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## Main Question or Discussion Point

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?

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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