on rotated states. To facilitate notation, write $$(p_1, p_2)$$ as $$\mathbf p.$$


 * $$\begin{align}

P_kR(\theta)\left|\mathbf p, \ldots\right\rangle &= R(\theta)R(\theta)^{-1}P_kR(\theta)\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[R(\theta)^{-1}P_kR(\theta)\right]\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[\sum_{l=1}^2R(-\theta)_{kl}P_l\right]\left|p_1, p_2, \ldots\right\rangle\\ &= R(\theta)\left[R(-\theta)_{k1}P_1 + R(-\theta)_{k2}P_2\right]\left|p_1, p_2, \ldots\right\rangle,\\ \end{align}$$

or


 * $$\begin{align}

P_1R(\theta)\left|\mathbf p, \ldots\right\rangle &= R(\theta)\left[R(-\theta)_{11}P_1 + R(-\theta)_{12}P_2\right]\left|p_1, p_2, \ldots\right\rangle = R(\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 \cos \theta + p_2 \sin \theta \right],\\ P_2R(\theta)\left|\mathbf p, \ldots\right\rangle &= R(\theta)\left[R(-\theta)_{21}P_1 + R(-\theta)_{22}P_2\right]\left|p_1, p_2, \ldots\right\rangle = R(\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 (-\sin \theta) + p_2 \cos \theta \right]. \end{align}$$

On infinitesimal form, this is


 * $$\begin{align}

P_1R(\delta\theta)\left|\mathbf p, \ldots\right\rangle &= R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 \cos \delta\theta + p_2 \sin \delta\theta \right] \approx R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 + p_2 \delta\theta \right],\\ P_2R(\delta\theta)\left|\mathbf p, \ldots\right\rangle &= R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 (-\sin \delta\theta) + p_2 \cos \delta\theta \right] \approx R(\delta\theta)\left|p_1, p_2, \ldots\right\rangle\left[p_1 (-\delta\theta) + p_2 \right]. \end{align}$$

Since this is deduced from the postulated behavior of $$P_1$$ and $$P_2$$ on a single vector, and the result are eigenvalues different from the postulated ones for the single vector, the only reasonable conclusion is that $$R(\theta)\left|\mathbf p\right\rangle$$ is a new eigenvector of $$P_1$$ and $$P_2$$ orthogonal to $$\mathbf p$$. Evidently,


 * $$\begin{align}R(\theta)\left|\mathbf p\right\rangle &= \left|\mathbf pR(\theta)\right\rangle\\J\left|\mathbf p\right\rangle &= \left|\mathbf pJ\right\rangle.\end{align}$$

Since elements are orthogonal matrices, the norm of $$\left|\mathbf pR(\theta)\right\rangle$$ is the same as the norm of $$\mathbf p.$$ Thus the eigenvalue of the Casimir operator remains the same, and an infinite-dimensional unitary representation of $$\mathrm E(2)$$ is characterized by this eigenvalue.

Zero vector: Rotation group SO(2)
Here the labeling is $$\left|0, 0\, \ldots\right\rangle$$ is introduced. The first two zeros refer to the eigenvalues of the $$P_i$$. They act by definition according to


 * $$\begin{align}P_1\left|0, 0, \ldots\right\rangle &= \left|0, 0, \ldots\right\rangle 0 = \emptyset,\\ P_2\left|0, 0, \ldots\right\rangle &= \left|0, 0, \ldots\right\rangle 0 = \emptyset.\end{align}$$

It remains to work out how the little group acts. If the representation is to be irreducible, it must be one-dimensional, since only one-dimensional irreducible representations of $$\mathbf{so}(2)$$ exist. In these representations, labeled by $$m \in \mathbb Z$$ the generator $$J$$ of $$\mathbf{so}(2)$$ acts by


 * $$J|0, 0, ...\rangle = |0, 0, \cdots\rangle m.$$

This suggests the labeling $$\left|0, 0, m\right\rangle, m \in \mathbb Z$$ for the basis vector.

At the group level, one obtains


 * $$\begin{align}e^{-i\mathbf a \cdot \mathbf P}|0, 0, m\rangle &= |0, 0, m\rangle,\\ e^{-i\theta J}|0, 0, m\rangle &= e^{-im\theta}|0, 0, m\rangle,\end{align}$$

As it happens, the actions of the abelian subgroup $$\mathrm T(2)$$ and of the little group $$\mathrm{SO}(2)$$ described so far exhausts the action of all of $$\mathrm E(2)$$.

Tautological one-form
The tautological one-form is the unique horizontal one-form that "cancels" a pullback. That is, let


 * $$\beta:Q\to T^*Q$$

be any 1-form on $$Q$$ and consider it as a map from $$Q$$ to $$T^*Q$$, that is


 * $$\beta(q) = (q, p(q)), \quad p(q) \in T^*_q Q,$$

since one-form are sections.

Let $$\beta^*$$ denote the operation of pulling back by $$\beta$$. Then $$\beta^*$$ elaluated at $$(q, p(q)) \in T^*Q$$, denoted $$\beta^*_{(q, p(q))}$$, is a linear map from the cotangent space of the range $$T^*Q$$ at $$(q, p(q))$$, i. e.,


 * $$\beta^*_{(q, p(q))}:T^*_{(q, p(q)) \in T^*Q}(T^*_{(q, p(q))}Q) \rightarrow T^*_q(Q)$$

in all generality. Now since a one-form is a section, one has


 * $$\beta^*_q:T^*_{q }(T^*_{q}Q) \rightarrow T^*_q(Q)$$


 * $$\beta^*\theta = \beta$$,

which can be most easily understood in terms of coordinates:


 * $$\beta^*\theta = \beta^*(\sum_i p_i\, dq^i) =

\sum_i \beta^*p_i\, dq^i = \sum_i \beta_i\, dq^i = \beta.$$

So, by the commutation between the pull-back and the exterior derivative,


 * $$\beta^*\omega = -\beta^*d\theta = -d (\beta^*\theta) = -d\beta$$.

Weyl spinor
In theoretical physics a Weyl spinor is a set of two complex quantities that transform under Lorentz transformations under the $$(1/2, 0)$$ or $$(0, 1/2)$$ representation of the Lorentz group. The former are called left-handed and the latter right-handed. In the simplest description, a Weyl spinor is a column matrix with two complex entries. Weyl spinors may also be though of sitting in a Clifford algebra, in which case they are represented by complex matrices. The connection between the two descriptions is that the complex matrix can be expanded in a certain basis with two elements. The coefficients in this basis then constitute the corresponding column vector.

Weyl spinors are related to the Dirac spinor $$\Psi$$ via


 * $$\Psi = \left( \begin{matrix} \psi \\ \chi \end{matrix} \right).$$

In this expression,

Minkowski diagram
The term Minkowski diagram is used in both a generic and particular sense. In general, a Minkowski diagram is a graphic depiction of a portion of Minkowski space, usually where space has been curtailed to a single dimension. These two-dimensional diagrams portray worldlines as curves in a plane that correspond to motion along the spatial axis. The vertical axis is usually temporal, and the units of measurement are taken such that the light cone at an event consists of the lines of slope plus or minus one through that event. The horizontal lines corresponds to the usual notion of simultaneous events, for a stationary observer at the origin.

A particular Minkowski diagram illustrates the result of a Lorentz transformation. The Lorentz transformation relates two inertial frames of reference, where an observer stationary at the event (0, 0) makes a change of velocity along the $I$-axis. The new time axis of the observer forms an angle α with the previous time axis, with α &lt; π/4. See the first figure on the right. In the new frame of reference the simultaneous events lie parallel to a line inclined by α to the previous lines of simultaneity. This is the new $SO$-axis. Both the original set of axes and the primes set of axes have the property that they are orthogonal with respect to the Minkowski inner product or relativistic dot product.

Basics


The starting point is a subset of the Poincaré transformations relating the inertial frames of special relativity. A special inertial frame is chosen by a suitable spacetime translation and a suitable rotation in space such that the origin of the coordinate system $E^{+}(2)$ is located appropriately, e.g. one imagines a Lorentz observer stationary in space there, and the coordinate axes are directed appropriately, e.g. such that dynamical changes of, for example, a moving particle, or the motion of the origin of another inertial system effects one coordinate only, typically taken to be the $IO(2)$-coordinate. According to this setup, the remaining freedom in choice of inertial coordinates are Lorentz boosts.

In order to derive the properties of the diagram, one starts with knowledge of the properties of Lorentz transformations. Let $ℝ^{2}$ and $z = 1$ be the coordinates of any two events in spacetime. Lorentz transformations leave the bilinear form


 * $$x \cdot y = x^0y^0 - x^1y^1 - x^2y^2 -x^3y^3$$

invariant. This means in particular that if $ℝ^{3}$ and $x ∈ ℝ^{2}$ are the coordinates of one event described in different inertial systems related by a Lorentz transformation. that


 * $$x \cdot x = x^2 = \left(x^0\right)^2 - \left(x^1\right)^2 - \left(x^2\right)^2 -\left(x^3\right)^2 = x^\prime \cdot x^\prime = x'^2 = \left({x'}^0\right)^2 - \left({x^\prime}^1\right)^2 - \left({x^\prime}^2\right)^2 -\left({x^\prime}^3\right)^2.$$

The invariance was used justify the third equality sign.

At this time it is convenient to assume for the coordinates of the events of interest the simple expression $(x, 1) ∈ ℝ^{3}$, with reference to the appropriate choice of origin and orientation of the coordinate system described in the first paragraph.

Now draw a 2-dimensional diagram with $x$ labeling the vertical axis and $$ labeling the horizontal axis. Each point in this diagram represents a point in spacetime with coordinates $x$ where $4$ and $(0, 0, 0, 0)$ can be read off from the diagram using orthogonal projection onto the vertical axis and orthogonal projection onto the horizontal axis respectively.

The above expression simply becomes


 * $$ (ct)^2 - x^2 = (ct')^2 - x'^2,$$

where the coordinates $x$ and $x = (x^{0}, x^{1}, x^{2}, x^{3})$ are related by a Lorentz boost in the $y = (y^{0}, y^{1}, y^{2}, y^{3})$-direction. Considering a fixed event inside the light cone, i.e. a timelike event ($x = (x^{0}, x^{1}, x^{2}, x^{3})$, and writing


 * $$c^2t^2 - x^2 = R^2 > 0$$

one may solve for $x&prime; = (x&prime;^{0}, x&prime;^{1}, x&prime;^{2}, x&prime;^{3})$ and write


 * $$ct = \pm\sqrt{R^2 + x^2}.$$

This is the equation of a hyperbola of one sheet (for either the positive or the negative solution). This says that an event $x = (ct, x, y, z) = (ct, x, 0, 0) ≡ (ct, x)$ with coordinates $ct$ will be described in any coordinates $(ct, x, 0, 0)$}} related by a Lorentz boost in the $ct$-direction to the original coordinates such that when plotted in the original coordinates, the point $$ expressed in the new cordinates lies on the hyperbola. That is


 * $$ct' = \pm\sqrt{R^2 + x'^2}.$$

To recapitulate, a single timelike event can be marked by two points in a Minkowski diagram provided the two points are joined by one of the two branches of the hyperbolas of two sheets. Likewise, single spacelike event can be marked by two points in a Minkowski diagram provided the two points are joined by one of the hyperbolas of one sheet defined by


 * $$c^2t^2 - x^2 = R^2 < 0.$$

Affine Minkowski space
The following may be taken as a postulate:
 * Spacetime of special relativity is a four-dimensional affine space $x$.

Affine spaces and vector spaces are similar, with one significant difference being that affine spaces have no distinguished point, whereas vector spaces do, the origin. An $(ct, x)$-dimensional affine space $(ct&prime;, x&prime;)$ has a set of bijections $x$ with the property that the composite maps


 * $$\kappa_{\alpha\beta} = \phi_a \circ \phi_b^{-1}:\mathbb R^n \rightarrow \mathbb R^n; v \in \mathbb R^n \mapsto Lv + a \in \mathbb R^n,$$

are bijective affine transformations, i. e. $ct^{2} > x^{2})$ is bijective linear transformation and $$ is a vector in $ct$. In this situation $E$ is called the standard vector space of the affine space $$. The set of charts $(ct_{E}, x_{E}$, called an affine atlas is a smooth atlas on $$ and can be extended to a maximal atlas, giving $$ the structure of a smooth manifold. The maps $(ct&prime;_{E}, x&prime;_{E}$ are called affine coordinate transformations.

The notion of translation in $$ is defined as a map


 * $$t_a: A \rightarrow A; x \mapsto y$$

such that


 * $$\phi_\alpha(y) = \phi_\alpha(x) + a.$$

This is independent of $$. For every $x$ define


 * $$(x_1, y_1) \approx (x_2, y_2)$$

if


 * $$(x_2, y_2) = (t_a(x_1), t_a(y_1))$$

for some translation $M$. This is similar to the concept of parallel transport in a vector space, allowing for vectors to be moved around freely. Equivalence holds if and only if


 * $$y_1 = t_b(x_1), \quad y_2 = t_b(x_2)$$

for some translation $n$. The relation $A$ is an equivalence relation, and an equivalence class in $Φ_{α}: A → ℝ^{n}$ is characterized by the vector $L$. A vector space structure results on $ℝ^{n}$ with origin the class with $ℝ^{n}$. The ordered pairs should be thought of as displacement vectors (attached at some point) rather than true vectors (that can be freely moved around).

Space and Time
In his seminal paper termed Space and Time, Minkowski notes the symmetries associated with Newtonian mechanics with space and time seen as separate entities. Minkowski keeps in the beginning a distinction between "change in position" (invariance under spatial translation and invariance under spatial rotation), and "change in motion" (invariance under Galilean boosts). He also notes invariance under translations in time separately. In modern terms, Newtonian mechanics is invariant under Galilean transformations. This is the topic of Galilean relativity.

He then introduces the world manifold consisting of the totality of world-points $Φ_{α}$. This is Minkowski space as a set (so far stripped of further structure) and the elements are today called events. Minkowski's terminology lives on in the higher dimensional analogues world line and world sheet. The motivation:

"Subjects of our perception are always places and times connected. No one has observed a place except at a particular time, or has observed a time except at a particular place."

- Minkowski

Minkowski assumes for the sake of discussion that at every world-point something exists: "In order not to allow any yawning gap to exist, we shall suppose that at every place and time, something perceptible exists. In order not to say either matter or electricity, we shall simply use the word substance for this something. We direct our attention to the substantial point located at world-point x, y, z, t, ..."

- Minkowski

He deduces the concept of world-line as the spacetime history of such a substantial point, parametrized by $κ_{αβ}$. This would today probably be referred to as a possible world line of a massive point particle with positive mass.

For simplicity, assumption is made that the world is endowed with an origin and that one has four initially orthogonal axes. But then ,by considering Galilean boosts in the plane as transformations preserving the form of Newtonian mechanics, the time axis, initially assumed orthogonal to the space axes, is free to point in any upward direction. In order to establish a connection, Minkowski considers (this is motivated only later) the upper sheet of the hyperbola $(x, y) ∈ A × A$ with $t_{a}, a ∈ ℝ^{n}$ being an adjustable parameter.

I.e., in his words, he asks what the requirement of orthogonality in space has to do with this perfect freedom of the time-axis towards the upper half. This question is hard to understand, since the answer in special relativity is the non-freedom of the time axis is connected to behavior under boosts.

