Gauge Transformations and the Covariant Derivative

[PeroK]

TL;DR

Looking for an explanation for this and whether I am misunderstanding something.

This is from QFT for Gifted Amateur, chapter 14.

We have a Lagrangian density: $$\mathcal{L} = (D^{\mu}\psi)^*(D_{\mu}\psi)$$
Where $$D_{\mu} = \partial_{\mu} + iq A_{\mu}(x)$$
is the covariant derivative.

And a global gauge transformation$$\psi(x) \rightarrow \psi(x)e^{i\alpha(x)}$$
We are looking for a condition on $A_{\mu}(x)$ to make the Lagrangian invariant under the global gauge transformation.

The solution given is that $A_{\mu}(x)$ transforms according to:$$A_{\mu}(x) \rightarrow A_{\mu}(x) - \frac 1 q \partial_{\mu}\alpha(x) $$

However, when I expand the second term in the Lagrangian, I get:
$$D_{\mu}\psi(x) \rightarrow (\partial_{\mu} + iq A_{\mu}(x) -i\partial_{\mu}\alpha)\psi(x)e^{i\alpha(x)}
= D_{\mu}\psi(x)e^{i\alpha(x)} + iqA_{\mu}(x)\psi(x)e^{i\alpha(x)}$$
And there is an extra term involving $A_{\mu}(x)$.

My only thought is that if we assume that for the original $\psi(x)$ we have $A_{\mu}(x) = 0$, then that term vanishes. But, I'm not convinced. Perhaps I'm misunderstanding what it means for the Lagrangian to be invariant in this case?

Thanks in advance for any help.

 

[PeterDonis]

Perhaps I'm misunderstanding what it means for the Lagrangian to be invariant in this case?

I think you are taking some shortcuts in notation which are confusing you. Try it this way:

We start out with $\psi(x)$ and $D_\mu = \partial_\mu + i q A_\mu$. The gauge transformation is:

$$
\psi(x) \rightarrow \psi^\prime(x) = \psi(x) e^{i \alpha (x)}
$$
$$
A_\mu(x) \rightarrow A^\prime_\mu(x) = A_\mu(x) - \frac{1}{q} \partial_\mu \alpha(x)
$$

which gives us

$$
D_\mu \rightarrow D^\prime_\mu = \partial_\mu + i q A^\prime_\mu = \partial_\mu + i q A_\mu - i \partial_\mu \alpha
$$

We want to prove that

$$
\left( \left[D^\prime\right]^\mu \psi^\prime \right)^* \left( D^\prime_\mu \psi^\prime \right) = \left( D^\mu \psi \right)^* \left( D_\mu \psi \right)
$$

Expanding out the two sides and the covariant derivatives gives:

$$
\left( e^{- i \alpha} \partial^\mu \psi - i \psi e^{- i \alpha} \partial^\mu \alpha - i q A^\mu \psi e^{- i \alpha} + i \psi e^{- i \alpha} \partial^\mu \alpha \right) \left( e^{i \alpha} \partial_\mu \psi + i \psi e^{i \alpha} \partial_\mu \alpha + i q A_\mu \psi e^{i \alpha} - i \psi e^{i \alpha} \partial_\mu \alpha \right) = \left( \partial^\mu \psi - i q A^\mu \psi \right) \left( \partial_\mu \psi + i q A_\mu \psi \right)
$$

Canceling terms and factoring the exponentials out of the LHS gives:

$$
\left( \partial^\mu \psi - i q A^\mu \psi \right) \left( \partial_\mu \psi + i q A_\mu \psi \right) e^{- i \alpha} e^{i \alpha} = \left( \partial^\mu \psi - i q A^\mu \psi \right) \left( \partial_\mu \psi + i q A_\mu \psi \right)
$$

And then the exponentials on the LHS cancel and we have an identity.

 

[PeroK]

Thanks Peter. I've just spotted where I went wrong. It was a shortcut that went awry, as you guessed!