Conserved quantities as symmetry generators

[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&#039;\:\left[q^0(x&#039;,t), \phi(x, t)\right]\\<br /> = \frac{i}{\hbar}\int dx&#039;\:\left(\left[\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}(x&#039;,t),\phi(x,t)\right]\delta \phi(x&#039;,t) + \frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}(x&#039;,t)\left[\delta\phi(x&#039;,t),\phi(x,t)\right] - \left[F^0(x&#039;,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&#039;,t)\right] = i\hbar\delta(x&#039;-x):

\frac{i}{\hbar}\left[Q,\phi(x,t)\right] = \delta \phi(x,t) + \frac{i}{\hbar}\int dx&#039;\:\left(\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}\left[\delta\phi(x&#039;,t),\phi(x,t)\right] - \left[F^0(x&#039;,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?

 

[samalkhaiat]

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&#039;\:\left[q^0(x&#039;,t), \phi(x, t)\right]\\<br /> = \frac{i}{\hbar}\int dx&#039;\:\left(\left[\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}(x&#039;,t),\phi(x,t)\right]\delta \phi(x&#039;,t) + \frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}(x&#039;,t)\left[\delta\phi(x&#039;,t),\phi(x,t)\right] - \left[F^0(x&#039;,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&#039;,t)\right] = i\hbar\delta(x&#039;-x):

\frac{i}{\hbar}\left[Q,\phi(x,t)\right] = \delta \phi(x,t) + \frac{i}{\hbar}\int dx&#039;\:\left(\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}\left[\delta\phi(x&#039;,t),\phi(x,t)\right] - \left[F^0(x&#039;,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?

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