Élie Cartan 1923–24 considered this question in the context of a spacetime manifold with Newtonian gravitation (like Minkowski, though Minkowski does not mention gravitation specifically), but using the tools of differential geometry. It turns out Newtonian spacetime is curved! . Using this fact, and considering the most general transformations of spacetime preserving the canonical equations are the Galilean transformations, but these are allowed to depend on time. One consequence is the existence of special coordinates (in the same sense that Cartesian coordinates are special for Lorentz frames),


 * $$(t, x_1, x_2, x_3), x^i \cdot x^j = \delta_{ij} = \overline g_(\mathbb R^3),$$

where $t_{b}, b ∈ ℝ^{n}$ is the Euclidean metric on space This should be compared to


 * $$(ct, x^1, x^2, x^3), x^i \cdot x^j = \eta_{\mu\nu}x^\mu x^\nu$$

of Minkowski space. Thus there is a Riemannian metric on space, but this does not allow for extension to all of spacetime in the Newtonian setting. This corresponds to the "perfect freedom" of the time-axis in the language of Minkowski.

With the aid of a figure, an early Minkowski diagram, new values $≈$ and $A/≈$ are constructed so that again $b ∈ ℝ^{n}$. The connection between the parameter $A/≈$ and the diagram is such that in the limit $b = 0$ the hyperbola flattens out and approaches the $x, y, z, t$-axis. The group of transformations for $$ finite (with translations in spacetime and space rotations also included) is denoted $(−∞ < t < ∞)$ and that for infinite $x$ $x$. It is, argues Minkowski, in this context reasonable to identify $c^{2}t^{2} − x^{2} − y^{2} − z^{2} = 1$ with the full symmetry group of Newtonian mechanics (the Galilean group). Minkowski notes that a mathematician very well could have asked if some natural phenomena could be invariant under $c$ with $E$ finie. He adds that it would have been an extraordinarily triumph for pure mathematics.

The justification given for these transformations is the, by then known, invariance of the laws of electrodynamics under $\overline{g}_{ℝ^{3}}|undefined$ with $a$ being the speed of light, in other words, $x$ is the Poincare group.

Axiom: "The substance existing at any world point can always be conceived to be at rest, if time and space are interpreted suitably."

- Minkowski

Corollary: "The axiom means, that in a world-point the expression


 * ${\displaystyle c^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2},\,}$

shall always be positive or what is equivalent to the same thing, every velocity $A$ should always be smaller than $A$."

- Minkowski

It is important to recall Minkowski's consideration of substantial world-points only, for else the conclusions are false. In modern terminology, the speed of a massive particle is less than the speed of light in any intertial frame of reference, and for every massive particle, there is an instantaneous rest frame whether the particle is accelerating or not. The axiom, in the form of the restatement here, is sometimes referred to as the third postulate of relativity. .

Suppose now that mechanics obeys Galilean relativity (symmetry group $t$), but electromagnetic theory obeys special relativity (symmetry group $c^{2}t&prime;^{2} − x&prime;^{2} − y^{2} − z^{2} = 1$). The concept of a rigid body no sense in special relativity. {{Quote|text=Now if we have an optics with $c$, and on the other hand if there were rigid bodies, it is easy to see that one $c → ∞$-direction is preferred by the two hyperboloidal shells belonging to the groups $x$}, and $G_{c}$, which would have got the further consequence, that by means of suitable rigid instruments in the laboratory, we can perceive a change in natural phenomena, in case of different orientations with regard to the direction of progressive motion of the earth.|author=Minkowski|source=Space and Time}}

This is ruled out by experiment. The null result result of the Michelson–Morley experiment had previously prompted researchers to postulate that rigid bodies suffers a length contraction in the direction of motion relative to the ether, the Lorentz–FitzGerald contraction hypothesis.

"According to Lorentz every body in motion, shall suffer a contraction in the direction of its motion, namely at velocity $A$ in the ratio


 * $\frac{1}{\sqrt {1-   \frac {v^{2}} {c^{2}}     }}.$

This hypothesis sounds rather fantastical. For the contraction is not to be thought of as a consequence of resistances in the ether, but purely as a gift from above, as a condition accompanying the state of motion.|author=Minkowski|source=Space and Time|undefined"

Minkowski shows, using his diagram, that the exact same length contraction follows from the transformations of coordinates under $G_{∞}$.

He then proceeds to consider an arbitrary event $C_{∞}$ with coordinates $C_{∞}$ and defines the cone


 * $$c^2t^2 - x^2 - y^2 - z^2 = 0.$$

The part with $G_{c}$ is called the fore-cone, and the part with $G_{c}$ the aft-cone. In modern terminology, one has the light cone, the past light cone and the future light cone. The regions enclosed by the cones are called the for-side (of $A$) and the aft-side, today absolute past and absolute future'' respectively. Time-like vectors (those sitting inside the light cone) and space-like vectors are defined. It is noted that any time-like vectors can be used to define a new time-axis. Likewise, any space-like vector $α$ can be used to define axes such that the events at $c$ and $c$ occur at either the same time or event $c$ either preceding or following $c$ in time.

Minkowski defines two vectors $G_{c}$ and $G_{c}$ as normal if


 * $$c^2tt_1 - xx_1 - yy_2 - zz_1 = 0,$$

where one vector is in a unit hyperboloid inside the light cone, and the other is the unit hyperboloid od space-like vectors, by which is meant they are orthonormal.

Minkowski goes on to define proper time:

"Let us now fix our attention upon the world-line of a substantial point running through the world-point $G_{c}$; then as we follow the progress of the line, the quantity


 * $d\tau ={\frac {1}{c}}\sqrt {c^2dt^2-dx^2-dy^2-dz^2},$

corresponds to the time-like vector-element $t$.

The integral


 * $\tau =\int d\tau$

of this sum, taken over the world-line from any fixed initial $G_{∞}$ to any variable endpoint $v$, may be called the "proper-time" of the substantial point in $c$.|author=Minkowski|source=Space and Time|undefined"

Two things may be noted: This applies to the "substantial world points" of Minkowski – massive point particles of positive mass. The definition of proper time is fully Lorentz invariant, since the quantity under he square root is (by definition of $G_{c}$ or ($G_{c}$)). Niether Parity (physics) nor time reversal, which are Lorentz transformations, has any effect on proper time. Thus it is a Lorentz scalar, defined along a time-like curve in spacetime.

It is clear that proper time can be used as a parameter of the curve a massive particle traces out in spacetime, i.e. $O$. By differentiation w r t $v$ one obtains


 * $$ \begin{align}c^2\dot t^2 - \dot x^2- \dot y ^2 - \dot z^2 &= c^2\\

c^2\dot t\ddot t -\dot x\ddot x - \dot y\ddot y - \dot z\ddot z & = 0\end{align}$$

The first equation says that the four-velocity is constant. The second equation says that the particles four-acceleration is always orthogonal to its four-velocity.

Outer measure
An outer measure on a set $$X$$ is a function $$\phi$$ from the power set $$2^X$$ of $$X$$ to the non-negative extended reals $$[0, \infty]$$, $$\phi: 2^X \rightarrow [0, \infty],$$ with the property that for $$A \subset X$$ and any countable set $$\mathcal F \subset 2^X$$

Remarks
 * It is understood that countable set means either finite set or countably infinite.
 * It is understood that there is a unique summation operator defined on all functions on all sets with range $$[0, \infty]$$. In fact, the summation operator is defined more generally, roughly speaking, it is defined whenever $$\infty - \infty$$ does not occur when the formal definition is applied to a function with range all of the extended reals.
 * The condition $$\mathcal F$$ can be paraphrased as $$\mathcal F$$ is a countable cover of $$A.$$

Basic properties

 * The empty collection $$\mathcal F = \emptyset$$ is a countable cover of $$\emptyset$$,


 * where and the last equality is by definition of the summation operator.


 * If $$A \subset B \subset X$$ then $$\mathcal F = \{B\}$$ covers $$A$$, hence the property of monotonicity,


 * Let $$A = \bigcup_{i=1}^\infty A_i.$$ Then $$\mathcal F = \{A_1, A_2, \ldots\}$$ covers $$A$$, whence


 * which is the property of countable subadditivity.

The properties $O$–$S$, here deduced, are sometimes taken as the defining properties for any function $$\phi: 2^X \rightarrow [0, \infty]$$ to be an outer measure. The property $O$ is then a derived property.

Lorentz transformation
Let an object be at rest at the origin a Lorentz frame (unprimed) and let the particle velocity be $$u'$$ in another frame (primed). The coordinates of the object in the two frames are related through


 * $$x'^\alpha = {\Lambda^\alpha}_\beta x^\alpha.$$

Taking differentials,


 * $$dx'^\alpha = {\Lambda^\alpha}_\beta dx^\alpha.$$

Since the object is at rest in the unprimed frame, $$d\mathbf x = 0 \Rightarrow \beta = 0$$, thus

$$\begin{align}dx'^i &= {\Lambda^i}_0 x^0 = {\Lambda^i}_0 cdt, \\dx'^0 = cdt' &= {\Lambda^0}_0 dx^0 = {\Lambda^0}_0 cdt.\end{align}$$

Divide the upper equation by the lower to obtain


 * $$\frac{dx'^i}{dx'^0} = \frac{{\Lambda^i}_0cdt}{{\Lambda^0}_0cdt} \Rightarrow \frac{{\Lambda^i}_0}{{\Lambda^0}_0} = \frac{u'^i}{c},$$

so

The Lorentz group is characterized by


 * $$\Lambda^{\mathrm T}\eta \Lambda = \eta, \quad \Lambda\eta^{-1}\Lambda^{\mathrm T} = \eta^{-1} (= \eta),$$

or in component form


 * $${\Lambda^\alpha}_\gamma {\Lambda^\beta}_\delta \eta_{\alpha\beta} = \eta_{\gamma\delta}, \quad {\Lambda^\mu}_\alpha {\Lambda^\nu}_\beta \eta^{\alpha\beta} = \eta^{\mu\nu}.$$

where


 * $$\eta = \eta^{-1} = \left( \begin{matrix}-1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{matrix}\right)$$

is read off from the bilienar form, see O(p, q). The matrix $$\eta$$ corresponds to the matrix $$I_{p,q}$$ with $$p=3$$ and $$q=1$$ in the linked article. Put $$\gamma = \delta = 0$$ in these relations. Then

Equations $S$ and $O$ can be solved to yield


 * $$\begin{align}{\Lambda^0}_0 &= \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}},\\{\Lambda^i}_0 &= {\Lambda^0}_i = \frac{u'^i}{c}\gamma \equiv \beta_i\gamma.\end{align}$$

For the standard configuration, where $$\mathbf u'$$ of the particle is taken in the negative $(0, 0, 0, 0$-direction (the particle is at rest in the unprimed frame and moves) and no rotation is involved, this means

$$\left(\begin{matrix}ct'\\x'\\y'\\z'\end{matrix}\right) = \left( \begin{matrix}\gamma & \beta\gamma & 0 & 0\\\beta\gamma & \gamma & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{matrix}\right)\left(\begin{matrix}ct\\x\\y\\z\end{matrix}\right).$$

The $${\Lambda^2}_2$$ entry was found using the condition $$\det \Lambda = 1$$.

From this one reads off


 * $$\begin{align}

x' &= \gamma(x+\beta ct)\\ ct' &= \gamma(ct+\beta x).\end{align} $$

Wave function
Using the Dirac bra-ket machinery, it is relatively simple to derive the Schrödinger equation in various representations. In abstract notation, it reads


 * $$i\hbar \frac{\partial}{\partial t}\Psi\rangle = \hat H \Psi\rangle.$$

In order to obtain a coordinate representation of it, one may use a resolution of the identity,


 * $$\int_{-\infty}^{+\infty} |x\rangle\langle x| = \mathrm{Id},$$

where $ct < 0$ is the identity operator on Hilbert space. The kets $ct > 0$ form, since the position operator is Hermitian, a complete set by the spectral theorem. Multiply the abstract Schrödinger equation from the left with the bra corresponding to $(ct, x, y, z)$, and insert a resolution of the identity in the right spot,


 * $$i\hbar \frac{\partial}{\partial t}\langle x|\Psi\rangle = \int_{-\infty}^{+\infty} \langle x|\hat H |x'\rangle\langle x'|\Psi\rangle dx',$$

or


 * $$i\hbar \frac{\partial}{\partial t}\Psi(x) = \hat H \Psi(x).$$

For a free particle one has in the coordinate representation


 * $$\hat H = \frac{\hat P^2}{2m} = \frac{(-i\hbar)^2}{2m}\frac{\partial^2}{\partial x^2},$$

which may be taken as a basic postulate, see canonical quantization.

Wigner's theorem
Wigner's theorem, proved by Eugene Wigner in 1931, is a cornerstone of the mathematical formulation of quantum mechanics. The theorem specifies how physical symmetries such as rotations, translations, and CPT are represented on the Hilbert space of states.

According to the theorem, any symmetry transformation of ray space is represented by a linear and unitary or antilinear and antiunitary transformation of Hilbert space. The representation of a symmetry group on Hilbert space is either an ordinary representation or a projective representation.

Invariance principles derived from symmetry transformations play an important role in physics and, as Wigner noted in [Houtappel et al., 1965, they serve as a guide in the search for new laws of nature and as tools for obtaining properties of the solutions of equations provided by the laws of nature. Moreover, representations of symmetry groups, in particular the Lorentz group, though not sufficient to replace the quantum mechanical equations entirely, can replace them to a large extent. One can say that representations can replace the equation of motion, it cannot replace, however, connections holding between operators at one instant of time. It may be mentioned, finally, that these developments apply not only in quantutm mechanics, but also to all linear theories, e.g., the Maxwell equations in empty space. The only difference is that there is no arbitrary factor in the description and the o, can be omitted in (3a) and one is led to real representations instead of representations up to a factor.

History
In his book in 1931 book, Wigner postulated that the transition probability between two states has an invariant physical sense. This assumption led him to consider transformations of the states of a physical system which preserve the transition probability associated to any pair of states. He proved that any such transformation, called a symmetry transformation, is induced either by a unitary or by an antiunitary operator on the Hilbert space associated to the physical system. This is the result that is known as Wigner's theorem.

From the mathematical point of view, the proof given by Wigner was incomplete, and, according to some, not rigorous, if not incorrect. More than thirty years after the publication of Wigner's book, the first complete proof of his theorem were given in and in. The alleged incorrectness or non-rigor in Wigner's proof is contested in, where Wigner's original proof is spelled in detail. According to Bargmann, Uhlhorn's result is interesting though, because the premises from which the theorem is proved are different, and it highlights the connection between Wigner's theorem and the first fundamental theorem of projective geometry.

In the last two decades, L. Molnár has renewed the subject, especially with regard to its mathematical aspects, and several new proofs and generalizations have been published in recent years. See Outline of proof of Wigner's theorem below.

