User:Maschen/Spherical basis


 * "Spherical tensor" redirects to here.

In pure and applied mathematics, a spherical basis is the basis used to express spherical tensors. While spherical polar coordinates are one orthogonal coordinate system for expressing vectors and tensors using polar and azimuthal angles and radial distance, the spherical basis are constructed from the standard basis and complex numbers. The form the spherical basis takes closely relates to the description of angular momentum in quantum mechanics, and spherical harmonic functions.

A spherical tensor is a special case of a Cartesian tensor.

Definition
The spherical basis vectors can be defined in terms of the Cartesian basis (ex, ey, ez) = (e1, e2, e3) using complex-valued coefficients in the xy plane:


 * $$\begin{align}

\mathbf{e}_+ & = -\frac{1}{\sqrt{2}} \mathbf{e}_x -\frac{i}{\sqrt{2}}\mathbf{e}_y \\ \mathbf{e}_{-} & = +\frac{1}{\sqrt{2}}\mathbf{e}_x - \frac{i}{\sqrt{2}}\mathbf{e}_y \\ \end{align} \quad \rightleftharpoons \quad \mathbf{e}_\pm = \mp\frac{1}{\sqrt{2}}\left(\mathbf{e}_x \pm i\mathbf{e}_y\right)\,$$

where i denotes the imaginary unit, and one normal to the plane in the z direction:


 * $$\mathbf{e}_0 = \mathbf{e}_z $$

The relations can be summarized by a change of basis:


 * $$\begin{pmatrix}

\mathbf{e}_+ \\ \mathbf{e}_{-} \\ \mathbf{e}_0 \end{pmatrix} = \begin{pmatrix} - \frac{1}{\sqrt{2}} & - \frac{i}{\sqrt{2}} & 0 \\ + \frac{1}{\sqrt{2}} & - \frac{i}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \mathbf{e}_x \\ \mathbf{e}_y \\ \mathbf{e}_z \end{pmatrix} $$

It is convenient to provide a name for this unitary matrix:


 * $$\mathbf{U} = \begin{pmatrix}

- \frac{1}{\sqrt{2}} & - \frac{i}{\sqrt{2}} & 0 \\ + \frac{1}{\sqrt{2}} & - \frac{i}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

in other words, its Hermitian conjugate (complex conjugate and matrix transposed) is also the inverse matrix:


 * $$\mathbf{U}^\dagger = \mathbf{U}^{\star \mathrm{T}} = \mathbf{U}^{-1} = \begin{pmatrix}

- \frac{1}{\sqrt{2}} & + \frac{1}{\sqrt{2}} & 0 \\ + \frac{i}{\sqrt{2}} & + \frac{i}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

Inverting the basis equations in matrix form:


 * $$\begin{pmatrix}

\mathbf{e}_x \\ \mathbf{e}_y \\ \mathbf{e}_z \end{pmatrix} = \mathbf{U}^{-1}\begin{pmatrix} \mathbf{e}_+ \\ \mathbf{e}_{-} \\ \mathbf{e}_0 \end{pmatrix}$$

or explictly:


 * $$\begin{align}

\mathbf{e}_x & = -\frac{1}{\sqrt{2}}\mathbf{e}_+ + \frac{1}{\sqrt{2}}\mathbf{e}_{-} \\ \mathbf{e}_y & = +\frac{i}{\sqrt{2}}\mathbf{e}_+ + \frac{i}{\sqrt{2}}\mathbf{e}_{-} \\ \end{align}$$


 * $$\mathbf{e}_z = \mathbf{e}_0 $$

Properties

 * Orthonormal basis

The spherical basis is an orthonormal basis, since the inner product ⟨, ⟩ of every pair vanishes meaning the basis vectors are all mutually orthogonal:


 * $$\left\langle \mathbf{e}_+, \mathbf{e}_{-} \right\rangle = \left\langle \mathbf{e}_{-} , \mathbf{e}_0 \right\rangle = \left\langle \mathbf{e}_0 , \mathbf{e}_+ \right\rangle = 0 $$

and each basis vector is a unit vector:


 * $$ \left\langle\mathbf{e}_+, \mathbf{e}_{+} \right\rangle = \left\langle\mathbf{e}_{-} , \mathbf{e}_{-} \right\rangle = \left\langle\mathbf{e}_0 , \mathbf{e}_0 \right\rangle = 1 $$

hence the need for the normalizing factor of 1/√2.

