Talk:Mathematical formulation of the Standard Model/Archive 1

Level of this article
Would it be possible to make this article a bit less hermetic? Many notations are not defined. It is for sure a big work I am not qualified for but I think one should create articles and link them to this article in order to allow the reader to understand (at least in principle) what it is all about. For example : what are the Dirac and Weyl notations? What are the $$\gamma$$ matrices? Could the editors think about making a bit more pedagogy? Thank you. 131.220.68.177 09:23, 20 July 2005 (UTC)


 * Quite right, and thanks for making the comment. When I started this article that is what I had in mind. That + a way of pulling together the "important" aspects of the SM. It will happen, but it will take time to write the background articles. --Bambaiah 09:39, July 20, 2005 (UTC)


 * Perhaps a better title for this article might be "Standard model (technical overview)" or "Standard model (technical details)". "Basic details" implies it's been written with the average, non-technical Wikipedia reader in mind, whom I guarantee will be lost within the first 2 paragraphs. If anything, the companion Standard model article is better suited to the moniker "basic details". 59.167.43.134 15:33, 3 August 2005 (UTC)

I was thinking "Standard Model (mathematical formulation)". But you know, this is supposed to be an encyclopedia. Should something that's so technical it's only accessible to advanced readers be moved to Wikibooks instead? This article could also be significantly expanded to actually make it possible for people who actually have had advanced college-level math understand what's being presented here. And after that's done, it really would be book length. -- Beland

Yes, think it is just the "basic details" in the title that is misleading. The actual article is at Standard Model, this is by all appearances supposed to be a quick summary of the facts for reference. Maybe a fitting title would be Standard Model Fact Sheet or something. Baad 08:27, 31 October 2005 (UTC)

Much work needed in this article
"This article uses the Dirac basis instead of the more appropriate Weyl basis for describing spinors. The Weyl basis is more convenient because there is no natural correspondence between the left handed and right handed fermion fields other than that generated dynamically through the Yukawa couplings after the Higgs field has acquired a vacuum expectation value"

Ok now without having to know what a Dirac vs a Weyl basis is, this comes across as saying "We're going to bake this cake with sawdust, what it really needs is eggs and flour because they bind well and provide nutrition, but we've decided for this article to use sawdust instead." Kuratowski&#39;s Ghost (talk) 23:46, 5 July 2008 (UTC)

It's technically more like saying "We're going to discuss bathroom water consumption in terms of flow and temperature, although we should really be discussing it in terms of hot water flow and cold water flow." Having that paragraph as the first in the body of the article wasn't very helpful though; I've moved it a bit further down.

A lot of the article text basically argues for the point of view that neutrinos are massless, despite experimental evidence to the contrary. That needs a rewrite to be more neutral. 213.21.117.168 (talk) 13:59, 27 March 2010 (UTC)

A big cleanup needed
This article in its current (June 2009) state is no more than a collection of some facts without any higher level of 'composition'. Instead of presenting in a logical way the mathematical background of the Standard Model, it just chooses some random facts in a random order. I like and really support the idea of having an article about the SM that summarizes the technical (let's say, mathematical) details of the theory (after all, I use Wikipedia quite a lot as a secondary source to prepare myself for my exams), but the current state of the article is really far from this. Since I'm not an expert of the field, nor a native English speaker, I don't find myself good enough to do this job, but please, somebody who has a good English textbook, it would be great if you could fix this article... Thanks! AdamSiska (talk) 02:58, 24 June 2009 (UTC)

Rewrite draft
The following is a draft for a new first section of the article, meant to replace the current one; section headings are one level below what they're intended to be. I don't think I'll get it finished today though. 130.239.119.207 (talk) 13:12, 9 April 2010 (UTC)
 * I think that this is excellently written, and should be moved to the main article as it stands. As it looks a little abandoned, unless there are any objections I'll do this myself in a few days. 2.27.105.116 (talk) 23:54, 17 July 2012 (UTC)

Quantum field theory
The standard model is a quantum field theory, meaning its fundamental objects are quantum fields which are defined at all points in spacetime: That these are quantum rather than classical fields have the mathematical consequence that they are operator-valued. In particular, values of the fields generally don't commute. As operators, they act upon the quantum state (ket vector).
 * the fermion field $$\psi$$, which accounts for "matter particles",
 * the electroweak boson fields $$W_1$$, $$W_2$$, $$W_3$$, and $$B$$,
 * the gluon field $$G_a$$, and
 * the Higgs field $$\phi$$.

The dynamics of the quantum state and the fundamental fields are determined by the Lagrangian density $$\mathcal{L}$$ (usually for short just called the Lagrangian). This plays a role similar to that of the Schrödinger equation in non-relativistic quantum mechanics, but a Lagrangian is not an equation — rather, it is an polynomial function of the fields and their derivatives. While it would be possible to derive a system of differential equations governing the fields from the Langrangian, it is more common to use other techniques to compute with quantum field theories.