Rays and ray space
It is a postulate of quantum mechanics that vectors in Hilbert space that are scalar nonzero multiples of each other represent the same pure state. A ray is a set
 * $$\underline{\Psi} = \{e^{i\alpha}\Psi|\alpha \in \mathbb{R}\}, \Psi \in \mathcal{H},$$

and a ray whose vectors have unit norm is called a unit ray. There is a one-to-one correspondence between physical pure states and unit rays. The space of all rays is called ray space. Ray space is sometimes awkward to work with. It is, for instance, not a vector space with well-defined linear combinations of rays. But a transformation of a physical system is a transformation of states, hence mathematically a transformation of ray space. In quantum mechanics, a transformation of a physical system gives rise to a bijective ray transformation $(ct_{1}, x_{1}, y_{1}, z_{1})$ of ray space,
 * $$T: \underline{\Psi} \subset \mathcal{H} \mapsto \underline{\Psi'} = T\underline{\Psi} \subset \mathcal{H'}.$$

Symmetry transformations
Intuitively, a symmetry transformation is a change in which "nothing happens" or a change of our view that does not change the outcomes of possible experiments. For example, translating a system in a homogeneous environment should have no effect on the system. Likewise for rotating a system in an isotropic environment. This becomes even clearer when one considers the mathematically equivalent passive'' transformations, i.e. simply changes of coordinates and let the system be. Usually, the domain and range Hilbert spaces are the same. An exception would be (in a non-relativistic theory) the Hilbert space of electron states that is subjected to a charge conjugation transformation. In this case the electron states are mapped to the Hilbert space of positron states and vice versa. To make this precise, introduce the ray product,
 * $$\underline{\Psi} \cdot \underline{\Phi} = |\langle \Psi, \Phi\rangle|,$$

where $P(x, y, z, t)$ is the Hilbert space inner product. A ray transformation is called a symmetry transformation if
 * $$T \underline{\Psi} \cdot T\underline{\Phi} = \underline{\Psi} \cdot \underline{\Phi}, \quad \forall \Psi, \Phi \in \mathcal{H}.$$

In particular, unit rays are taken to unit rays. The significance of this definition is that transition probabilities are preserved. In particular the Born rule, another postulate of quantum mechanics, will predict the same probabilities in the transformed and untransformed systems,
 * $$P(\Psi \rightarrow \Phi) = |\langle\Psi, \Phi\rangle|^2 = [\underline{\Psi} \cdot \underline{\Phi}]^2 = [T\underline{\Psi} \cdot T\underline{\Phi}]^2 = |\langle\Psi', \Phi'\rangle|^2 = P(\Psi' \rightarrow \Phi'), \quad \Psi' \in T\underline{\Psi}, \Phi' \in T\underline{\Phi}.$$

It is clear from the definitions that this is independent of the representatives of the rays chosen.

Symmetry groups
Some facts about symmetry transformations that can be verified using the definition:
 * The product of symmetry two transformations, i.e. two symmetry transformations applied in concession, is a symmetry transformation.
 * Any symmetry transformation has an inverse.
 * The identity transformation is a symmetry transformation.
 * Multiplication of symmetry transformations is associative.

The set of symmetry transformations thus forms a group, the symmetry group of the system. Some important frequently occurring subgroups in the symmetry group of a system are realizations of These groups are also referred to as symmetry groups of the system.
 * The symmetric group with its subgroups. This is important on the exchange of particle labels.
 * The Poincaré group. It encodes the fundamental symmetries of spacetime.
 * Internal symmetry groups like SU(2) and SU(3). They describe so called internal symmetries, like isospin and color charge peculiar to quantum mechanical systems.

Statement of Wigner's theorem
A linear transformation $S$ of Hilbert space is unitary if
 * $$\langle U \Psi, U \Phi\rangle = \langle \Psi, \Phi\rangle,$$

and an antilinear transformation is antiunitary if
 * $$\langle A \Psi, A \Phi\rangle = \langle \Psi, \Phi\rangle^* = \langle  \Phi, \Psi\rangle.$$

Given a unitary transformation $P$ of Hilbert space, define
 * $$T: \underline{\Psi} = \{e^{i\alpha}U\Psi| \alpha \in \mathbb{R}\} \mapsto \underline{\Psi'} = \{e^{i\beta}U\Psi| \beta \in \mathbb{R}\}.$$

This is a symmetry transformation since
 * $$T\underline{\Psi}\cdot T\underline{\Psi}=\underline{\Psi'}\cdot\underline{\Psi'}=|\langle e^{i\alpha}U\Psi, e^{i\beta}U\Phi\rangle| = |\langle \Psi, \Phi\rangle| = \underline{\Psi}\cdot\underline{\Phi}.$$

In the same way an antilinear antiunitary transformation of Hilbert space induces a symmetry transformation. One says that a transformation $dx, dy, dz, dt$ of Hilbert space is compatible with the transformation $P$ of ray space if for all $P_{0}$,
 * $$T\underline{\Psi} = \{e^{i\alpha}U\Psi|\alpha \in \mathbb R\}.$$

Transformations of Hilbert space by either a unitary linear transformation or an antiunitary antilinear operator are obviously then compatible with the transformations or ray space they induce per above. Wigner's theorem states a converse of this:
 * Wigner's theorem (1931): If $τ$ and $$ are Hilbert spaces and if
 * $$T:\underline{\Psi} \subset \mathcal H \mapsto \underline{\Psi'} \subset \mathcal K$$
 * is a symmetry transformation, then there exists a transformation $O(1, 3)$ which is compatible with $$ and such that $$ is either unitary and linear or antiunitary and antilinear if $O(3, 1)$. If $t = t(τ), x = x(τ), y = y(τ), z = z(τ)$ there exists a unitary transformation $x$ and an antiunitary transformation $Id$, both compatible with $$''.

Proofs can be found in, and.

Representations and projective representations
A transformation compatible with a symmetry transformation is not unique. One has the following.
 * Theorem: If $$ and $$ are two additive transformations of $$ onto $$, both compatible with the ray transformation $$ with $|x⟩$, then
 * $$V = Ue^{i\alpha}, \alpha \in \mathbb R.$$

If $$ is a symmetry group in this latter sense, and if $|x⟩$ with $T$, then
 * $$T(f)T(g) = T(h),$$

where the $$ are ray transformations. From the last theorem, one has for the compatible representatives $U$,
 * $$U(f)U(g) = \omega(f, g)U(fg) = e^{i\xi(f,g)}U(fg),$$

where $⟨Ψ,Φ⟩$ is a phase factor. The function $U$ is called a $U$-cocycle or Schur multiplier. A map $Ψ$ satisfying the above relation for some vector space $T$ is called a projective representation or a ray representation. If $V:H → K$, then it is called a representation. One should note that the terminology differs between mathematics and physics. In the linked article, term projective representation has a slightly different meaning, but the term as presented here enters as an ingredient and the mathematics per se is the same. Applying the last relation (several times) to the product $H$ and appealing to the known associativity of multiplication of operators on $K$, one finds
 * $$\begin{align}\omega(f,g)\omega(fg,h) &= \omega(g,h)\omega(f,gh),\\

\xi(f,g) + \xi(fg,h) &= \xi(g,h) + \xi(f,gh) \quad (\operatorname{mod} 2\pi).\end{align}$$ Upon redefinition of the phases,
 * $$U(g) \mapsto \hat{U}(g) = \eta(g)U = e^{i\zeta(g)}U(g),$$

which is allowed by last theorem, one finds
 * $$\begin{align}\hat{\omega}(g,h) &= \omega(g,h)\eta(g)\eta(h)\eta(gh)^{-1},\\

\hat{\xi}(g,h) &= \xi(g,h) + \zeta(g) + \zeta(h) - \zeta(gh) \quad (\operatorname{mod} 2\pi),\end{align}$$ where the hatted quantities are defined by
 * $$\hat{U}(f)\hat{U}(g) = \hat{\omega}(f, g)\hat{U}(fg) = e^{i\hat{\xi}(f,g)}\hat{U}(fg).$$

This freedom of choice of phases can be used to simplify the phase factors. In the case of the Lorentz group and its subgroup the rotation group SO(3), phases can be chosen such that $dim H ≥ 2$. For their respective universal covering groups SL(2,C) and Spin(3), it is even possible to have $dim H = 1$. The study of of redefinition of phases involves group cohomology. Two functions related as the hatted and unhatted versions of $U:H → K$ above are said to be cohomologous. They belong to the same second cohomology class, i.e. they are represented by the same element in $A:H → K$, the second cohomology group of $dim H ≥ 2$. If a an element of $f, g, h ∈ ''G$ contains the trivial function $fg = h$, then it is said to be trivial. The topic can be studied at the level of Lie algebras and Lie algebra cohomology as well.

Outline of proof of Wigner's theorem
Like for many foundational theorems, Wigner's theorem has been proved in a variety of ways. Approaches include a direct manipulation in Hilbert space, an algebraic approach allowing for generalization,, a complex analysis approach, , and variants. It has also been generalized in different directions. Generalizations of Wigner's theorem to indefinite inner product spaces, Hilbert modules, type I1 factors,, Banach spaces, quaternionic Hilbert spaces, etc. exist. The premises of the theorem have been altered, preservation of orthogonality (a weaker assumption than Wigner's assumption of preserved inner product for all vectors) suffices, and the requirement of bijectivity of the ray transformation and separability of the Hilbert space (the latter is equivalent to dropping the requirement of a countable basis) have been dropped.

Preliminaries
Among all proofs, Bargmann's is probably the most accessible. Knowledge of basics of inner product spaces and very basic theory of complex numbers suffice for comprehension. Wigner adopts definition $T$ for ray space.

Real multiplication of rays
The ray product is as before and, in addition, multiplication of rays by real numbers is defined by
 * $$\mathbf f\rho = \{e^{i\alpha}\mathbf f\rho\}, \quad \rho \in \mathbb R.$$

This multiplication satisfies
 * $$\begin{align}(\mathbf f\rho)\sigma &= (\mathbf f)\rho\sigma,\\

\mathbf f\rho \cdot \mathbf g\sigma &= (\mathbf f \cdot \mathbf g)\rho\sigma. \end{align}$$ Every ray may then be expressed as the product of a real number and a unit ray.

Symmetry transformations
Symmetry transformations are defined on all unit rays (but initially on unit rays only). They are demanded to preserve the ray product. The facts that symmetry transformations are one-to-one and onto follow at once. For unit rays $ω(f, g)$,
 * $$T\mathbf e_1 = T\mathbf e_2 \Rightarrow 1 = T\mathbf e_1\cdot T\mathbf e_2 = \mathbf e_1 \cdot \mathbf e_2 = (e_1, e_2) = ||e_1||||e_2|| = 1$$

for representatives $2$ of $U:G → GL(V)$. The second to last equality is equality in the Cauchy-Schwartz inequality, which holds only if $ω(f, g) = 1$ is a multiple of $ω(g, h) = ± 1$, necessarily here of modulus $ω(g, h) = 1$. Hence $ω$ and $V$ is injective. Then $T$ has a left inverse $H^{2}(G)$, a symmetry transformation required to preserve the ray product as well. Let $G$ be a maximal orthonormal set in $U$. Then $H^{2}(G)$ is an orthonormal set. It is complete, for if not, then there is a nonzero $V$ in the range orthogonal to all $ω = 0$. Under the left inverse, $e_{1}, e_{2}$ then has zero ray product with every $e_{1}, e_{2}$ contradicting the completeness of $e_{1}, e_{2}$. Thus $e_{1}$ is complete and $H$ is surjective.

Conclusion of theorem
The theorem guarantees the existence of an operator $K$ with the following properties:

and

for $T$ being either the identity or complex conjugation. Only property $G$ is used in the construction, together with consistency with previous definitions. The properties in $T$ will follow.

Independence of unitary-antiunitary character of U on realization
The proof amounts to constructing an operator with these properties. The unitarity or antiunitarity of the operator the theorem asserts the existence of is determined by $U$. To see this, define a map from $e_{2}$ to $1$,
 * $$\Delta(\mathbf a_1, \mathbf a_2, \mathbf a_3) = (a_1,a_2)(a_2,a_3)(a_3,a_1),$$

which is, using properties of the inner product on $ω$, seen to be independent of representatives. Now
 * $$\begin{align}\Delta(T \mathbf a_1, T \mathbf a_2, T \mathbf a_3) &= (a'_1, a'_2)(a'_2, a'_3)(a'_3, a'_1) = (Ua_1, Ua_2)(Ua_2, Ua_3)(Ua_3, Ua_1)\\

&= \chi(a_1, a_2)\chi(a_2, a_3)\chi(a_3, a_1) = \chi((a_1, a_2)(a_2, a_3)(a_3, a_1)) = \chi(\Delta(a_1,a_2,a_3)), \end{align}$$ where $V$ and properties of $fgh$ are used. This can hold only if $H$ is unitary when $e_{1} = e_{2}$ is the identity and antiunitary when $T^{−1}$ is complex conjugation. This means that the character of $$ in this respect does not depend on any arbitrary choice in the proof and can be determined without construction of $T$.

Extension of T
Before construction of $T$ the unit ray map $H$ is extended from unit rays to all of ray space by
 * $$T(\mathbf e\lambda) = (T\mathbf e)\lambda, \quad \lambda \in \mathbb R.$$

One has

the first two following from the definitions, and the third from the second.

Lemma 1
Suppose now that $f$ has been constructed in a subspace (at least one-dimensional). Then, for $T$ in this subspace,

Since $$ by $U$, the first equality follows from $$. Likewise, $To$, where the phase factor was absorbed into $e ' _{α} Te_{α} ⊂ H '$. Since $T^{−1}f$, the second assertion is true provided $e_{α}$ satisfies the two assertions in $$. Equation $χ$ is evident from $$ and $$.

Expansion coefficients in the range
Consider an orthonormal set,
 * $$\mathbf f_\rho, 1 \le \rho \le m, \quad \mathbf f_\rho \cdot \mathbf f_\sigma = \delta_{\rho\sigma},\quad \mathbf f'_\rho = T\mathbf f_\rho,\quad f'_\rho \in \mathbf f'_\rho.$$

with $T$ finite, and let Let
 * $$a = \sum f_\rho \alpha_\rho = \sum f_\rho (f_\rho, a).$$

Then if $o$ is any element in $To$,

Proof:
 * $$\begin{align}||a' - \sum_{\rho= 1}^m f'_\rho(f'_\rho, a')||^2 &= ||a'||^2

- \left(a', \sum_{\rho = 1}^m f'_\rho(f'_\rho, a')\right) - \left(\sum_{\rho = 1}^m f'_\rho(f'_\rho, a' ),a'\right) + \sum_{\rho=1}^m \sum_{\sigma=1}^m (f'_\rho(f'_\rho,a'), f'_\sigma(f'_\sigma,a'))\\ &=||a'||^2 - \sum_{\rho = 1}^m (a', f'_\rho)(f'_\rho, a') - \sum_{\rho = 1}^m (f'_\rho, a)(f'_\rho, a') + \sum_{\rho=1}^m (f'_\rho,a') (f'_\rho,a')\\ &=||a'||^2 - \sum_{\rho=1}^m |(a', f'_\rho)|^2 = ||a||^2 - \sum_{\rho=1}^m |(a, f_\rho)|^2 = ||a - \sum_{\rho= 1}^m f_\rho(f_\rho, a)||^2 = 0. \end{align}$$ The second to last equality, while not entirely obvious, follows by comparison with the far left and the first term in the last row. Orthonormality of the $R × R × R$ and the fact that $H$ preserves absolute values of inner products have been used. Hence $$ holds and $ℂ$ since
 * $$|(f'_\rho, a')| = \mathbf f_\rho \cdot \mathbf a' = T\mathbf f_\rho \cdot T\mathbf a = \mathbf f_\rho \cdot \mathbf a = |(f_\rho, a)| = |\alpha_\rho|.$$

Construction of U
Fix a unit ray $χ$ with $χ$. Fix $Ua ∈ Ta$ and define

This is arbitrary up to a factor of $U(aλ) = U(a|λ|e^{iarg λ}) ∈ Tae^{iarg λ}|λ| = Ta|λ|$. This is the only arbitrary choice made in the proof, in accord with that the final complete operator is unique up to the same factor.

