Is the QED Action Invariant Under Gauge Transformation?

[Cirrus79]

Hello,

I don't understand two steps in solution to the problem:

I. Homework Statement

Show that QED action is invariant under gauge transformation.

II. Relevant equations

QED action:

$S= \int{d^{4} x \left[\overline{\Psi}\left(i\gamma^{\mu} D_{\mu} -m \right)\Psi -\frac{1}{4}F_{\mu \nu}F^{\mu \nu}\right]}$

Gauge transformation:

$\Psi\rightarrow e^{-iQ\chi}\Psi$
$A_{\mu}\rightarrow A_{\mu}+\frac{1}{e}\partial_{\mu}\chi$

III. The attempt at a solution

1. First I show that $D_{\mu}\Psi\rightarrow e^{-iQ\chi}D_{\mu}\Psi$

$D_{\mu}\Psi=(\partial_{\mu}+ieQA_{\mu})\Psi \rightarrow \left[\partial_{\mu}+ieQ\left(A_{\mu}+\frac{1}{e}\partial_{\mu}\chi\right)\right] e^{-iQ\chi}\Psi =$

$=\left(\partial_{\mu}+ieQA_{\mu}+iQ\partial_{\mu}\chi\right) e^{-iQ\chi}\Psi=$

$=e^{-iQ\chi}\left(-iQ\partial_{\mu}\chi+ieQA_{\mu}+iQ\partial_{\mu}\chi+\partial_{\mu}\right) \Psi=$
$=e^{-iQ\chi}\left(\partial_{\mu}+ieQA_{\mu}\right) \Psi=e^{-iQ\chi}D_{\mu}\Psi$

The problem is in the third line. Where does $\partial_{\mu}$ come from?
I get:

$(iQ\partial_{\mu}\chi) e^{-iQ\chi}\Psi=\left(e^{-iQ\chi}iQ\partial_{\mu}\chi -e^{-iQ\chi}i^{2} Q^{2}\chi\partial_{\mu}\chi\right)\Psi$

What am I doing wrong?

2. Then I show that $F^{\mu \nu}\rightarrow F^{\mu \nu}$ and $\overline{\Psi}\rightarrow e^{iQ\chi}\overline{\Psi}$

3. And finally:

$\overline{\Psi}(i\gamma^{\mu} D_{\mu} -m)\Psi\rightarrow e^{iQ\chi}\overline{\Psi}(i\gamma^{\mu}e^{-iQ\chi}D_{\mu}-m)e^{-iQ\chi}\Psi=$

$=e^{iQ\chi}\overline{\Psi}e^{-iQ\chi}(i\gamma^{\mu} D_{\mu} -m)\Psi=\overline{\Psi}(i\gamma^{\mu} D_{\mu} -m)\Psi$

Here is the second problem. I get:

$\overline{\Psi}(i\gamma^{\mu} D_{\mu} -m)\Psi\rightarrow e^{iQ\chi}\overline{\Psi}(i\gamma^{\mu}e^{-iQ\chi}D_{\mu}-m)e^{-iQ\chi}\Psi=$

$=e^{iQ\chi}\overline{\Psi}(i\gamma^{\mu}e^{-iQ\chi} D_{\mu}e^{-iQ\chi} -e^{-iQ\chi}m)\Psi =e^{iQ\chi}\overline{\Psi}e^{-iQ\chi}(i\gamma^{\mu} D_{\mu}e^{-iQ\chi} -m)\Psi$

I can't figure out what happens here.

I will be very grateful for your help.

 

[sgd37]

that is one weird way of doing this

I don't understand your first problem since the derivative is there in the second line as well and all other previous lines. First thing to understand is that $$\chi$$ is a coordinate dependent function so its derivative is non trivial. The line you put up of your solution I'm sorry to say is nonsense

the second problem seems to be orthographic in nature or arising from bad definitions, it should be

$\overline{\Psi}(i\gamma^{\mu} D_{\mu} -m)\Psi\rightarrow e^{iQ\chi}\overline{\Psi}(i\gamma^{\mu}e^{-iQ\chi}D_{\mu}-me^{-iQ\chi}) \Psi$

since $$\overline{\Psi }' (i\gamma^{\mu} D'_{\mu} -m)\Psi ' = e^{iQ\chi}\overline{\Psi}(i\gamma^{\mu}D'_{\mu}-m)e^{-iQ\chi} \Psi$$

and you have already shown that $$D'_{\mu} \Psi ' = e^{-iQ\chi} D_{\mu} \Psi$$

 

[Cirrus79]

that is one weird way of doing this

Can you provide a reference to better solution for this problem? This one is from lecture notes.

First thing to understand is that $$\chi$$ is a coordinate dependent function so its derivative is non trivial.

This made me think, and i realized that I should use (fg)'=f'g+fg' to the first $\partial _{\mu}$ not the second one. Now I get it.

the second problem seems to be orthographic in nature or arising from bad definitions, it should be

$\overline{\Psi}(i\gamma^{\mu} D_{\mu} -m)\Psi\rightarrow e^{iQ\chi}\overline{\Psi}(i\gamma^{\mu}e^{-iQ\chi}D_{\mu}-me^{-iQ\chi}) \Psi$

since $$\overline{\Psi }' (i\gamma^{\mu} D'_{\mu} -m)\Psi ' = e^{iQ\chi}\overline{\Psi}(i\gamma^{\mu}D'_{\mu}-m)e^{-iQ\chi} \Psi$$

and you have already shown that $$D'_{\mu} \Psi ' = e^{-iQ\chi} D_{\mu} \Psi$$

Of course. I tried to transform $$\Psi$$ twice.

Thank you very much!

 

[dextercioby]

With all due respect, but the problem you're trying to solve is logically circular. The Noether procedure to couple the free fields specifically uses the invariance of the overall coupled action to derive the coupling term. Otherwise, the $j^{\mu}A_{\mu}$ coupling couldn't be derived.

 

[sgd37]

I think you're turning a simple exercise into something it was never intended to demonstrate