The standard model is furthermore a gauge theory, which means there are degrees of freedom in the mathematical formalism which do not correspond to changes in the physical state. The gauge group of the standard model is $$\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$$, where U(1) acts on $$B$$ and $$\phi$$, SU(2) acts on $$W$$ and $$\phi$$, and SU(3) acts on $$G$$. The fermion field $$\psi$$ also transforms under these symmetries, although all of them leave some parts of it unchanged.

The role of the quantum fields
In classical mechanics, the state of a system can usually be captured by a small set of variables, and the dynamics of the system is thus determined by the time evolution of these variables. In classical field theory, the field is part of the state of the system, so in order to describe it completely one effectively introduces separate variables for every point in spacetime (even though there are many restrictions on how the values of the field "variables" may vary from point to point, for example in the form of field equations involving partial derivatives of the fields).

In quantum mechanics, the classical variables are turned into operators, but these do not capture the state of the system, which is instead encoded into a wavefunction $$\psi$$ or more abstract ket vector. If $$\psi$$ is an eigenstate with respect to an operator $$P$$, then $$ P \psi = \lambda \psi $$ for the corresponding eigenvalue $$\lambda$$, and hence letting an operator $$P$$ act on $$\psi$$ is analogous to multiplying $$\psi$$ by the value of the classical variable to which $$P$$ corresponds. By extension, a classical formula where all variables have been replaced by the corresponding operators will behave like an operator which, when it acts upon the state of the system, multiplies it by the analogue of the quantity that the classical formula would compute. The formula as such does however not contain any information about the state of the system; it would evaluate to the same operator regardless of what state the system is in.

Quantum fields relate to quantum mechanics as classical fields do to classical mechanics, i.e., there is a separate operator for every point in spacetime, and these operators do not carry any information about the state of the system; they are merely used to exhibit some aspect of the state, at the point to which they belong. In particular, the quantum fields are not wavefunctions, even though the equations which govern their time evolution may be deceptively similar to those of the corresponding wavefunction in a semiclassical formulation. There is no variation in strength of the fields between different points in spacetime; the variation that happens is rather one of phase factors. —Preceding unsigned comment added by 130.239.119.132 (talk) 16:01, 14 February 2011 (UTC)

Vectors, scalars, and spinors
Mathematically it may look as though all of the fields are vector-valued (in addition to being operator-valued), since they all have several components, can be multiplied by matrices, etc., but physicists assign a more specific meaning to the word: a vector is something which transforms like a four-vector under Lorentz transformations, and a scalar is something which does not transform under Lorentz transformations. The $$B$$, $$W_j$$, and $$G_a$$ fields are all vectors in this sense, so the corresponding particles are said to be vector bosons. The Higgs field $$\phi$$ is a scalar.

The fermion field $$\psi$$ does transform under Lorentz transformations, but not like a vector should; rotations will only turn it by half the angle a proper vector should. Therefore these constitute a third kind of quantity, which is known as a spinor.

It is common to make use of abstract index notation for the vector fields, in which case the vector fields all come with a Lorentzian index $$\mu$$, like so: $$B^\mu\,$$, $$W_j^\mu$$, and $$G_a^\mu$$. If abstract index notation is used also for spinors then these will carry a spinorial index and the Dirac gamma will carry one Lorentzian and two spinorian indices, but it is more common to regard spinors as column matrices and the Dirac gamma $$\gamma^\mu$$ as a matrix which additionally carries a Lorentzian index. The Feynman slash notation can be used to turn a vector field into a linear operator on spinors, like so: $$ {\not}B = \gamma^\mu B_\mu $$; this may involve raising and lowering indices.

Alternative presentations of the fields
As is common in quantum theory, there is more than one way to look at things. In particular, there are several alternative presentations of the basic fields which in particular contexts may be more appropriate than those that are given above.

The barred fermion field $$ \bar{\psi} $$ is defined to be $$ \psi^\dagger \gamma^0 $$, where $$ \dagger $$ denotes the Hermitian adjoint and $$\gamma^0$$ is the zeroth gamma matrix. If $$\psi$$ is thought of as an n ×1 matrix then $$\bar{\psi}$$ should be thought of as a 1×n matrix.

Rather than having one fermion field $$\psi$$, it can be split up into separate components for each type of particle. This mirrors the historical evolution of quantum field theory, since the electron component $$\psi_e$$ (describing the electron and its antiparticle the positron) is then the original $$\psi$$ field of quantum electrodynamics, which was later accompanied by $$\psi_\mu$$ and $$\psi_\tau$$ fields for the muon and tauon respectively (and their antiparticles). Electroweak theory added $$\psi_{\nu_e}$$, $$\psi_{\nu_\mu}$$, and $$\psi_{\nu_\tau}$$ for the corresponding neutrinos, and the quarks add still further components. In order to be four-spinors like the electron and other lepton components, there must be one quark component for every combination of flavour and colour, bringing the total to 24 (3 for charged leptons, 3 for neutrinos, and 2·3·3 = 18 for quarks).