In general, for two vectors with complex coefficients in the same real-valued orthonormal basis ei, with the property ei·ej = δij, we have:


 * $$\left\langle \mathbf{a}, \mathbf{b} \right\rangle = \mathbf{a} \cdot \mathbf{b}^\star = \sum_j a_j b_j^\star $$

where · is the usual dot product and the complex conjugate * must be used to keep the magnitude (or "norm") of the vector positive definite.


 * Coordinates in the Cartesian and spherical basis

A vector A can be written:


 * $$\mathbf{A} = A_x \mathbf{e}_x + A_y \mathbf{e}_y + A_z \mathbf{e}_z = A_+ \mathbf{e}_{+} + A_{-} \mathbf{e}_{-} + A_0 \mathbf{e}_0 $$

where the coordinates of A can be expressed easily in the standard basis, in Cartesian notation:


 * $$A_x = \left\langle \mathbf{A}, \mathbf{e}_x \right\rangle \,, \quad A_y = \left\langle\mathbf{A} , \mathbf{e}_y \right\rangle \,, \quad A_z = \left\langle\mathbf{A} , \mathbf{e}_z \right\rangle \quad \rightleftharpoons \quad A_j = \left\langle\mathbf{A} , \mathbf{e}_j\right\rangle $$

where the index j takes the values x, y, z, or 1, 2, 3, and this translates to the corresponding components for the spherical basis like so:



\begin{align} A_+ & = \left\langle \mathbf{A}, \mathbf{e}_+ \right\rangle = -\frac{A_x}{\sqrt{2}} + \frac{iA_y}{\sqrt{2}} \\ A_{-} & = \left\langle \mathbf{A}, \mathbf{e}_{-} \right\rangle = +\frac{A_x}{\sqrt{2}} + \frac{iA_y}{\sqrt{2}} \\ \end{align} \quad \rightleftharpoons \quad A_\pm = \left\langle \mathbf{A}, \mathbf{e}_\pm \right\rangle = \frac{1}{\sqrt{2}} \left( \mp A_x + iA_y \right) $$


 * $$A_0 = \left\langle\mathbf{A}, \mathbf{e}_0 \right\rangle = \left\langle \mathbf{A} , \mathbf{e}_z \right\rangle = A_z $$

or in matrix form:


 * $$\begin{pmatrix}

A_+ \\ A_{-} \\ A_0 \end{pmatrix} = \begin{pmatrix} - \frac{1}{\sqrt{2}} & + \frac{i}{\sqrt{2}} & 0 \\ + \frac{1}{\sqrt{2}} & + \frac{i}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} A_x \\ A_y \\ A_z \end{pmatrix} =\mathbf{U}^\mathrm{T} \begin{pmatrix} A_x \\ A_y \\ A_z \end{pmatrix} $$

Inverting these gives in matrix form:


 * $$\begin{pmatrix}

A_x \\ A_y \\ A_z \end{pmatrix} = \begin{pmatrix} - \frac{1}{\sqrt{2}} & + \frac{1}{\sqrt{2}} & 0 \\ - \frac{i}{\sqrt{2}} & - \frac{i}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} A_+ \\ A_{-} \\ A_0 \end{pmatrix} = (\mathbf{U}^\mathrm{T})^{-1} \begin{pmatrix} A_+ \\ A_{-} \\ A_0 \end{pmatrix} $$

or explicitly:


 * $$A_x = \frac{1}{\sqrt{2}} ( - A_+ + A_{-} ) $$
 * $$A_y = \frac{i}{\sqrt{2}} ( A_+ + A_{-} ) $$
 * $$A_z = A_0 $$


 * Inner product

The inner product between two vectors A and B in the spherical basis follows from the above definition of the inner product:


 * $$\left\langle \mathbf{A}, \mathbf{B} \right\rangle = A_+ B_+^\star + A_{-}B_{-}^\star + A_0 B_0^\star $$

The state space of angular momentum eigenkets
The rotation operator about the unit vector n (defining the axis of rotation) through angle θ is


 * $$U[R(\theta, \mathbf{n})] = \exp\left(-\frac{i\theta}{\hbar}\mathbf{n}\cdot\hat{\mathbf{J}}\right)$$

where J = (Jx, Jy, Jz) are the total angular momentum matrices:


 * $$J_x = \begin{pmatrix}

0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{pmatrix}\,\quad J_y = \begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{pmatrix}\,\quad J_z = \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

and let R = R(θ, n) is a rotation matrix. The orthonormal basis set for total angular momentum is $|j, m\rangle$, where j is the total angular momentum quantum number and m is the magnetic angular momentum quantum number, which takes values −j, −j + 1, ..., j − 1, j. A general state


