# Conserved quantities as symmetry generators

1. Oct 14, 2013

### Chopin

Suppose we have a Lagrangian $\mathcal{L(\phi, \partial_\mu \phi)}$ over a field $\phi$, and some variation on the field $\delta \phi$. If this variation induces a variation $\delta \mathcal{L} = \partial_\mu F^\mu$ for some function $F^\mu$, then Noether's Theorem tells us that if we construct the quantity $q^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi - F^\mu$, then $\partial_\mu q^\mu = 0$.

In quantum field theory, it is apparently also the case that this quantity can be used to form a generator for the symmetry. I'd like to understand how this is so, but am having some issues proving it. Here's what I understand:

1. Since $\partial_\mu q^\mu = 0$, if we construct $Q(t) = \int dx\:q^0(x,t)$, then $\frac{d}{dt}Q(t) = 0$, so $Q(t) = Q$.

2. Showing that $Q$ generates the symmetry means I need to show that $\frac{i}{\hbar}\left[Q, \phi\right] = \delta \phi$. Expanding that out, we have:
$$\frac{i}{\hbar}\left[Q,\phi(x,t)\right] = \frac{i}{\hbar}\int dx'\:\left[q^0(x',t), \phi(x, t)\right]\\ = \frac{i}{\hbar}\int dx'\:\left(\left[\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}(x',t),\phi(x,t)\right]\delta \phi(x',t) + \frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}(x',t)\left[\delta\phi(x',t),\phi(x,t)\right] - \left[F^0(x',t), \phi(x, t)\right]\right)$$

3. Now we impose the canonical commutation relations $\left[\phi(x, t), \frac{\partial \mathcal{L}}{\partial(\partial_0\phi)}(x',t)\right] = i\hbar\delta(x'-x)$:

$$\frac{i}{\hbar}\left[Q,\phi(x,t)\right] = \delta \phi(x,t) + \frac{i}{\hbar}\int dx'\:\left(\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}\left[\delta\phi(x',t),\phi(x,t)\right] - \left[F^0(x',t), \phi(x, t)\right]\right)$$

The first term is exactly what I want. Therefore, I need to somehow get the last two terms to cancel, but I'm not exactly sure how to go about doing this. Can anybody illustrate how I can make this cancellation happen?

Last edited: Oct 14, 2013
2. Apr 27, 2014

### samalkhaiat

Obviously, if you don’t have an explicit expression for $F^{ \mu }$, you will not be able to find it’s commutator with $\phi$. But, if you can derive the Noether current directly from the action integral, you find that
$$F^{ \mu } = - \mathcal{L} \delta x^{ \mu } .$$
So, you will have
$$-[ F^{ 0 } ( t , y ) \ , \ \phi ( t , x ) ] = \delta x^{ 0 } [ \mathcal{ L } ( t , y ) \ , \ \phi ( t , x ) ] . \ \ \ (1)$$
Now, the local density
$$\mathcal{L} ( y ) = \mathcal{L} ( \phi , \dot{ \phi } , \nabla \phi ) ,$$
depends on the conjugate field $\pi ( x )$ only through the “velocity” $\partial_{ 0 } \phi ( y )$. So, we can calculate the following “functional” derivative
$$\frac{ \delta \mathcal{L} ( y ) }{ \delta \pi ( x ) } = \frac{ \partial \mathcal{L} ( y ) }{ \partial \dot{ \phi } ( y ) } \frac{ \delta \dot{ \phi } ( y ) }{ \delta \pi ( x ) } = \pi ( y ) \ \frac{ \delta \dot{ \phi } ( y ) }{ \delta \pi ( x ) } .$$
In bracket language, this simply means
$$[ \mathcal{ L } ( t , y ) \ , \ \phi ( t , x ) ] = \pi ( y ) \ [ \dot{ \phi } ( t , y ) , \phi ( t , x ) ] . \ \ \ (2)$$
Ok, we haven’t finished yet. Now, use
$$\delta \phi = \bar{ \delta } \phi - \delta x^{ \mu } \partial_{ \mu } \phi ,$$
where
$$\bar{ \delta } \phi_{ a } ( x ) = \bar{ \phi }_{ a } ( \bar{ x } ) - \phi_{ a } ( x ) = J_{ a b } \phi_{ b } ( x ) ,$$
and $J_{ a b }$ are constant transformation matrices. So, now we have
$$[ \delta \phi ( t , y ) \ , \ \phi ( t , x ) ] = - \delta x^{ 0 } [ \dot{ \phi } ( t , y ) \ , \ \phi ( t , x ) ] . \ \ \ (3)$$
Substitute (1), (2) and (3) in your last equation and get
$$\delta \phi ( x ) = [ i Q \ , \ \phi (x) ] .$$

Sam

Last edited: Apr 27, 2014