Now decompose $a$. For any $U(a) ∈Ta$,
 * $$a = e\alpha + z, z \in \mathcal P, \quad \alpha = (e, a), \quad z = a - e\alpha.$$

First consider $χ$ for which $U$ = $χ_{a}$ or $a '$. Set $a ' = Ta$ and define $f_{ρ}$, so that $U$ is a unit vector, and let $α ' _{ρ} = (f ' _{ρ}, a ' )$ and $e$ be the corresponding rays. Accordingly, $Te = e '$, $e ∈ e, e ' ∈ e '$. For any $e^{iφ}, φ ∈ ℝ$ and $H = P ⊕ span e$,
 * $$a' = e'\alpha'_0 + f'\alpha'_1, \quad |\alpha'_0| = 1$$

according to $U$. But then
 * $$a'' = e' + f'\alpha'_1/\alpha'_0 = f'\frac{(a', f')}{(a', e')}\in \mathbf a'$$

as well. Now pick instead $a ∈ H$ and $0$ and apply the same procedure;
 * $$b'' = e' + g'\beta'_1/\beta'_0 = g'\frac{(b', g')}{(b', e')}\in \mathbf a'.$$

But, necessarily, $1$ and $a = e + z$ so
 * $$g'\frac{(b',g')}{(b',e')} = f'e^{i\delta}\frac{e^{i\gamma}}{e^{i\gamma}} \frac{(a',f')e^{-i\delta}}{(a',e')}= f'\frac{(a', f')}{(a', e')},$$

whence $f = z/||z||$. This means that $a, f$ contains a unique vector of the form $z$ with $z = f||z||$. Set
 * $$Vz = f'\beta'$$

and define

Since $||f|| = 1$ it is permissible to set

Analysis of the map V
Let $a ' ∈ a ' = Ta$ and let definition $U$ apply to both. From $T$ and $$ follows

and from $U$ and $a$ follows
 * $$|(e'+Vw,e'+Vx)|^2 = |(U(e+w),U(e+x)|^2 = |(e+w,e+x)|^2,$$

or, which is the same thing after expansion of both sides,
 * $$|1+(Vw,Vx)|^2 = |1+(w,x)|^2.$$

But
 * $$|1+(Vw,Vx)|^2 = 1 + 2\operatorname{Re}(Vw,Vx) + |(Vw,Vx)|^2 = 1 + 2\operatorname{Re}(Vw,Vx) + |(w,x)|^2 = 1 + 2\operatorname{Re}(w,x) + |(w,x)|^2,$$

whence

The second to last step uses $$ and the last step is just expansion of the RHS of the equation above. Also,

if $f ' ∈ f ' = Tf$ is real. An imaginary part in $b ' ∈ a '$ would violate $$.

Real part of the function χ
Now fix two vectors $g ' ∈ f '$ and define $b ' = a ' e^{iγ}$ and a second orthogonal unit vector $g ' = f ' e^{iδ}$ such that

and let $b = a$. By the same procedure as above, $a '$, and $e ' + f ' β '$ and $|β ' | = |z|$ are orthogonal. By $$,

Equation $$ gives
 * $$\operatorname{Re}(V(f_\rho\alpha),V(f_\rho\beta)) = \operatorname{Re}(U(f_\rho\alpha),U(f_\rho\beta)) = \operatorname{Re}(Uf_\rho\chi_{f_\rho}(\alpha),Uf_1\chi_{f_\rho}(\beta)) =\operatorname{Re} \chi_{f_\rho}(\alpha)^*\chi_{f_\rho}(\beta) = \operatorname{Re}\alpha^*\beta.$$

With $$,

and with $Vz = f ' β ' ∈ Tf||z|| = Tz$ real, using $$,

Coefficients α in terms of the function χ
Now set
 * $$x = \sum_{\rho=1}^2f_\rho\alpha_\rho.$$

Then, by $$,
 * $$x' = Vx = \sum_{\rho=1}^2f'_\rho\alpha'_\rho, \quad |\alpha'_\rho| = |\alpha_\rho|.$$

Put $w, x ∈ P$. Then
 * $$(f_\rho\gamma_\rho,f_\rho\alpha_\rho) = (f_\rho\gamma_\rho,x) = 1 \in \mathbb R,$$

hence by $m$, $$ and orthonormality,
 * $$\chi_\rho(\gamma_\rho)^*\chi_\rho(\alpha_\rho) = (Vf_\rho\gamma_\rho,Vf_\rho\alpha_\rho) = (Vf_\rho\gamma_\rho,x') = \chi_\rho(\gamma)^*(f'_\rho, x')

= \chi_\rho(\gamma_\rho)^*\alpha'_\rho,$$ and thus

Independence of χ
Next set
 * $$w = \sum_{\rho=1}^2 f_\rho \Rightarrow Vw = \sum_{\rho=1}^2 f'_\rho, \quad V(w\alpha) = \sum_{\rho=1}^2 f'_\rho \chi_{f_\rho}(\alpha) = (Vw)\chi_w(\alpha) = \left(\sum_{\rho=1}^2 f'_\rho\right)\chi_w(\alpha),$$

which can hold only if
 * $$\chi_{f_1}(\alpha) = \chi_{f_2}(\alpha) = \chi_{f_w}(\alpha),$$

whence

Imaginary part of the function χ
Then set $(w,x)$. By $T$, $(Vw,Vx)$, and by $$, $y, z ∈ P$. Hence $f_{1} = y/|y|$. By equation $$,
 * $$\operatorname{Im}\chi_1(\beta) = \operatorname{Re}\chi_1(i^*\beta) = \eta\operatorname{Re}(\chi_1(i)\chi_1(\beta)) = \eta\operatorname{Re}(\chi_1(i)\chi_1(\beta)) = \eta \operatorname{Re}(i^*\beta) = \eta\operatorname{Im}(\beta).$$

Combination with $a$ yields
 * $$\chi_1(\beta) = \beta, \eta = 1, \quad \chi_1(\beta) = \beta^*, \eta = -1.$$

Whether $α$ is complex conjugation or the identity, it satisfies

The right properties of V
Finally, let $f_{2}$ with
 * $$w = \sum_{\rho=1}^2 f_\rho\alpha_\rho, \quad x = \sum_{\rho=1}^2 f_\rho\beta_\rho$$.

One has

Parts $f$ and $$ follow from $$ and observations $$ and $$ respectively. Part $$ is a consequence of $$ and observation $$.

Construction of U completed
Lastly, consider general vectors $L = span { f_{1}, f_{2} }$ for $f ' _{ρ} = Vf_{ρ} ∈ Tf_{ρ}$. Define $f ' _{1}$. Then $f ' _{2}$ and $β$. $γ_{ρ} = α*_{ρ}^{−1}$ is defined as before. But then $β = i$ and one may define

This definition coincides with the previous ones in case $|χ_{1}(i)| = 1$. Using $$ and $$ one confirms that $$ satisfies all advertized properties in $$,
 * $$\begin{align}U(a_1 + a_2) &= U(e(\alpha_1 + \alpha_2) + z_1 + z_1) = U\left(e + \frac{z_1 + z_2}{\alpha_1 + \alpha_2}\right)\chi(\alpha_1 + \alpha_2)\\

&= e'\chi(\alpha_1) + e'\chi(\alpha_2) + Vz_1 + Vz_2 = U(a_1) + U(a_2),\end{align}$$
 * $$\begin{align}U(a\lambda) &= U(e\alpha\lambda + z\lambda) = U\left(e + \frac{z\lambda}{\alpha\lambda}\right)\chi(\alpha\lambda)\\

&= \left(e' + V\frac{z\lambda}{\alpha\lambda}\right)\chi(\alpha\lambda) = e'\chi(\alpha)\chi(\lambda) + Vz\chi(\lambda) = U(a)\chi(\lambda),\end{align}$$
 * $$\begin{align}(Ua_1,Ua_2) &= (e'\chi(\alpha_1) + Vz_1, e'\chi(\alpha_2) + Vz_2) = \chi^*(\alpha_1)\chi(\alpha_2) + (Vz_1, Vz_2)\\

&= \chi^*(\alpha_1)\chi(\alpha_1) + \chi((z_1, z_2)) = \chi((e\alpha_1 + z_1,e\alpha_2 + z_2)) = \chi((a,b)).\end{align}$$

Dirac's expansors
To make concrete what a representation space of a unitary infinite-dimensional representation of the Lorentz group may look like, one may look at the publication. In this paper first an infinite-dimensional unitary representation of $Reχ_{1}(i) = 0$ is constructed and then extended to an infinite-dimensional unitary representation of $χ_{1}(i) = ±i ≡ ηi$. The $w, x ∈ L$ representation is not irreducible, it couldn't be because all its irreducible unitary representations are finite-dimensional. On the contrary, the $a = eα + z, z ∈ P$ representation is not reducible to finite-dimensional unitary representations because its irreducible unitary representations are all infinite-dimensional. While Dirac's proposed expansors found little relevance in physics, they were of considerable mathematical import. Its analysis of infinite unitary representations was later developed further by Harish-Chandra and Bargmann.

O(3)
Begin by considering power series
 * $$a_0 + a_1\xi_1 + a_2\xi_1^2 + \cdots = \sum_{r=0}^\infty a_r\xi_1^r$$

in the real variable $α ≠ 0, 1$ with real coefficients. Let $b = e + z/α$ be the inner product space of sequences $a = bα$ such that
 * $$\sum_{r=0}^\infty r! a_r^2$$

converges. Then
 * $$\langle (a_r)|(b_r)\rangle= \sum_{r=0}^\infty r! a_r b_r$$

is an inner product on this space and
 * $$ ||(a_r)|| = \sqrt{\sum_{r=0}^\infty r! a_r^2}$$

defines the norm. Define similar vector spaces $Ta = (Tb)|α|$ and $Ub ∈ Tb$ and define
 * $$V = V_1 \otimes V_2 \otimes V_3.$$

Also, define product operations
 * $$(V_1, V_2, V_3) \rightarrow V; ((a_r), (b_s), (c_t)) \mapsto (a_r) \otimes (b_s) \otimes (c_t) \in V.$$

By using bases,
 * $$\begin{align}e_0 &= (1,0,0,\ldots), e_1 = (0, 1, 0, \ldots), \ldots\\

f_0 &= (1,0,0,\ldots), f_1 = (0, 1, 0, \ldots), \ldots\\ g_0 &= (1,0,0,\ldots), g_1 = (0, 1, 0, \ldots), \ldots,\end{align}$$ one may write
 * $$(a_r) \otimes (b_s) \otimes (c_t) = \sum_{r, s, t \ge 0} A_{rst} e_r \otimes f_s \otimes g_t,$$

with
 * $$A_{rst}=a_rb_sc_t.$$

The elements of the "triple indexed sequence" $Ubχ(α) ∈ Ta$ may then be though of as coefficients in a power series
 * $$P = \sum_{r = 0}^\infty \sum_{s = 0}^\infty \sum_{t = 0}^\infty A_{rst}\xi_1^r\xi_2^s\xi_3^t,$$

the inner product being defined by
 * $$\langle (A_{rst})|(B_{rst})\rangle=\sum_{r = 0}^\infty \sum_{s = 0}^\infty \sum_{t = 0}^\infty r!s!t!A_{rst}B_{rst}.$$

Now apply an element of $α = 0, 1$ to the triple $O(3)$, $O(3, 1)$. Then
 * $$P = \sum_{r = 0}^\infty \sum_{s = 0}^\infty \sum_{t = 0}^\infty {A'}_{rst}{\xi'}_1^r{\xi'}_2^s{\xi'}_3^t,$$

with $O(3)$ ($O(3, 1)$ fixed) some linear combination of $ξ_{1}$ (as $V_{1}$ varies).

It turns out that
 * $$||(A'_{rst})|| = ||(A_{rst})||,$$

implying that the space $(a_{r})$ is a representation space of a unitary representation of $V_{2}$. Consider first the transformation
 * $$\xi_1 = \xi'_1 + \epsilon \xi'_2, \xi_2 = \xi'_2 - \epsilon \xi'_1, \xi_3 = \xi'_3.$$

Plug this into the expression for $V_{3}$,
 * $$P = \sum A_{rst}({\xi'_1}^r{\xi'_2}^s + r\epsilon{\xi'_1}^{r-1}{\xi'_2}^{s+1} - s\epsilon{\xi'_1}^{r+1}{\xi'_2}^{s-1}){\xi'_3}^t,$$

so
 * $$A'_{rst} = A_{rst} + (r+1)\epsilon A_{r+1,s-1,t} - (s+1)\epsilon A_{r-1,s+1,t}.$$

The squared norm is thus
 * $$\sum r!s!t!{A'}_{rst}^2 = \sum r!s!t![A_{rst}^2 + 2(r+1)\epsilon A_{rst}A_{r+1,s-1,t} - 2(s+1)\epsilon A_{rst}A_{r-1,s+1,t}] = \sum r!s!t!{A}_{rst}^2,$$

the last step by virtue of changing summation variable $(A_{rst})$ in the second term and $O(3)$ in the third.

The space $ξ ≡ (ξ_{1}, ξ_{2}, ξ_{3})$ is clearly infinite-dimensional, but by considering homogeneous polynomials of a fixed degree, say $ξ' = Rξ, R ∈ O(3)$, then the space is finite. In this case the $A'_{rst}$ can be seen as to equal certain components of a symmetrical tensor of rank $rst$. To see this, write
 * $$P = \sum A^{\iota_1, \iota_2, \ldots \iota_u} \xi_{\iota_1} \xi_{\iota_2} \cdots \xi_{\iota_u}.$$

Then there are
 * $$\frac{r!s!t!}{u!}$$

terms with $$ indices equal to $A_{rst}$, $$ indices equal to $rst$ and $$ indices equal to $V$. These terms are identical since the $O(3)$ commute, any part not symmetric in all indices would drop out, and one may write
 * $$A_{rst} = \frac{u!}{r!s!t!}A_{\iota_1\iota_2\ldots\iota_u}, \quad\iota_1, \ldots \iota_r = 1, \iota_{r + 1}, \ldots, \iota_{r + s} = 2, \iota_{r + s + 1}, \ldots, \iota_{r + s + t} = 3.$$

Due to the invariance of $P$, the latter expression for it implies that the quantity $r → r − 1$ must transform as a tensor of rank $$ because each $s → s + 1$ transforms as a vector.

The total space is clearly an infinite sum of such finite-dimensional spaces, so that the infinite-dimensional representation of $V$ decomposes as a direct sum of finite-dimensional unitary representations that can, in turn, be broken down to irreducibles.

Generalizations
By considering more $u$-touples of variables, say $A_{rst}$, $u$ and $1$, one obtains more unitary representations. However, by restricting these to homogeneous polynomials of degree $$, the resulting tensor of rank $$ is not symmetric.