 * $$ | \alpha \rangle = \sum_m |j,m\rangle $$

in the space rotates to a new state $|\overline{α}\rangle$ by:


 * $$ |\bar{\alpha} \rangle = U(R)|\alpha\rangle $$

using the completeness condition:


 * $$ I = \sum_m |j, m' \rangle \langle j, m' | $$

we have


 * $$ | \bar{\alpha} \rangle = I U(R)|\alpha\rangle = \sum_{mm'} | j, m' \rangle \langle j, m' | U(R) |j , m \rangle $$

introducing the Wigner matrix elements:


 * $$D^{(j)}_{m'm} = \langle j, m' | U(R) |j,m \rangle$$

gives the matrix multiplication:


 * $$ | \bar{\alpha} \rangle = \sum_{mm'} D^{(j)}_{m'm} | j, m \rangle $$

For one basis ket:


 * $$ | \overline{j, m} \rangle = \sum_{mm'} D^{(j)}_{m'm} | j , m \rangle $$

An operator Ω is invariant under a unitary transformation U if:


 * $$\hat{\Omega} = {U}^\dagger \hat{\Omega} U $$

in this case for the rotation U(R):


 * $$\hat{\Omega} = {U(R)}^\dagger \hat{\Omega} U(R) = \exp\left(\frac{i\theta}{\hbar}\mathbf{n}\cdot\hat{\mathbf{J}}\right) \hat{\Omega} \exp\left(-\frac{i\theta}{\hbar}\mathbf{n}\cdot\hat{\mathbf{J}}\right) $$

The infinitesimal rotation operator is:


 * $$U[R(\theta, \mathbf{n})] = 1 - \frac{i\theta}{\hbar}\mathbf{n}\cdot\hat{\mathbf{J}} $$

Scalar operators
A scalar operator is invariant under rotations:


 * $$ U(R)^\dagger \hat{S} U(R) = \hat{S}$$

Vector operators
A vector operator can be rotated according to:


 * $$ U(R)^\dagger \hat{V}_i {U(R)} = R_{ij} \hat{V}_j $$

from this one can derive the commutation relation:


 * $$ \left[ \hat{V}_a, \hat{J}_b \right] = i \hbar \varepsilon_{abc} \hat{V}_c $$

where εijk is the Levi-Civita symbol, which all vector operators must satisfy, by construction.

Tensor operators
A tensor operator can be rotated according to:


 * $$ U(R)^\dagger \hat{T}_{pqr\cdots} U(R) = R_{pi}R_{qj}R_{rk}\cdots \hat{T}_{ijk\cdots}$$

Consider a dyadic tensor with components T = aibj, this rotates infinitesimally according to:


 * $$ U(R)^\dagger \hat{T}_{pq} U(R) = R_{pi} R_{qj} \hat{T}_{ij} = R_{pi} \hat{a}_i R_{qj} \hat{b}_j $$

Cartesian dyadic tensors are reducible, which means they can be re-expressed in terms of the two vector operators


 * $$\hat{\mathbf{a}} = \mathbf{e}_i \hat{a}_i \,,\quad \hat{\mathbf{b}} = \mathbf{e}_j \hat{b}_j $$

into a rank 0 tensor (scalar), a rank 1 tensor (an antisymmetric tensor), and a rank 2 tensor (a symmetric tensor with zero trace):


 * $$\mathbf{T} = \mathbf{T}^{(1)} + \mathbf{T}^{(2)} + \mathbf{T}^{(3)} $$

where the first term


 * $$T^{(1)}_{ij} = \frac{\hat{a}_k \hat{b}_k}{3}\delta_{ij} $$

includes just one component, a scalar equivalently written (a·b)/3, the second


 * $$T^{(2)}_{ij} = \frac{1}{2}[\hat{a}_i \hat{b}_j - \hat{a}_j \hat{b}_i] = \hat{a}_{[i} \hat{b}_{j]} $$

includes three independent components, equivalently the components of (a×b)/2, and the third


 * $$T^{(3)}_{ij} = \frac{1}{2}(\hat{a}_i \hat{b}_j + \hat{a}_j \hat{b}_i) - \frac{\hat{a}_k \hat{b}_k}{3}\delta_{ij} = \hat{a}_{(i} \hat{b}_{j)} - T^{(1)}_{ij}$$

includes five independent components. Throughout, δij is the Kronecker delta, the components of the identity matrix. The number in the superscripted brackets denotes the tensor rank. These three terms are irreducible, which means they cannot be decomposed further and still be tensors satisfying the defining transformation laws under which they must be invariant. These also correspond to the number of spherical harmonic functions 2$\ell$ + 1 for ℓ = 0, 1, 2, the same as the ranks for each tensor. Each of the irreducible representations T(1), T(2) ... transform like angular momentum eigenstates according to the number of independent components.

