Conserved quantities as symmetry generators

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

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 [itex]\partial_\mu q^\mu = 0[/itex], if we construct [itex]Q(t) = \int dx\:q^0(x,t)[/itex], then [itex]\frac{d}{dt}Q(t) = 0[/itex], so [itex]Q(t) = Q[/itex].

2. Showing that [itex]Q[/itex] generates the symmetry means I need to show that [itex]\frac{i}{\hbar}\left[Q, \phi\right] = \delta \phi[/itex]. Expanding that out, we have:
[tex]\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)[/tex]

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

[tex]\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)[/tex]

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

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 [itex]\partial_\mu q^\mu = 0[/itex], if we construct [itex]Q(t) = \int dx\:q^0(x,t)[/itex], then [itex]\frac{d}{dt}Q(t) = 0[/itex], so [itex]Q(t) = Q[/itex].

2. Showing that [itex]Q[/itex] generates the symmetry means I need to show that [itex]\frac{i}{\hbar}\left[Q, \phi\right] = \delta \phi[/itex]. Expanding that out, we have:
[tex]\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)[/tex]

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

[tex]\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)[/tex]

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

Sam
 
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