Standard Model Lagrangian and Euler-Lagrange Equations

Question

First off, note that I only know physics through quantum mechanics, so forgive me if this is a foolish question. I've seen the Standard Model's Lagrangian density written out in full. My question is, could one send this through the Euler-Lagrange equations to get a system of equations describing the evolution of the fields, or does quantum field theory have different ways to extract information from it?

Answer 1

There are several formalisms in which you can work out quantum field theory, and one of them does indeed involve using the Euler-Lagrange equation in some capacity. This is the Schwinger–Dyson equation.

As all quantum formalisms should be roughly equivalent, you can probably prove this quite generally, but it is simplest to show using the path integral formulation (quite simply because it's the one where the action appears directly). Recall that in the path integral formalism of QFT, we have, roughly speaking,

\begin{equation} \langle 0 | F[\phi(x)] | 0 \rangle = \int \mathscr{D}\phi(x) F[x] e^{i S[\phi(x)]} \end{equation}

Using some manner of equivalent of the fundamental theorem of calculus, we have that the path integral of this total derivative is zero :

\begin{equation} \int \mathscr{D}\phi(x) \frac{\delta}{\delta \phi(x)} \left[ F[x] e^{i S[\phi(x)]} \right] = 0 \end{equation}

And therefore, we get that

\begin{equation} \int \mathscr{D}\phi(x) e^{i S[\phi(x)]} \left[ \frac{\delta F[x]}{\delta \phi(x)} + i F[x] \frac{\delta S}{\delta \phi(x)} \right] = 0 \end{equation}

This leads to the Schwinger-Dyson equation :

\begin{equation} \left\langle \frac{\delta F[x]}{\delta \phi(x)} \right\rangle = - i \left\langle F[x] \frac{\delta S}{\delta \phi(x)} \right\rangle \end{equation}

In particular, if we consider the raw quantity for $F = 1$, we simply get

\begin{equation} \left\langle \frac{\delta S}{\delta \phi(x)} \right\rangle = 0 \end{equation}

This is the equivalent of the Euler-Lagrange equation for quantum field theory, and this is why we can use the Klein-Gordon or whatever else equation to work out the field operators of our theory.

Answer 2

That is exactly what you do.

Note that you don't need quantum field theory till the very end.

You've got your nice Lagrangian density $\mathcal{L}$, and apply the classical field theory toolkit of symmetries (global, local), gauge covariant derivatives and Euler-Lagrange equations.

You then get the equation of motion describing your (classical) field.

The quantumness of the problem arises when you want to write you classical field $\psi(x)$ in terms of creation and annihilation operators, $\psi(x) \propto a + a^\dagger$, because you want to quantise the theory. I.e. write in terms of discrete excitations, be it waves on a string or particles.

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Note: for the Dirac equation, for example, you'll find that you need at least a $(4\times1)$ object (called a spinor) to have a non-zero solution. Though you don't need to quantise the theory to get to this result, quantum mechanics allows you to interpret this as the presence of spin.