O(3, 1)
Now consider a decending power series in a real variable $2$,
 * $$k_0/\xi_o + k_1/\xi_0^2 + \cdots = \sum_{k=0}^\infty k_n\xi^{-n-1}.$$

Go through the same steps as in the $3$ case and define $ξ_{i}$ with the inner product
 * $$\langle (k_{n})|(f_{n})\rangle = \sum_{n=0}^\infty \frac{k_nf_n}{n!}.$$

In the same fashion as above, define
 * $$V = V_0 \otimes V_1 \otimes V_2 \otimes V_3,$$

with inner product
 * $$ \langle(A_{nrst})|(B_{nrst})\rangle = \sum_{n=0}^\infty\sum_{r=0}^\infty\sum_{s=0}^\infty\sum_{t=0}^\infty \frac{r!s!t!}{\Gamma(n + 1)!}A_{nrst}B_{nrst}.$$

The usage of the Gamma function is motivated by that $P$ for $A$ and anticipating occurrences of some positive powers of $ξ_{i}$. When accepting nonzero coefficients for negative values of $α = 1$, the inner product is, strictly speaking, a pseudo-inner product space. By going through the same procedure as in the $O(3)$ case, one finds that if $3$, $ξ ≡ (ξ_{1}, ξ_{2}, ξ_{3})$. Then
 * $$Q = \sum_{n = 0}^\infty\sum_{r = 0}^\infty \sum_{s = 0}^\infty \sum_{t = 0}^\infty {A'}_{brst}{\xi'}_0^{-n-1}{\xi'}_1^r{\xi'}_2^s{\xi'}_3^t,$$

with $η ≡ (η_{1}, η_{2}, η_{3})$ ($ζ ≡ (ζ_{1}, ζ_{2}, ζ_{3})$ fixed) some linear combination of $ξ_{0}$ (as $O(3)$ varies). It turns out that
 * $$||(A'_{nrst})|| = ||(A_{nrst})||,$$

so that one obtains a unitary representation of the Lorentz group with representation space $V_{0}$. The fact that this representation is unitary implies that it is not a direct sum of finite-dimensional unitary representations, since no such exist.

The coefficients
 * $$A \equiv (A_{nrst}) \in V$$

(required to have finite norm) were termed expansors by Dirac. The name derives from the connection with binomial expansions (actually multinomial) under Lorentz transformations. If the power series associated to $Γ(n + 1) = n!$ is homogeneous of fixed degree, they are termed homogeneous expansors. The analogy with the $n ∈ ℕ$-dimensional case suggests that these can be regarded as symmetrical tensors in spacetime with the suffix $ξ_{0}$ occuring a negative number of times. By contrast to the $O(3)$-dimensional case, homogeneous expansors are not finite.

Details
In detail, apply a Lorentz transformation to the $ξ ≡ (ξ_{0}, ξ_{1}, ξ_{2}, ξ_{3})$,
 * $$\xi_\mu = \Lambda_\mu^\nu\xi'_\nu.$$

This forces
 * $$\xi_1^r\xi_2^s\xi_3^rt = p(\xi'_0, \xi'_1, \xi'_2, \xi'_3),$$

where $$ is a finite polynomial in the $ξ' = Λξ, Λ ∈ O(3, 1)$, and
 * $$\xi_0^{-n-1} = (\Lambda_0^0\xi'_0 + \Lambda_0^1\xi'_1 + \Lambda_0^2\xi'_2 + \Lambda_0^3\xi'_3)^{-n-1}.$$

The latter may be expressed as a series with ascending powers in $A'_{nrst}$ and descending powers in $nrst$. For a brief proof put
 * $$\begin{align}(\Lambda_0^0\xi'_0 + \Lambda_0^1\xi'_1 + \Lambda_0^2\xi'_2 + \Lambda_0^3\xi'_3)^{-n-1} &\equiv \left[a(1 + \frac{b}{a} + \frac{c}{c} + \frac{d}{a})\right]^{-n-1}\\

&\equiv a^{-n-1}\frac{1}{[1 (b' + c' + d')]^{n+1}}\\ &\equiv A^{-n-1}\frac{1}{1 + e}\cdots\frac{1}{1 + e}\\ &= A^{-n-1}(1-e+e^2 - \cdots)\cdots(1-e+e^2 - \cdots)\end{align}.$$ Make re-substitutions and the result follows.

While there may be positive powers of $A_{nrst}$ in $nrst$ (originating from $$), these contributions will be killed off by the gamma function when calculating the norm.

Consider first
 * $$\xi_0 = \xi'_0 + \epsilon \xi'_1, \xi_1 = \xi'_1 + \epsilon \xi'_0, \xi_2 = \xi'_2, \xi_3 = \xi'_3.$$

Substitution yields
 * $$Q = \sum A_{nrst}\left[{\xi'_0}^{-n-1}{\xi'_1}^{r} - (n+1)\epsilon{\xi'_0}^{-n-2}{\xi'_1}^{r+1} + r\epsilon{\xi'_0}^{-n}{\xi'_1}^{r-1}\right]{\xi'_2}^{s}{\xi'_3}^{t}.$$

This yields
 * $$A'_{nrst} = A_{nrst} - n\epsilon A_{n-1, r-1,s,t} + (r+1)\epsilon A_{n+1, r+1, s, t},$$

and
 * $$\begin{align}\sum \frac{r!s!t!}{\Gamma(n+1)}{A'_{nrst}}^2 &= \sum \frac{r!s!t!}{\Gamma(n+1)}\left[A_{nrst}^2 - 2n\epsilon A_{nrst}A_{n-1,r-1s,t} + 2(r+1)\epsilon A_{n+1,r+1,s,t}A_{nrst}\right]\\ &= \sum \frac{r!s!t!}{\Gamma(n+1)}{A_{nrst}}^2,\end{align} $$

since the second and third terms cancel when substituting $V$ in the former and $A$ in the latter.

Generalizations
By considering more $3$-touples of variables, say $0$, $3$ and $ξ_{μ}$, one obtains more unitary representations. The introduction of the gamma function opens up for another generalization. Instead of considering $ξ_{μ}$, one my consider $ξ ' _{1}, ξ ' _{2}, ξ ' _{3}$ for $ξ ' _{0}$ any real integer. If $ξ'_{0}$ then the norm is still positive definite. This is the case if the expression for $ξ_{0}^{−n − 1}ξ_{1}^{r}ξ_{2}^{s}ξ_{3}^{t}$ is homogeneous of negative degree. If $n → n + 1$ is homogeneous of positive degree, there will be a finite number of negative terms in the expression for the norm. Dirac calls such representations nearly unitary.

Convergence
Suppose $r → r − 1$ is homogeneous of degree $4$. Then write
 * $$Q = \sum_{n=0}^\infty \xi_0^{n-1}\sum_{r+s+t = u + n, r,s,t \ge 0}A_{nrst}\xi_1^r\xi_2^s\xi_3^t,$$

and
 * $$||Q||^2 = \sum_{n=0}^\infty \frac{1}{n!} \sum_{r+s+t = u + n, r,s,t \ge 0} r!s!t!A_{nrst}^2.$$

Now apply,
 * $$(x_1y_1 + x_2y_2 + \cdots) \le (x_1^2 + x_2^2 + \cdots)(y_1^2 + y_2^2 + \cdots)$$

this is the Cauchy–Schwartz inequality in sequence space with the inner product
 * $$\langle(x_n)|(y_n)\rangle = \sum_{n=1}^\infty x_n y_n.$$

Fix $$, subject to the condition above. Put
 * $$x_i = \xi_1^r\xi_2^s\xi_2^t, \quad y_i = A_{nrst}\frac{r!s!t!}{(r+s+t)!}, \quad 1 \le i \le \frac{(r+s+t)!}{r!s!t!},$$

and Cauch–Schwartz inequality yields
 * $$\left[\sum_i \xi_1^r\xi_2^s\xi_3^t A_{nrst}\frac{r!s!t!}{(r + s + t)!}\right]^2 = \left[\xi_1^r\xi_2^s\xi_3^t A_{nrst}\right]^2 \le \sum_i x_i^2 \sum_i y_i^2.$$

One finds
 * $$\sum_i y_i^2 = \frac{r!s!t!}{r+s+t}A_{nrst}^2,$$

and
 * $$\sum_i x_i^2 = \frac{(r+s+t)!}{r!s!t!}\xi_1^{2r}\xi_2^{2s}\xi_3^{2t}.$$

Now sum over the allowable values for $ξ ≡ (ξ_{0}, ξ_{1}, ξ_{2}, ξ_{3})$ for a fixed $η ≡ (η_{0}, η_{1}, η_{2}, η_{3})$, indexed by $$,
 * $$\left[\Sigma_{S}\xi_1^r\xi_2^s\xi_3^t A_{nrst}\right]^2 \le \sum_{S} \frac{(r+s+t)!}{r!s!t!}\xi_1^{2r}\xi_2^{2s}\xi_3^{2t} \sum_{S} \frac{r!s!t!}{r+s+t}A_{nrst}^2,$$

and employ the multinomial theorem to find
 * $$\left[\Sigma_{S}\xi_1^r\xi_2^s\xi_3^t A_{nrst}\right]^2 \le (\xi_1^2 + \xi_2^2 + \xi_1^3)^{n+u} \frac{1}{(n+u)!}\sum_{S} r!s!t!A_{nrst}^2.$$

Suppose now that
 * $$\sum_{n=0}^\infty \frac{1}{n!} \sum_{S} r!s!t!A_{nrst}^2$$

converges. Then the terms are bounded,
 * $$\frac{1}{n!} \sum_{S} r!s!t!A_{nrst}^2 < \kappa \quad \forall n.$$

It follows that
 * $$|\xi_0^{-n-1}\sum_SA_{nrst}\xi_1^rxi_2^sxi_3^t| < \kappa^{\frac{1}{2}}|\xi_0|^{u-1}\left\{ \frac{\xi_1^2 + \xi_2^2 + \xi_3^2}{\xi_0^2} \right\}^{\frac{1}{2}(n+u)}\left\{ \frac{n!}{(n+u)!} \right\}^{\frac{1}{2}},$$

so if
 * $$\xi_0^2 > \xi_1^2 + \xi_2^2 + \xi_3^2,$$

then $ζ ≡ (ζ_{0}, ζ_{1}, ζ_{2}, ζ_{3})$ will converge whenever it is homogeneous and $n ∈ ℕ$ converges.

Special expansors
Expansors being coefficients of
 * $$Q = (\xi_0^2 - \xi_1^2 - \xi_2^2 - \xi_3^2)^{-l}$$

go over into themselves under Lorentz transformations. This is obvious from the the definition of the Lorentz transformation,
 * $$\sum P_{nrst}\xi_0^{-n-1}\xi_1^r\xi_2^s\xi_3^t = (\xi_0^2 - \xi_1^2 - \xi_2^2 - \xi_3^2)^{-l} = ({\xi'_0}^2 - {\xi'_1}^2 - {\xi'_2}^2 - {\xi'_3}^2)^{-l} = \sum P'_{nrst}{\xi'_0}^{-n-1}{\xi'_1}^r{\xi'_2}^s{\xi'_3}^t.$$

To see what they look like, write
 * $$\begin{align}Q &= \frac{1}{[\xi_0^2(1-T)]^l}, \quad T = \frac{\xi_1^2 + \xi_2^2 + \xi_3^2}{\xi_0^2}\\

&= \xi_0^{-2l} \frac{1}{1-T}\times\cdots\times\frac{1}{1-T}\\ &= \xi_0^{-2l}(1+T+T^2+\cdots)^l,\end{align}$$ the last step by virtue of the geometric series. Now consider the case $n ∈ { n_{0}, n_{0} + 1, &hellip; }$. Then
 * $$\begin{align}Q &= \xi_0^{-2}\left(1 + \frac{\xi_1^2 + \xi_2^2 + \xi_3^2}{\xi_0^2} + \left(\frac{\xi_1^2 + \xi_2^2 + \xi_3^2}{\xi_0^2}\right)^2 + \cdots\right)\\

&= \xi_0^{-2}\sum_{n=0}^\infty\left(\frac{\xi_1^2 + \xi_2^2 + \xi_3^2}{\xi_0^2}\right)^n\\ &= \xi_0^{-2}\sum_{n=0}^\infty\sum_{k_1+k_2+k_3=n} {n \choose k_1k_2k_3}\prod_{1\le i\le 3}\left(\frac{\xi_i^2}{\xi_0^2}\right)^{k_i}\\ &= \sum P_{nrst}\xi_0^{-n-1}\xi_1^r\xi_2^s\xi_3^t,\end{align}$$ the last manipulation using the multinomial expansion. The coefficients are then found by matching coefficients.

If $$ is a homogeneous polynomial in $n_{0}$, then
 * $$f(\xi_0, \xi_1, \xi_2, \xi_3)(\xi_0^2 - \xi_1^2 - \xi_2^2 - \xi_3^2)^{-l}$$

transforms like $$, i.e. like a tensor of order equal to the degree of $$. This expansor is infinite. Write
 * $$f(\xi_0, \xi_1, \xi_2, \xi_3) = g_u + \xi_0g_{u-1} + (\xi_0^2 - \mathbf{\xi}^2)g_{u-2} + \xi_0(\xi_0^2 - \mathbf{\xi}^2)g_{u-3} + (\xi_0^2 - \mathbf{\xi})^2g_{u-4} + \cdots,$$

where the $$ are homogeneous polynomials in $n_{0} > −1$ of the degree given by the index. It is not entirely obvious that this can be done. To see that it can, create the $$'s by considering $$ term by term. For each term, begin by constructing the $$ of lowest degree. This construction will yield residual terms that go into the $Q$'s of the next higher degree. For example
 * $$f(\xi_0, \xi_1, \xi_2, \xi_3) = \xi_0^2\xi_1\xi_2\xi_3 = (\xi_0^2 - \xi_1^2 - \xi_2^2 - \xi_3^2)\xi_1\xi_2\xi_3 + \xi_1^3\xi_2\xi_3 + \xi_1\xi_2^3\xi_3 + \xi_1\xi_2\xi_3^3,$$

so that in this case
 * $$g = \xi_1^3\xi_2\xi_3 + \xi_1\xi_2^3\xi_3 + \xi_1\xi_2\xi_3^3, \quad g_2 = \xi_1\xi_2\xi_3.$$