Consider the case where a is the position vector x and b the momentum vector p for a particle. The reduction of T into the three parts above will contain terms with orbital angular momentum and action.

Spherical harmonic functions and Spherical tensors
The spherical basis is closely related to the algebra of orbital angular momentum in quantum mechanics, and the eigenstates $|l, m\rangle$ of the orbital angular momentum operator L are spherical harmonics:


 * $$ Y_\ell^m( \theta, \varphi ) = \langle \theta,\phi | m,\ell \rangle = \sqrt  \, P_\ell^m ( \cos{\theta} ) \, e^{i m \phi } $$

where Pℓm is an associated Legendre polynomial. The formalism of spherical harmonics which have wide applications in applied mathematics; solutions of Laplace's equation on a 3d sphere are also spherical harmonics.

A spherical harmonic state rotates according to:


 * $${U(R)}^\dagger T_{k}^{q'} U(R) = \sum_{q} D^{(k)}_{qq'} T_{k}^{q} $$

A spherical tensor Tkq of rank k is defined to rotate according to:


 * $${U(R)}^\dagger T_{k}^{q'} U(R) = \sum_{q} D^{(k)}_{qq'} T_{k}^{q} $$

where q = ... k, k − 1, ..., −k + 1, −k. For spherical tensors, k and q are analogous labels for ℓ and m respectively.

from which the commutation relations can be derived:


 * $$\left[ J_\pm, T_{k}^{q} \right] = \pm \hbar \sqrt{( k \mp q )(k \pm q + 1)} T_{k}^{q \pm 1} $$
 * $$\left[ J_z, T_{k}^{q} \right] = \hbar q T_{k}^{q} $$

A spherical harmonic state can be constructed out of the Clebsch–Gordan coefficients:

similarly for the spherical tensor:

Orbital angular momentum and spherical harmonics
Orbital angular momentum operators have the ladder operators:


 * $$L_\pm = L_x \pm i L_y = \pm ( \pm L_x + i L_y ) $$

which raise or lower the orbital magnetic quantum number m by one unit, m takes values −ℓ, −ℓ + 1, ... ℓ − 1, ℓ. (Some authors place a factor of 1/2 on the left hand side of the equation). This has almost exactly the same form as the spherical basis, aside from constant multiplicative factors.

The angular momentum operators satisfy the commutation rules:


 * $$\left[ L_x, L_y \right] = i\hbar L_z $$

and cyclic permutations of x, y, z components.

Spherical tensor operators and quantum spin
Spherical tensors are formed from algebraic combinations of the spin matrices Sx, Sy, Sz, for a spin system with total quantum number j = ℓ + s (and ℓ = 0). Tensor operators can be constructed from two perspectives.

One way is to specify how spherical tensors transform under a physical rotation - a group theoretical definition. A rotated spin eigenstate can be decomposed into a linear combination of the initial eigenstates: the coefficients in the linear combination consist of Wigner rotation matrix entries. Spherical tensor operators are sometimes defined as the set of operators that transform just like the spin eigenkets under a rotation.

Another related procedure requires that the spherical tensors satisfy certain commutation relations with respect to the basic operators Sx, Sy, Sz - a differential or algebraic definition, and arises from the behavior of tensor operators under infinitesimally small rotations.

Since Sx, Sy, Sz are usually introduced algebraically through the angular momentum commutation rules


 * $$\left[S_x, S_y \right] = i\hbar S_z $$

and cyclic permutations of x, y, z components. The commutation approach is a popular way of introducing spherical tensors. Additional physical and practical motivations for utilizing spherical tensors include:


 * aid in the calculation of matrix elements when rotational symmetry is present.

Applications
Spherical bases have broad applications in pure and applied mathematics and physical sciences where spherical geometries occur.

Magnetic resonance
The spherical tensor formalism provides a common platform for treating coherence and relaxation in nuclear magnetic resonance. In NMR and EPR, spherical tensor operators are employed to express the quantum dynamics of particle spin, by means of an equation of motion for the density matrix entries, or to formulate dynamics in terms of an equation of motion in Liouville space. The Liouville space equation of motion governs the observable averages of spin variables. When relaxation is formulated using a spherical tensor basis in Liouville space, insight is gained because the relaxation matrix exhibits the cross-relaxation of spin observables directly.