Equations of motion and Noether's theorem in the functional formalism
This post has been migrated from my old blog, the math-physics learning seminar.
First, let us recall the derivation of the equations of motion and Noether’s theorem in classical field theory. We have some action functional (S[\phi]) defined by some local Lagrangian:
[ S[\phi] = \int L(\phi, \partial \phi) dx. ]
The classical equations of motion are just the Euler-Lagrange equations
[ \frac{\delta S}{\delta \phi(x)} = 0
\iff \partial_\mu \left( \frac{\partial L}{\partial(\partial_\mu\phi)} \right)
= \frac{\partial L}{\partial \phi} ]
Now suppose that (S) is invariant under some transformation (\phi(x) \mapsto \phi(x) + \epsilon(x) \eta(x)), so that (S[\phi] = S[\phi+\epsilon \eta]). Here we treat (\eta) as a fixed function but (\epsilon) may be an arbitrary infinitesimal function. The Lagrangian is not necessarily invariant, but rather can transform with a total derivative:
[ L(\phi+\epsilon \eta) = L(\phi)
-
\frac{\partial L}{\partial (\partial_\mu \phi)} \eta \partial_\mu \epsilon
-
\epsilon \partial_\mu f^\mu ]
For some unknown vector field (f^\mu) (which we could compute given any particular Lagrangian). So let’s compute
\begin{align}\delta_\epsilon S &= \int \delta_\epsilon L \\
&= \int \frac{\partial L}{\partial (\partial_\mu \phi)}
\eta \partial_\mu \epsilon + \epsilon \partial_\mu f^\mu \\
&= \int \partial_\mu \left(f^\mu - \frac{\partial L}{\partial (\partial_\mu \phi)} \eta \right) \epsilon
\end{align}
Let us define the Noether current (J^\mu) by
[ J^\mu = \frac{\partial L}{\partial (\partial_\mu \phi)} \eta - f^\mu. ]
Then the previous computation showed that
[ \frac{\delta S}{\delta \epsilon} = -\partial_\mu J^\mu. ]
If (\phi) is a solution to the Euler-Lagrange equations, then the variation (dS) vanishes, hence we obtain:
Theorem (Noether’s theorem) The Noether current is divergence free, i.e.
[ \partial_\mu J^\mu = 0.]
Functional Version
First, we derive the functional analogue of the classical equations of motion. Consider an expectation value
[ \langle \mathcal{O(\phi)} \rangle
= \int \mathcal{O}(\phi) e^{\frac{i}{\hbar} S} \mathcal{D}\phi ]
We’ll assume that (\phi) takes values in a vector space (or bundle). Then we can perform a change of variables (\psi = \phi + \epsilon), and since (\mathcal{D}\phi = \mathcal{D}\psi) we find that
[ \int \mathcal{O}(\phi+\epsilon) \exp\left(\frac{i}{\hbar} S[\phi] \right) \mathcal{D}\phi ]
is independent of (\epsilon). Expanding to first order in (\epsilon), we have
[ 0 = \int \left(\frac{\delta\mathcal{O}}{\delta \phi}
- \frac{i \mathcal{O}}{\hbar} \frac{\delta S}{\delta \phi} \right)
\exp \left( \frac{i}{\hbar} S \right) \mathcal{D}\phi ]
So we find the quantum analogue of the equations of motion:
[ \left\langle \frac{\delta \mathcal{O}}{\delta \phi} \right\rangle
- \frac{i}{\hbar} \left\langle \mathcal{O} \frac{\delta S}{\delta \phi} \right\rangle = 0]
Next, we move on to the quantum version of Noether’s theorem. Suppose there is a transformation (Q) of the fields leaving the action invariant. Assuming the path integral measure is invariant, we obtain
[ \left\langle QF \right \rangle + \frac{i}{\hbar} \left\langle F QS \right\rangle = 0]
To compare with the classical result, consider (Q) to be the (singular) operator
[ Q = \frac{\delta}{\delta \epsilon(x)} ]
Then by the previous calculations,
[ Q S = -\delta_\mu J^\mu, ]
so we obtain
[ \left\langle \frac{\delta \mathcal{O}}{\delta \epsilon(x)} \right\rangle
= \frac{i}{\hbar} \left\langle \mathcal{O} \partial_\mu J^\mu \right\rangle. ]
This is the Ward-Takahashi identity, the quantum analogue of Noether’s theorem.