An independent decomposition of $$\psi$$ is that into chirality components
 * "Left" chirality: $$\psi^L = \frac{1}{2}(1-\gamma_5)\psi$$
 * "Right" chirality: $$\psi^R = \frac{1}{2}(1+\gamma_5)\psi$$

where $$\gamma_5$$ is the fifth gamma matrix. This is important because the weak interaction only interacts with the left component of the field. It is furthermore only $$\psi^L$$ which is transformed by the SU(2) gauge group (in a sense because the weak isospin of $$\psi^R$$ is zero) and U(1) acts differently on $$\psi^L_e$$ than on $$\psi^R_e$$ (because they have different weak hypercharges).

The physically observed Z bosons and photons rather correspond to the fields
 * $$\begin{align}

Z={}& \cos \theta_W W_3 - \sin \theta_W B \text{,}\\ A ={}& \sin \theta_W W_3 + \cos \theta_W B \text{,} \end{align}$$ where $$\theta_W$$ is the Weinberg angle. This $$A$$ field has as a classical analogue the electromagnetic four-potential.

The charged W bosons similarly correspond to linear combinations
 * $$W^{\pm} = \frac1\sqrt2\left(W_1 \pm i W_2\right).$$

Finally, the quantum fields are sometimes decomposed into "positive" and "negative" energy parts: $$ \psi = \psi^{+} + \psi^{-} $$. This is not so common when a quantum field theory has been set up, but often features prominently in the process of quantizing a field theory. —Preceding unsigned comment added by 94.255.156.147 (talk) 23:47, 14 February 2011 (UTC)

Perturbative QFT and the interaction picture
Much of the qualititative descriptions of the standard model in terms of "particles" and "forces" comes from the perturbative quantum field theory view of the model. In this, the Langrangian is decomposed as $$ \mathcal{L} = \mathcal{L}_0 + \mathcal{L}_\mathrm{I} $$ into separate free field and interaction Langrangians. The free fields care for particles in isolation, whereas processes involving several particles arise through interactions. The idea is that the state vector should only change when particles interact, meaning a free particle is one whose quantum state is constant. This corresponds to the interaction picture in quantum mechanics.

In the more common Schrödinger picture, even the states of free particles change over time: typically the phase changes at a rate which depends on their energy. In the alternative Heisenberg picture, state vectors are kept constant, at the price of having the operators (in particular the observables) be time-dependent. The interaction picture constitutes an intermediate between the two, where some time dependence is placed in the operators (the quantum fields) and some in the state vector. In QFT, the former is called the free field part of the model, and the latter is called the interaction part. The free field model can be solved exactly, and then the solutions to the full model can be expressed as pertubations of the free field solutions, for example using the Dyson series.

It should be observed that the decomposition into free fields and interations is in principle arbitrary. For example renormalization in QED modifies the mass of the free field electron to match that of a physical electron (with an electromagnetic field), and will in doing so add a term to the free field Lagrangian which must be cancelled by a counterterm in the interaction Lagrangian, that then shows up as a two-line vertex in the Feynman diagrams. This is also how the Higgs field is thought to give particles mass: the part of the interaction term which corresponds to the (nonzero) vacuum expectation value of the Higgs field is moved from the interaction to the free field Lagrangian, where it looks just like a mass term having nothing to do with Higgs.

Free fields
Under the usual free/interaction decomposition, which is suitable for low energies, the free fields obey the following equations: These equations can be solved exactly. One usually does so by considering first solutions that are periodic with some period $$L$$ along each spatial axis; later taking the $$L \rightarrow \infty$$ limit will lift this periodicity restriction.
 * The fermion field $$\psi$$ satisfies the Dirac equation; $$ (i \hbar {\not}\partial - m_f c) \psi_f = 0 $$ for each type $$f$$ of fermion.
 * The photon field $$A$$ satisfies the wave equation $$ \partial_\mu \partial^\mu A^\nu = 0 $$.
 * The Higgs field $$\phi$$ satisfies the Klein–Gordon equation.
 * The weak interaction fields $$Z$$, $$W^+$$, and $$W^-$$ also satisfy the Klein–Gordon equation.