Looking at the contributions, one finds
 * $$\begin{align}

g_u(\xi_0^2 - \mathbf{\xi}^2)^{-l} &= \sum_{m=0}^\infty\frac{l(l+1)(l+2)\cdots(l+m-1)}{m!}\frac{g_u\mathbf{\xi}^{2m}}{\xi_0^{2(m+l)}},\\ g_{u-1}\xi_0(\xi_0^2 - \mathbf{\xi}^2)^{-l} &= \sum_{m=0}^\infty\frac{l(l+1)(l+2)\cdots(l+m-1)}{(m+l)!}\frac{g_{u-1}\mathbf{\xi}^{2m}}{\xi_0^{2(m+l) - 1}},\\ g_{u-2}(\xi_0^2 - \mathbf{\xi}^2)^{-l+1} &= \sum_{m=0}^\infty\frac{l(l+1)(l+2)\cdots(l+m-1)}{m!}\frac{g_{u-2}\mathbf{\xi}^{2(m+1)}}{\xi_0^{2(m+l)}},\\ g_{u-3}(\xi_0^2 - \mathbf{\xi}^2)^{-l} &= \sum_{m=0}^\infty\frac{l(l+1)(l+2)\cdots(l+m-1)}{(m+l)!}\frac{g_{u-3}\mathbf{\xi}^{2(m+1)}}{\xi_0^{2(m+l)-1}}.\end{align}$$ To prove this, extract $Q$ from the denominator and use the geometric series. Then use
 * $$\left(\sum_{n=0}^\infty x^n\right)^l = \sum_{n=0}^\infty \frac{(n+1)(n+2)\cdots(n+l-1)}{(l-1)!}x^n = \sum_{n=0}^\infty\frac{(l+n-1)!}{n!(l-1)!)}x^n = \frac{l(l+1)(l+2)\cdots(l+n-1)}{n!}x^n, $$

which, in turn, is proved by induction on $Q$. The observation here is that, for $u − 1$ large, the $r, s, t$'th term of $n$ is of order $Q$ smaller than the corresponding term for $||(A_{nrst})||$. For convergence issues then, consideration of $l = 1$ and $ξ_{0}, ξ_{1}, ξ_{2}, ξ_{3}$ suffices. For this purpose, write
 * $$g_u = S_u + \mathbf{\xi^2}S_{u-2} + \mathbf{\xi^4}S_{u-4} + \cdots, \quad g_{u-1} = S_{u-1 }+ \mathbf{\xi^2}S_{u-3} + \mathbf{\xi^4}S_{u-5} + \cdots,$$

the existence motivated by the same reasoning as that for the $ξ_{1}, ξ_{2}, ξ_{3}$'s. These each contribute
 * $$\begin{align}S_{u-2r}\mathbf{\xi}^{2r}(\xi_0^2 - \mathbf{\xi}^2)^{-l} &= S_{u-2r}\sum_{m=0}^\infty \frac{(n+l-1)!\mathbf{\xi}^{2(m+r)}}{m!(l-1)!\xi_0^{2(m+r)}},\\

S_{u-2r-1}\xi_0^2\mathbf{\xi}^{2r}(\xi_0^2 - \mathbf{\xi}^2)^{-l} &= S_{u-2r-1}\sum_{m=0}^\infty \frac{(n+l-1)!\mathbf{\xi}^{2(m+r)}}{m!(l-1)!\xi_0^{2(m+r)-1}}.\end{align}$$ The squared length of expansors whose components are the coefficients of the above expressions are series on the form
 * $$\begin{align}&c\sum_m,\frac{(m+l+1)!^2}{m!^2(l-1)!^2}\frac{4^{m+r}(m+r)!(m+u-r + \frac{1}{2})!}{2m+2l-1)!}\\

&c\sum_m \frac{(m+l+1)!^2}{m!^2(l-1)!^2}\frac{4^{m+r}(m+r)!(m+u-r - \frac{1}{2})!}{2m+2l-2)!}.\end{align}$$ For large $g$, the ratio of one term and the next in the series is $ξ_{0}^{2}$, implying that the series diverges. The corresponding expansor is thus infinite. Moreover, expansors of this form are orthogonal meaning that the total expansor corresponding to $l$ diverges, hence is of infinite length.

Transformation of variables
Expansors can be put into correspondence with another set of functions. For this purpose, introduce a set of operators on $m$-functions,


 * $$\begin{align}X_0 &= \frac{1}{\sqrt{2}}(\xi_0 - \frac{\partial}{\partial \xi_0}), \quad X_r = \frac{1}{\sqrt{2}}(\xi_r + \frac{\partial}{\partial \xi_r}) \quad (r = 1,2,3), \\

\frac{\partial}{\partial x_0} &= \frac{1}{\sqrt{2}}(\xi_0 + \frac{\partial}{\partial \xi_0}), \quad \frac{\partial}{\partial X_0} = \frac{1}{\sqrt{2}}(-\xi_r + \frac{\partial}{\partial \xi_r}).\end{align}$$

Now consider the $m$-function $g_{u - 2}$. It vanishes when operated on by
 * $$\frac{\partial}{\partial \xi_r} = \frac{1}{\sqrt{2}}(x_r + \frac{\partial}{\partial x_r}),$$

and effectively (it is killed off by the gamma function in the norm) also when multiplied by


 * $$\xi_0 = \frac{1}{\sqrt{2}}(x_0 + \frac{\partial}{\partial x_0}).$$

Put this $m^{l}$-function in association with an $g_{u}$-function that is killed off by the right hand sides. From the theory of differential equations, this is uniquely (up to the constant)


 * $$\xi_0^{-1} \equiv \frac{1}{\pi}e^{-\frac{1}{2}(x_0^2 + \mathbf{x}^2)}.$$

Multiply this equation by


 * $$(-\frac{\partial}{\partial \xi_0})^n\xi_1^r\xi_2^s\xi_3^t$$

on the left and the operator equal to it on the right. This yields


 * $$\xi_0^{-n-1}\xi_1^r\xi_2^s\xi_3^t \equiv \pi^{-1}n!2^{-\frac{1}{2}(n+r+s+t)}

\left(x_0 - \frac{\partial}{\partial x_0}\right)^n \left(x_1 - \frac{\partial}{\partial x_1}\right)^r \left(x_2 - \frac{\partial}{\partial x_2}\right)^s \left(x_3 - \frac{\partial}{\partial x_3}\right)^t e^{-\frac{1}{2}(x_0^2 + \mathbf{x}^2)} = F_{nrst}(x_0, x_1, x_2, x_3).$$

Thus a general expansor can be identified with both a $g_{u}$-function and an $χ$-functions and


 * $$\sum A_{nrst}\xi_0^{-n-1}\xi_1^r\xi_2^s\xi_3^t \equiv \sum A_{nrst}F_{nrst}(x_0, x_1, x_2, x_3).$$

The chief virtue of this identification is that the inner product of two expansors become


 * $$\langle(A_{nrst})|(B_{nrst})\rangle = \sum\int\int\int\int F_{nrst}G_{nrst}dx_0dx_1dx_2dx_3.$$

Proof

 * $$\begin{align}I &= \int_{-\infty}^{\infty} \left(z-\frac{d}{dz}\right)^{m}e^{-\frac{1}{2}z^2} \cdot \left(z-\frac{d}{dz}\right)^{m'}e^{-\frac{1}{2}z^2}dz

=\int_{-\infty}^{\infty} \left(z-\frac{d}{dz}\right)\left(z-\frac{d}{dz}\right)^{m-1}e^{-\frac{1}{2}z^2} \cdot \left(z-\frac{d}{dz}\right)^{m'}e^{-\frac{1}{2}z^2}dz\\&=\int_{-\infty}^{\infty} \left(z-\frac{d}{dz}\right)^{m-1}e^{-\frac{1}{2}z^2} \cdot z\left(z-\frac{d}{dz}\right)^{m'}e^{-\frac{1}{2}z^2}dz - \int_{-\infty}^{\infty} \frac{d}{dz}\left(z-\frac{d}{dz}\right)^{m-1}e^{-\frac{1}{2}z^2} \cdot \left(z-\frac{d}{dz}\right)^{m'}e^{-\frac{1}{2}z^2}dz\\ &=\int_{-\infty}^{\infty} f \cdot zgdz - \int_{-\infty}^{\infty} \frac{df}{dz} \cdot gdz\\ &=\int_{-\infty}^{\infty} f \cdot zgdz - \left[f\cdot g\right]_{-\infty}^\infty + \int_{-\infty}^{\infty} f\cdot \frac{dg}{dz} dz\\ &= \int_{-\infty}^{\infty} f \cdot \left(z + \frac{d}{dz}\right)gdz\\ &=\int_{-\infty}^{\infty} \left(z-\frac{d}{dz}\right)^{m-1}e^{-\frac{1}{2}z^2} \cdot \left(z + \frac{d}{dz}\right)\left(z-\frac{d}{dz}\right)^{m'}e^{-\frac{1}{2}z^2}dz\\ &=\int_{-\infty}^{\infty} \left(z-\frac{d}{dz}\right)^{m-1}e^{-\frac{1}{2}z^2} \cdot \left\{\left(z-\frac{d}{dz}\right)^{m'}\left(z + \frac{d}{dz}\right) + 2m'\left(z-\frac{d}{dz}\right)^{m'-1}\right\}e^{-\frac{1}{2}z^2}dz\\ &= 2m'\int_{-\infty}^{\infty} \left(z-\frac{d}{dz}\right)^{m-1}e^{-\frac{1}{2}z^2} \cdot \left(z-\frac{d}{dz}\right)^{m'-1}e^{-\frac{1}{2}z^2}dz.\end{align} $$

Apply this $g_{u − 1}$ times. If $g$, then the prefactor of the integral on the LHS will yield zero. Using the standard integral for the Gaussian, the result becomes
 * $$I = 2^mm!\sqrt{\pi}\delta{mm'}.$$

This result applied to two expansors in the guise of $m$ functions is


 * $$\int\int\int\int F_{nrst}F_{n'r's't'}dx_0dx_1dx_2dx_3 = \frac{r!s!t!}{n!}\delta{nn'}\delta{rr'}\delta{ss'}\delta{tt'}.$$

The conclusion is that the set of all $1 + u⁄m$-functions may be seen as a unitary representation of the Lorentz group. The transformation rule is determined by that of the expansors. Naturally, the $f$ may be interpreted as spacetime coordinates.

Physical applications
Let the $ξ$ denote the coordinates of a four-dimensional harmonic oscillator. Then $ξ$ are interpreted as momentum operators and the energy of the operator can be taken as


 * $$\frac{1}{2}[x_1^2 + x_2^2 + x_3^2 - x_0^2 + (p^1)^2 + (p^2)^2 + (p^3)^2 - (p^0)^2].$$

The state of the oscillator characterized by $ξ_{0}^{−1}$ is then represented by the function $ξ$; in the $x$-representation it is represented by the expansor $ξ$.

Sandbox
$m$, $m > m'$ , $$, $x$, $x$, $x_{μ}$, $x_{μ}$

S-matrix
Consider
 * $$\langle\Psi_\beta^\pm, V\Psi_\alpha^\pm\rangle = \langle\Psi_\beta^\pm, V\Psi_\alpha^\pm\rangle.$$

Use
 * $$\Psi_\alpha^\pm = \Phi_\alpha + (E_\alpha - H_0 \pm i\epsilon)^{-1}V\Psi_\alpha^\pm$$

on both sides. Then
 * $$\langle\Psi_\beta^\pm, V(\Phi_\alpha + (E_\alpha - H_0 \pm i\epsilon)^{-1}V\Psi_\alpha^\pm)\rangle = \langle\Phi_\beta + (E_\beta - H_0 \pm i\epsilon)^{-1}V\Psi_\beta^\pm, V\Psi_\alpha^\pm\rangle,$$

or
 * $$\langle\Psi_\beta^\pm, V\Phi_\alpha\rangle + \langle\Psi_\beta^\pm, V(E_\alpha - H_0 \pm i\epsilon)^{-1}V\Psi_\alpha^\pm)\rangle = \langle\Phi_\beta, V\Psi_\alpha^\pm\rangle + \langle(E_\beta - H_0 \pm i\epsilon)^{-1}V\Psi_\beta^\pm, V\Psi_\alpha^\pm\rangle.$$

But Weinberg gets
 * $$\langle\Psi_\beta^\pm, V\Phi_\alpha\rangle + \langle\Psi_\beta^\pm, V(E_\alpha - H_0 \pm i\epsilon)^{-1}V\Psi_\alpha^\pm)\rangle = \langle\Phi_\beta, V\Psi_\alpha^\pm\rangle + \langle\Psi_\beta^\pm, V(E_\beta - H_0 \mp i\epsilon)^{-1}V\Psi_\alpha^\pm\rangle,$$

or
 * $$\begin{align}\langle\Psi_\beta^\pm, V\Phi_\alpha\rangle - \langle\Phi_\beta, V\Psi_\alpha^\pm\rangle  &= -[\langle\Psi_\beta^\pm, V(E_\alpha - H_0 \pm i\epsilon)^{-1}V\Psi_\alpha^\pm)\rangle - \langle\Psi_\beta^\pm, V(E_\beta - H_0 \mp i\epsilon)^{-1}V\Psi_\alpha^\pm\rangle]\\

&= -\int d\gamma[\langle\Psi_\beta^\pm, V(E_\alpha - H_\gamma \pm i\epsilon)^{-1}|\Phi_\gamma\rangle\langle\Phi_\gamma| V\Psi_\alpha^\pm)\rangle - \langle\Psi_\beta^\pm, V(E_\beta - H_\gamma \mp i\epsilon)^{-1}|\Phi_\gamma\rangle\langle\Phi_\gamma| V\Psi_\alpha^\pm\rangle]\\ &= -\int d\gamma[T_{\gamma\beta}^{\pm^*}T_{\gamma\alpha}(E_\alpha - H_\gamma \pm i\epsilon)^{-1} - T_{\gamma\beta}^{\pm^*}T_{\gamma\alpha}(E_\beta - H_\gamma \mp i\epsilon)^{-1}]\\ T_{\alpha\beta}^{\pm^*} - T_{\beta\alpha} &= -\int d\gamma T_{\gamma\beta}^{\pm^*}T_{\gamma\alpha}[(E_\alpha - H_\gamma \pm i\epsilon)^{-1} - (E_\beta - H_\gamma \mp i\epsilon)^{-1}]\end{align}.$$ Obviously, use has been made of
 * $$\langle\Psi_\beta^\pm, V\Phi_\alpha\rangle = \langle V\Psi_\beta^\pm, \Phi_\alpha\rangle = \langle\Phi_\alpha,V\Psi_\beta^\pm\rangle^*.$$

We have
 * $$\lim_{\tau \rightarrow \mp \infty} \int d\alpha g(\alpha)e^{-i\tau E_\alpha}\Psi_\alpha^\pm = \lim_{\tau \rightarrow \mp \infty} \int d\alpha g(\alpha)e^{-i\tau E_\alpha}\Phi_\alpha,$$

or
 * $$\lim_{\tau \rightarrow \mp \infty} e^{-iH\tau}\int d\alpha g(\alpha)\Psi_\alpha^\pm = \lim_{\tau \rightarrow \mp \infty} e^{-iH_0\tau}\int d\alpha g(\alpha)\Phi_\alpha.$$

Thus
 * $$\lim_{\tau \rightarrow \mp \infty} \int d\alpha g(\alpha)e^{-i\tau E_\alpha}V\Psi_\alpha^\pm = \lim_{\tau \rightarrow \mp \infty} \int d\alpha g(\alpha)e^{-i\tau E_\alpha}V\Phi_\alpha$$

and
 * $$\lim_{\tau \rightarrow \mp \infty} (\int d\beta g^*(\beta)e^{iE_\beta\tau}\Psi^\pm,\int d\alpha e^{-i\tau E_\alpha}V\Psi_\alpha^\pm) = \lim_{\tau \rightarrow \mp \infty} (\int d\beta g^*(\beta)e^{iE_\beta\tau}\Phi_\beta, \int d\alpha e^{-i\tau E_\alpha}V\Phi_\alpha^\pm),$$

i.e.
 * $$\int \int d\beta d\alpha g^*(\beta)g(\alpha)e^{i(E_\beta - E_\alpha)\tau}(\Psi_\beta^\pm, V\Psi_\alpha^\pm) =