In the periodic case, the solution for a field $$F$$ (any of the above) can be expressed as a Fourier series of the form
 * $$ F(x) = \beta \sum_{\mathbf{p}} \sum_r E_{\mathbf{p}}^{-1/2} \left( a_r(\mathbf{p}) u_r(\mathbf{p}) e^{-ipx/\hbar} + b^\dagger_r(\mathbf{p}) v_r(\mathbf{p}) e^{ipx/\hbar} \right)$$

where: In the $$L \rightarrow \infty$$ limit, the sum would turn into an integral with help from the $$V$$ hidden inside $$\beta$$. The numeric value of $$\beta$$ also depends on the normalization chosen for $$u_r(\mathbf{p})$$ and $$v_r(\mathbf{p})$$.
 * $$\beta$$ is a normalization factor; for the fermion field $$\psi_f$$ it is $$ \sqrt{ m_f c^2 / V} $$, where $$ V = L^3 $$ is the volume of the fundamental cell considered; for the photon field $$ A^\mu $$ it is $$ \hbar c / \sqrt{2V} $$.
 * The sum over $$\mathbf{p}$$ is over all momenta consistent with the period $$L$$, i.e., over all vectors $$ \frac{2\pi\hbar}{L}(n_1,n_2,n_3)$$ where $$n_1,n_2,n_3$$ are integers.
 * The sum over $$r$$ covers other degrees of freedom specific for the field, such as polarization or spin; it usually comes out as a sum from $$1$$ to $$2$$ or from $$1$$ to $$3$$.
 * $$E_{\mathbf{p}}$$ is the relativistic energy for a momentum $$\mathbf{p}$$ quantum of the field, $$ = \sqrt{ m^2 c^4 + c^2 \mathbf{p}^2 } $$ when the rest mass is $$m$$.
 * $$a_r(\mathbf{p})$$ and $$b^\dagger_r(\mathbf{p})$$ are annihilation and creation respectively operators for "a-particles" and "b-particles" respectively of momentum $$\mathbf{p}$$; "b-particles" are the antiparticles of "a-particles". Different fields have different "a-" and "b-particles". For some fields, $$a$$ and $$b$$ are the same.
 * $$u_r(\mathbf{p})$$ and $$v_r(\mathbf{p})$$ are non-operators which carry the vector or spinor aspects of the field (where relevant).
 * $$p = (E_{\mathbf{p}}/c, \mathbf{p})$$ is the four-momentum for a quanta with momentum $$\mathbf{p}$$. $$px = p_\mu x^\mu$$ denotes an inner product of four-vectors.

Technically, $$a^\dagger_r(\mathbf{p})$$ is the Hermitian adjoint of the operator $$a_r(\mathbf{p})$$ in the inner product space of ket vectors. The identification of $$a^\dagger_r(\mathbf{p})$$ and $$a_r(\mathbf{p})$$ as creation and annihilation operators comes from comparing conserved quantities for a state before and after one of these have acted upon it. $$a^\dagger_r(\mathbf{p})$$ can for example be seen to add one particle, because it will add $$1$$ to the eigenvalue of the a-particle number operator, and the momentum of that particle ought to be $$\mathbf{p}$$ since the eigenvalue of the vector-valued momentum operator increases by that much. For these derivations, one starts out with expressions for the operators in terms of the quantum fields. That the operators with $$\dagger$$ are creation operators and the one without annihilation operators is a convention, imposed by the sign of the commutation relations postulated for them.

An important step in preparation for calculating in perturbative quantum field theory is to separate the "operator" factors $$a$$ and $$b$$ above from their corresponding vector or spinor factors $$u$$ and $$v$$. The vertices of Feynman graphs come from the way that $$u$$ and $$v$$ from different factors in the interaction Lagrangian fit together, whereas the edges come from the way that the $$a$$s and $$b$$s must be moved around in order to put terms in the Dyson series on normal form.

Possible typo - "The Gauge Field Lagrangian"
The formula for the Lagrangian of the electroweak gauge fields differs from that given in Electroweak Interaction, in the second instance of W the term a appears as superscript instead of a subscript and an extra comma appears (indicating taking the derivative with respect to mu?)

If this is a typo, could someone correct it; if not, please explain the apparent discrepancy. Thanks. Harryjohnston (talk) 02:33, 25 May 2011 (UTC)
 * I didn't find any typos. The comma here is just a spacer to separate different kinds of superscript. I don't particularly like that notation but it is not incorrect. Dauto (talk) 04:07, 25 May 2011 (UTC)


 * I replaced the comma by a space to avoid the confusion. $$a$$ is not a Lorentz index, so it may raised or lowered as wished. (The position of an index only tells us how the quantity transforms under Lorentz transformations. $$a$$ numbers different fields which Lorentz-transform independently, the transformation acting only on the Greek indices like $$A^{\mu'}_a = {\Lambda^{\mu'}}_{\nu} A^{\nu}_a$$.)&thinsp;&mdash;&thinsp;Pt&thinsp;(T) 08:17, 26 May 2011 (UTC)