\int \int d\beta d\alpha g^*(\beta)g(\alpha)e^{i(E_\beta - E_\alpha)\tau}(\Phi_\beta, V\Phi_\alpha).$$

Asymptotic operators
Associated to the in and out states are the in and out field operators (fields) or asymptotic fields, $p^{μ} = i∂/∂x_{μ}$. To the full interacting theory there corresponds the interacting field $(n, r, s, t)$. In the context of axiomatic field theory, it is called the interpolating field. Again attention is focused on a neutral Klein-Gordon field in order to keep the notation as clean as possible. It is postulated to satisfy the same equal time commutation relations (ETCR) as the free field.
 * $$\begin{align}{[\phi(x), \pi(y)]}_{x_0=y_0} &= i\delta(\mathbf{x} - \mathbf{y}),\\

{[\phi(x), \phi(y)]}_{x_0=y_0} &= {[\pi(x), \pi(y)]}_{x_0=y_0} = 0.\end{align}$$

This is assuming that there are no gradient couplings modifying the canonically conjugate field $Fnrst$. In contrast to the non-interacting case, these commutation relations cannot be extended to unequal times. The commutator between the field operator and the momentum operator retains the free field behavior,
 * $$[P^\mu, \phi] = -i\partial^\mu\phi,$$

with formal solution

Like for the interacting states and the asymptotic states, there are connections between the interacting field and the asymptotic fields. The field equation for the interacting field is given by

where $$ is the physical mass and the source term in general depends on $ξ$. One term in the source is, again assuming no derivative couplings, partly given by
 * $$j(x) = \frac{\partial \mathcal{L}_1}{\partial\phi(x)},$$

with
 * $$\mathcal{L} = \mathcal{L}_0 + \mathcal{L}_1,$$

the free and interacting parts of the Lagrangian density. There is an additional term that accounts for the use of the physical mass $ξ^{−n − 1}ξ'r ξ's ξt$ in the Klein-Gordon equation instead of the bare mass, $ƒ̂$ appearing in the Lagrangian density,
 * $$j_{\delta m}(x) = (m^2 - m_0^2)\phi^2(x).$$

The field equations are formally solved by using Green's function techniques,

These are the Yang–Feldman equations. Here $f&#770;$ is the Pauli–Jordan function and $&and; f$ is the Heaviside step function. The in and out fields solve the homogeneous equation, while $&and; f$ solve
 * $$\begin{align}(\Box - m^2)\Delta_{R} &= \delta(x - x'),\\

(\Box - m^2)\Delta_{A} &= \delta(x' - x).\end{align}$$

One might naïvely expect from these equations that (under typical scattering conditions)

in analogy with the states. This will hold if $&#285;$ for some finite $&#293;$. But this is not the case since self-interaction cannot be "turned off". A slightly less naïve guess is to assume

where $Φ_{i}, Φ_{o}$ is a field renormalization constant. Its interpretation is that it gives the probability of $Φ$ creating a one-particle state when applied to the vacuum,
 * $$\langle 1| \phi(x)|0\rangle = \sqrt{Z}\langle 1| \phi_i(x)|0\rangle = \sqrt{Z}.$$

This probability is assumed to be less than one since the interacting field can create many-particle states. But this is inconsistent as speculated so far without more qualification of what is meant. Consider the string of equalities

These equations imply $π$, and nothing has changed. Implicit in $$ and the equation above is that $Φ$ converges in norm to $m$. Convergence in norm to $m_{0}$ holds, for a sequence of operators, if
 * $$\lim_{i \rightarrow \infty}||\phi -\phi_i|| \rightarrow 0.$$

This requirement is too strong, and one can instead demand

corresponding to demanding weak operator convergence or convergence in WOT, the weak operator topology. There is also an intermediate topology, the SOT, strong operator topology, where $Δ$ is demanded for all $Θ$. While $$ imply $$, $$ do not, and this opens up the possibility for a rigorous description of the asymptotic behavior. Next problem: The respective RHS of $$ have time dependence, see $$, valid for the asymptotic fields as well, so the limits aren't well-defined. This can be remedied by considering smeared field operators, analogous to wave packets. The scalar product in the Hilbert space of the scalar theory is given by
 * $$(u, v) = i\int d^3x u^*(x)\overleftrightarrow{\partial_0}v(x) \equiv \int d^3x (u^*(x)(\partial_0v(x)) - (\partial_0u^*(x))v(x)).$$

Use this to write
 * $$\phi(\mathbf{x}, t) = \sum_\alpha u_\alpha(\mathbf{x}, t)\phi^\alpha(t),$$

for
 * $$| \phi^\alpha(t)| a \rangle = i\int d^3x u_\alpha^*(\mathbf{x}, t)\overleftrightarrow{\partial_0}\phi(\mathbf{x}, t)|a\rangle \equiv (u_\alpha, \phi )| a \rangle, \quad a \in \mathcal{H}, $$

or, since $Δ_{R, A}$ is time-independent,

where the $j(x) = 0, x_{0} > |T|$ satisfy
 * $$(\Box - m^2)u_\alpha(x) = 0,$$

and constitute a complete set of localized wave functions. They can be expressed as
 * $$u_\alpha(\mathbf{x}, t) = \int d\alpha g_\alpha(\mathbf{p})u_{\mathbf{p}}(\mathbf{x}, t),$$

where $T$ is a smooth function. See the previous section for a definition of $Z$. The completeness can be expressed as
 * $$ \sum_\alpha u_\alpha^*(x)u_\alpha(x) = 1$$

and
 * $$ \chi = \sum_\alpha (u_\alpha, \chi)u_\alpha, \chi \in \mathcal{B(H)}.$$

Equivalently, write

it can be shown that $Φ$ are time-independent. The interpretation is that $Z = 1$ applied to the vacuum operates on wave packages described by $Φ$ as opposed to plane waves (associated to the creation and annihilation operators) or distinctly localized particles (associated to the field operator). The interpretation for the interacting operators is harder, since these are not known in detail. The asymptotic condition is now written . If one uses, instead of wave packets $√ZΦ_{i}$ uses plane waves $Φ$, then
 * $$\phi^\mathbf{p}(t) = (u_\mathbf{p}, \phi) = i\int d^3x u_\mathbf{p}^*(\mathbf{x}, t)\overleftrightarrow{\partial_0}\phi(\mathbf{x}, t) = a_\mathbf{p}(t),$$
 * $$\phi^\mathbf{p}_{i/o}(t) = (u_\mathbf{p}, \phi_{i/o}) = i\int d^3x u_\mathbf{p}^*(\mathbf{x}, t)\overleftrightarrow{\partial_0}\phi_{i/o}(\mathbf{x}, t) = a_{\mathbf{p}, i/o}(t),$$

the last equality following from the usual expansion of free fields in terms of creation and annihilation operators. The interpretation of $Φv → Φ_{i}v$ is more intricate due to the interaction. The $$ become

which are weaker (follow from, but does not imply) $$. By postulating $$, a weak version of $$ can be derived,

Crossing the Tsirelson bound
Crossing the Tsirelson bound amounts to, at least, leaving the confines of ordinary quantum mechanics. Crossing has been achieved (non-local-boxes or PR-boxes), but only purely mathematically, with no directly supporting physical principles, but still with the no-signalling condition imposed by special relativity satisfied. By incorporating the, physically speculative, idea of supersymmetry, that has received much attention the past decades as perhaps the most promising candidate for physics beyond the standard model, it is possible that the bound is crossed. But this comes with a price. One set of conditions that suffices in a minimal supersymmetric extension of qubits for breaking the bound, as outlined below, is the following:
 * Supersymmetry
 * A certain Grassmann-valued transition probability between states. The chosen definition of transition probability reduces to the usual Born rule when the states have no Grassmann-component, and the the transition probabilities add up to 1.
 * A particular choice of map from Grassmann-valued probabilities to real probabilities. This map must respect the $v$ symmetry of below, just like the usual transition probabilities respect the $a$ symmetry, but otherwise this choice has no counterpart in ordinary quantum mechanics.

In one "simple" case, the two-dimensional complex Hilbert space whose normalized elements are qubits, with symmetry group $u_{α}$, is replaced by a (2|1)-dimensional super Hilbert space with the orthosymplectic group $g_{α}$ as symmetry group, inhabited by superqubits. In this scenario, Lie algebras are replaced with Lie superalgebras, the Bloch sphere with the super-sphere $dα$ and so on.

The space of superqubits, here initially thought of as a Z2-graded vector space over the complex numbers, is spanned by the kets
 * $$|0\rangle, |1\rangle \quad \mathrm{and} \quad |*\rangle$$,

where the first two are of even grade, and the last one is odd. The space of linear transformations, denoted $Φ^{α}_{i, o}$ on any super vector space is (in a basis) just the set of $Φ^{α}_{i, o}$-matrices with entries from the ground field $g_{α}$, here $u_{α}$. These decompose uniquely as a sum of a grade-preserving and a grade-reversing transformation.

The (real) orthosymplectic Lie superalgebra $u_{p}$ is defined as
 * $$\mathfrak{osp}(p, q) = \left\{ X \in \mathfrak{gl}(p|q)|X^{ST}H - (-1)^{|X|}HX = 0\right\}.$$

Here $a_{p}$ is a matrix representing a bilinear form with both a symmetric and antisymmetric part. This is in almost complete analogy with how the Lie algebras of the classical groups are defined, but with two exceptions. There is the alternating sign depending on whether the transformations are even (grade-preserving) or odd, and the supertranspose is used as opposed to the ordinary Hermitian adjoint.

In the current application, the notion of a super vector space is generalized to a supermodule over a Grassmann algebra, so that the coefficients of a general vector expressed in the basis kets of above are Grassmann numbers rather than ordinary complex numbers. Likewise, the linear transformations are generalized, so that the entries in the representative matrices are Grassmann numbers.

The superqubit is not itself a candidate for foundational supersymmetric physics beyond the standard model because the space it lives in is not a representation space of the super-Poincaré group. But the superqubit and the Lie superalgebra finds supersymmetric applications in the field of condensed matter physics.

The CHSH inequality can be rephrased as a limit on the probability of winning a certain non-local game, the CHSH game,. The CHSH game is played as follows. The participants are Alice and Bob, who are competing against a referee. Alice and Bob are allowed to cooperate in the sense that they can agree upon a strategy before the game is played. But, they are not allowed to communicate during the game (the no-signaling condition). The referee sends to each of Alice and Bob a number, 0 or 1, denoted $UOSp(1|2)$ and $U(2)$ respectively, with a probability of $$ for each combination. Alice and Bob each sends a number, $U(2)$ and $UOSp(1|2)$ respectively, in return to the referee. The winning conditions for Alice and Bob are shown in the table below on the left.

On the face of it, one optimal strategy for Alice and Bob is to send 0 in return, regardless of what they receive. The only way they can lose is if the referee sends $S^{2|2} = UOSp(1|2)/U(0|1)$. Thus the maximized winning probability is 0.75, and this is the CHSH bound in this case. But if Alice and Bob share a so-called non-local resource, they can improve their winning chances. One such resource is an entangled state, such as a 2-qubit, e.g $End(K^{p, q})$, where $(p × q)$ and $K$ may (for instance) be interpreted as a spin up and spin down of the components of the spin of a spin $U$ particle with respect to some direction. (This is only a factor in a complete physical 2-particle state, but it is all that is needed for the discussion.) By using an agreed upon measurement strategy and utilizing the results of the measurements on their respective "ends" of the qubit, Alice and Bob can improve their chances of winning in the game, up to, and including, a limit, the Tsirelson bound, which in this case is $K = ℂ$. Loosely speaking, the reason that this is possible is that the measurement results are correlated due to the entanglement.

Now suppose Alice an Bob share a non-local resource in the form of a entangled state of a 2-superqubit. The amplitudes in super Hilbert space are Grassmann numbers. Results will therefore depend on how real probabilities are extracted. A Grassmann supernumber $osp(p|q)$ may always be written as
 * $$z = z_{\mathrm{body}} + z_{\mathrm{soul}}$$,

where the first term is a complex number and the second term is a sum over the Grassmann algebra generators with complex coefficients. The terms are referred to as the body and the soul. For a start, a counterpart of the Born rule must be chosen. A reasonable, but not canonical, choice is
 * $$p(\phi, \psi) = \langle\phi|\psi\rangle\langle\phi|\psi\rangle^\sharp$$.

The hash operator denotes graded involution, similar to complex conjugation. This choice reduces to the usual Born rule when there is no soul. Secondly, a map from the Grassmann-valued transition probabilities to the real numbers must be chosen. In, the authors consider three plausible ways of doing this. The first alternative considered is to simply ignore the "soul". The resulting probabilities are those of ordinary quantum mechanics, and then an optimal strategy for Alice and Bob reaches the Tsirelson bound. The second case considered is a trigonometric map. In this case too, the Tsirelson bound can be reached. The third case, the Modified Rogers case, as they call it, exceeds the bound. All states used in the games involve probabilities lying between 0 and 1, but the third interpretation of Grassmann transition probabilities in terms of real numbers permits changes of basis inducing negative transition probabilities. The authors write in the conclusion,
 * Violating Tsirelson's bound was always destined to involve paying a price and it remains to be seen whether the existence of negative transition probabilities is too high price to pay (even though we did not invoke them in the CHSH game). In fact, one might ask whether it is this feature alone, with or without supersymmetry, that is responsible for exceeding the bound. 

They finally present arguments in the direction that existence of negative transition probabilities alone is not sufficient for violation of Tsirelson's bound.

The general case
The formula will be proved by finding a general expression for
 * $$e^{-C(t)}\frac{d}{dt}e^{C(t)},$$

where $H$ is a curve in the Lie algebra. A Lie group is endowed with a smooth Lie algebra valued form $i$ such that for each point $j$ $a$ is a linear map. It is defined as $b$, the pushforward of left multiplication by $ij$. It satisfies
 * $$d\omega + \omega \wedge \omega = 0,$$

the Maurer-Cartan equations. See Vector valued differential form for the definitions of exterior derivative and wedge product for vector valued forms. In addition, there is a Lie bracket for Lie algebra valued forms. In terms of this, the Maurer–Cartan's equations read
 * $$d\omega + \frac{1}{2}[\omega, \omega] = 0.$$

In case $a + b (mod 2)$ is a matrix Lie group, then the Maurer–Cartan form cam be expressed as
 * $$\omega_g = g^{-1}dg.$$

The Lie algebra valued forms also pull back under smooth maps. Thus if $i = j = 1$ denotes the pulled back $Ψ = (u ⊗ u) + (d ⊗ d)$ under some map $u$, it too will satisfy the Maurer–Cartan's equations. Given coordinates $d$ on $cos^{2} π/8 ≈ 0.8536$ one may write
 * $$\omega^* = g^{-1}\frac{\partial g}{\partial s}ds + g^{-1}\frac{\partial g}{\partial t}dt \equiv \alpha(s,t)ds + \beta(s,t)dt.$$

The inverse of $z$ on the right hand side is the group inverse. Since the pullback commutes with exterior derivatives and commutators, the Maurer–Cartan equations read in these coordinates
 * $$\frac{\partial\beta}{\partial s} - \frac{\partial\alpha}{\partial t} + [\alpha,\beta] = 0.$$

Let
 * $$g(s, t) = e^{sC(t)},$$

where $C(t)$ is a $ω$ path in the Lie algebra. Then
 * $$\begin{align}\alpha(s, t) &= g^{-1} \frac{\partial g}{\partial s} = e^{-sC(t)} e^{sC(t)}C(t) = C(t),\\

\beta(s, t) &= g^{-1}\frac{\partial g}{\partial t} = e^{-sC(t)}\frac{\partial}{\partial t}e^{sC(t)},\end{align}$$ so that the Maurer–Cartan equations are

The desired general expression is thus $g ∈ G$. For fixed $ω_{g}:TG_{g} → TG_{e}$, make an ansatz for $ω = L_{g^{−1}∗}$ as a formal power series in $g^{−1}$, with coefficients in the Lie algebra,


 * $$\beta(s, t) = a_1s + a_2s^2 + \cdots,$$

and one finds


 * $$\begin{align}a_0 &= C'(t)\\

na_n &= -\mathrm{ad}_{C}a_{n-1}.\end{align}$$ This leads to
 * $$\beta(s, t) = sC'(t) + \frac{1}{2}s(-\mathrm{ad}C(t))C'(t) + \cdots + \frac{1}{n!}s^n(-\mathrm{ad} C(t))^{n-1}C'(t) + \cdots .$$

Define
 * $$\phi(z)=\frac{e^z-1}{z} = 1 + \frac{1}{2!}z + \frac{1}{3!}z^2 + \cdots,$$

and the general expression for $G$ becomes
 * $$\exp(C(t))\frac{d}{dt}\exp(C(t)) = \phi(-\mathrm{ad}C(t))C'(t)$$

or
 * $$\frac{d}{dt}\exp(C(t)) = \exp(-C(t))\frac{1-e^{\mathrm{ad}_C}}{\mathrm{ad}_C}\frac{dC(t)}{dt}.$$

Proof
Here $ω^{∗}$ is to be considered as a matrix of maps $ω$ of $g:ℝ^{2} → G$ variables, where $(s, t)$ is the dimension of the group. In exponential coordinates, the $ℝ^{2}$-tuples of variables $g$ can be thought of as elements of the Lie algebra with basis $C(t)$. Then
 * $$dg = (dg)_{ij},$$

where
 * $$(dg)_{ij} = \sum_{1 \le i \le k} \frac{\partial g_{ij}}{\partial X^i}dX_i.$$

In this expression,
 * $$\frac{\partial g_{ij}}{\partial X^i} \in \mathrm{T}G_g, \quad dX_i \in \mathrm{T^*}G_g.$$

Gamma matrices
Tong footnote.

Weinberg footnote.

For an explicit statement of the disputed sentence, please see. I consider David Tong a reliable reference.

About dependencies in odd spacetime dimension: In spaces or spacetimes with with odd dimensionality, the totally antisymmetric tensors of rank $$ can be linearly related by the conditions
 * $$\gamma^{[\mu_1}\gamma^{\mu_2}\cdots\gamma^{\mu_r]} = \epsilon^{\mu_1\mu_2\cdots\mu_d} \gamma_{[\mu_{r+1}}\gamma_{\mu_{r+2}}\cdots\gamma_{\mu_{d}]}, \quad r = 0, 1, \ldots, d-1,$$

where $C^{1}$ is the totally antisymmetric symbol and the left hand side is taken as the identity matrix for $β(1, t)$. Under these conditions there are only $t$ independent tensors, requiring $β$-matrices of dimensionality at least $s$.

So, yes, we can have dependencies. There is nothing in the definition of a Clifford algebra ruling it out. I consider Steven Weinberg a reliable reference.

About the representations in general: D even
 * $β$ ($g$) are $(g)_{ij}, 0 ≤ i, j ≤ n$×$k$ matrices, which are unique modulo a similarity transformation.
 * All $k$ are linearly independent.
 * All $k$×$(X^{i})$ matrices can be decomposed into $X_{i}, 1 ≤ i ≤ k$ ($γ^{5}$).

D odd
 * $(Γ^{A}) = (γ^{μ}, iγ^{5})$ ($A = (0, 1, 2, 3, 4)$) are ${ Γ^{A}, Γ^{B} } = 2η^{AB}$×${ γ^{0}, γ^{1}, γ^{2}, γ^{3}, γ_{5} }$ matrices, which are not unique; there are two representations.
 * Not all $ε$ are linearly independent.
 * All $r = 0$×$2^{d−1}$ matrices can be decomposed into $γ$ ($2^{(d−1)/2}$).

This wraps it up. I don't know personally about the authors reliability, the reference is new to me, but someone I trust ensures me that it is good enough. You may also want to see Higher-dimensional gamma matrices that tells exactly the same story.


 * Appendix E
 * Appendix E

Witt algebra and Polynomial loop algebra combined
Let $r$ be a polynomial loop group with simple compact Lie group $s$ and Lie algebra $γ_{a} = γ_{a}^{&dagger;}$, and let $a=1, ...,D$ be the Witt group. Then, if $2^{D/2}$ is a faithful representation of $t$, actions of both $u$ and $u$ may be defined and subsequently combined. The set on which the action is to be defined is


 * $$F = C^\infty[S^1, V],$$

the set of smooth functions from the circle to $u$. $2^{D/2}$ acts on $n$ by


 * $$\Pi_G(g)f(z) = \Sigma(g)f(z), \quad g \in G, f \in F, z \in S^1.$$

This extends in a natural way to an action of $p$ on $p$, since elements $n, r, s, t$ of $S$ are maps from $Γ_{a}$ into $f$,


 * $$\Pi_L(l)f(z) = \Pi_G(l(z))f(z), \quad l \in L, l(z) \in G, f \in F, z \in S^1.$$

The Witt group acts by


 * $$\Pi_W(w)f(z) = f(w^{-1}(z))\quad w \in W, f \in F, z \in S^1.$$

Both of these last two constructs yield representations (and not merely group actions). Now consider the ordered pair $2^{D/2}$ and define its action on $f$ to be


 * $$(w, l)f(z) = ((w \circ l)f)(z) = w(lf(z)) = w(l(z)f(z)) = l(w^{-1}(z))f(w^{-1}(z)).$$

Here, all action takes place in $2^{D/2}$, the linear operators on $f$. In order to save bandwidth and the eyes of the reader, it is necessary to identify elements of $g$ and $g$ with their images in $γ_{a_{1}a_{1}...a_{n}}|undefined$. Examination of $0 ≤ n ≤ D$ suggests the multiplication law


 * $$(w_1, l_1)(w_2, l_2) = (w_1 \circ w_2, (l_1 \circ w_2)\cdot l_2).$$

By this definition


 * $$(w_1, l_1)(w_2, l_2)f(z) = (w_1, l_1)w_2l_2f(z) = (w_1, l_1)w_2l_2(z)f(z) = (w_1, l_1)l_2(w_2^{-1}(z))f(w_2^{-1}(z)) \equiv (w_1, l_1)g(z).$$

Then


 * $$\begin{align}(w_1, l_1)g(z) &= l_1(w_1^{-1}(z))g(w_1^{-1}(z)) = l_1(w_1^{-1}(z))l_2(w_2^{-1}(w_1^{-1}(z)))f(w_2^{-1}(w_1^{-1}(z)))\\

&= l_1(w_1^{-1}(z))l_2((w_1 \circ w_2)^{-1}(z))f((w_1 \circ w_2)^{-1}(z)).\end{align}$$

Now


 * $$(l_1 \circ w_2)(z) = l_1(w_2(z)),$$

hence


 * $$(w_1 \circ w_2)(l_1 \circ w_2)(z) = w_1l_1(w_2(w_2^{-1}(z)) = w_1l_1(z) = l_1(w_1^{-1}(z)).$$

With this replacement,


 * $$(w_1, l_1)g(z) = (l_1 \circ w_2)((w_1 \circ w_2)^{-1}(z))l_2((w_1 \circ w_2)^{-1}(z))f((w_1 \circ w_2)^{-1}(z)).$$

In other words,


 * $$(w_1, l_1)(w_2, l_2)f(z) = (w_1 \circ w_2)(l_1 \circ w_2)\cdot l_2 f(z),$$

and the tentative group multiplication law


 * $$(w_1, l_1)(w_2, l_2) = (w_1 \circ w_2, (l_1 \circ w_2)\cdot l_2)$$

suggests itself. It might be noted that $γ_{a} = γ_{a}^{&dagger;}$.

This is indeed a group operation, with inverse operation


 * $$(w, l)^{-1} = (w^{-1}, (l \circ w^{-1})^{-1}).$$

For the inverse,


 * $$\begin{align}(w, l)(w^{-1}, (l \circ w^{-1})^{-1}) &= (w \circ w^{-1}, (l \circ w^{-1})\cdot (l \circ w^{-1})^{-1})) = (e_W, e_L)\\

(w^{-1}, (l \circ w^{-1})^{-1})(w, l) &= (w^{-1} \circ w, ((l \circ w^{-1})^{-1} \circ w)\cdot l) = (e_W, ((w \circ l^{-1}) \circ w)\cdot l) = (e_W, e_L)\end{align}$$

For associativity,
 * $$[(w_1, l_1)(w_2, l_2)](w_3, l_3) = (w_1 \circ w_2, (l_1 \circ w_2)\cdot l_2)(w_3, l_3) = (w_1 \circ w_2 \circ w_3, (((l_1 \circ w_2)\cdot l_2) \circ w_3) \cdot l_3),$$

and
 * $$(w_1, l_1)[(w_2, l_2)(w_3, l_3)] = (w_1, l_1)(w_2 \circ w_3, (l_2 \circ w_3)\cdot l_3) = (w_1 \circ w_2 \circ w_3, (l_1 \circ w_2 \circ w_3)\cdot (l_2 \circ w_3)\cdot l_3).$$

By using $a=1, ...,D$ in the first equations last expression, one finds that these are indeed equal.

By now, it is established that the pairs $2^{(D−1)/2}$) do constitute a group with the above multiplication rule. The multiplication rule makes no mention of $f$, the representations are faithful, and hence one may define the operations directly on pairs $2^{(D−1)/2}$ and forget about $g$. $x$ is normal in $&and; f$,


 * $$(w, l)(1_W, l_0)(w, l)^{-1} = (w, l)(1_W, l_0)(w^{-1}, (l \circ w^{-1})^{-1}) = (1_W, \ldots),$$

and $$ and $$ intersect in $Γ_{a}$. Thus $m$ is the semidirect product


 * $$E = W \otimes_s L.$$

The Lie algebra then is the set $2^{(D−1)/2}$. The Lie bracket between elements of $2^{(D−1)/2}$ and $γ_{a_{1}a_{1}...a_{n}}|undefined$ follow from the explicit expressions


 * $$ d_m = -z^{m+1}\frac{d}{dz}, T^a_n = z^nT^a,$$

from which it follows

Conformal group

 * $$\begin{align}x'^\mu &= \frac {k^2x^\mu}{x^\mu x_\mu},\\

x''^\mu &= x'^\mu - a^\mu,\\ x'^\mu &= \frac {k^2x^\mu}{x^\mu x_\mu}.\end{align}$$

Together,
 * $$x'^\mu = \frac{x^\mu - a^\mu x^2}{\sigma(x)}, \quad \sigma(x) = 1 - 2a^\mu x_\mu + a^2x^2, \quad a^2 = a^\mu a_\mu, x^2 = x^\mu x_\mu.$$

Relativistic string
The coordinates $$ of a string,


 * $$X (\tau, \sigma) = (X^0(\tau,\sigma), X^1(\tau,\sigma), X^2(\tau,\sigma), \ldots, X^d(\tau,\sigma)),$$

take on their values in generalized Minkowski space of dimension $$D$$. For each $$(\sigma, \tau)$$ they comprise a $$D$$-vector and reduce to a $$-vector for $$D=4$$. Define the coordinate partial derivatives


 * $$\begin{align}\dot{X} &\equiv \frac{\partial X}{\partial \tau},\\

X^\prime &\equiv \frac{\partial X}{\partial \sigma}.\end{align}$$

Define also the differential forms


 * $$dX^\mu = \frac{\partial X^\mu}{\partial \sigma}d\sigma + \frac{\partial X^\mu}{\partial \tau}d\tau$$

and


 * $$dF^{\mu\nu} = dX^\mu \wedge dX^\nu = \left(\frac{\partial X^\mu}{\partial \sigma}d\sigma + \frac{\partial X^\mu}{\partial \tau}d\tau\right) \wedge \left( \frac{\partial X^\nu}{\partial \sigma}d\sigma + \frac{\partial X^\nu}{\partial \tau}d\tau \right)

= \left(\frac{\partial X^\mu}{\partial \sigma} \frac{\partial X^\nu}{\partial \tau} - \frac{\partial X^\mu}{\partial \tau} \frac{\partial X^\nu}{\partial \sigma}\right) d\sigma \wedge d\tau $$

as well as


 * $$dF_{\mu\nu} = \partial_\mu \wedge \partial_\nu = \left(\frac{\partial X_\mu}{\partial \sigma}\partial_\sigma + \frac{\partial X_\mu}{\partial \tau}\partial_\tau\right) \wedge \left( \frac{\partial X_\nu}{\partial \sigma}\partial_\sigma + \frac{\partial X_\nu}{\partial \tau}\partial_\tau \right)

= \left(\frac{\partial X_\mu}{\partial \sigma} \frac{\partial X_\nu}{\partial \tau} - \frac{\partial X_\mu}{\partial \tau} \frac{\partial X_\nu}{\partial \sigma}\right) \partial_\sigma \wedge \partial_\tau .$$

Then


 * $$dF^{\mu\nu}dF_{\mu\nu} = \left( \frac{\partial X^\mu}{\partial \sigma} \frac{\partial X^\nu}{\partial \tau} - \frac{\partial X^\mu}{\partial \tau} \frac{\partial X^\nu}{\partial \sigma} \right)^2

= \left( \frac{\partial X^\mu}{\partial \sigma} \frac{\partial X^\nu}{\partial \tau} - \frac{\partial X^\mu}{\partial \tau} \frac{\partial X^\nu}{\partial \sigma} \right)^2$$

Metric
The metric on the world sheet of a string can be obtained in several ways. One "intuitive" way is to simply compute the line element. Thus, consider an infinitesimal displacement


 * $$ds = \frac{\partial X^\mu}{\partial \sigma}d\sigma + \frac{\partial X^\mu}{\partial \tau}d\tau.$$

Then the line element is


 * $$ds^2 = ds \cdot ds = $$

Static gauge
One obvious possibility is to put $$X^0(\tau, \sigma) = ct$$ in some Lorentz frame. Then

$$X^0(t, \sigma) = ct, X^i(\tau, \sigma) = X^i(t, \sigma), \quad 1 \le i \le D$$

for fixed $$t$$ comprises a string in space and a two-dimensional surface in spacetime called the world sheet for non-fixed $$t$$. This is called the static gauge. The tangent vector of a path traced out by a point particle is timelike or lightlike since it travels with a speed less than or equal to the speed of light. Thus require in the neighborhood of each of its points there exists an infinitesimal displacement along the surface which points in the time-like or light-like direction.

One may impose the constraints that the tangent vectors should be orthogonal,


 * $$\dot{X} \cdot X^\prime = 0,$$

and normalized,,


 * $$\dot{X}^2 + {X^\prime}^2 = 0,$$

at any point $$(\sigma, \tau)$$. One consequence of these conditions is that the ends of the string move at the speed of light. Another is the fact that only the motion of the string perpendicular to itself is dynamically significant.

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