# Is a local redefinition $q\to-q$ of "charge" $\to$ "minus charge" a gauge transformation?

This is claimed in the Scientific American article "Q&A: Lawrence Krauss on The Greatest Story Ever Told" published on March 21, 2017.

It says, “I could change the sign of each electric charge in nature locally. But I have to have a rule book.” What's the rule book? In this case, it’s the electromagnetic field.

However looking at explicit $U(1)$ gauge transformations $$\psi \to e^{ia(x)} \psi ,$$ $$A\to A+ \partial_\mu a(x).$$ I fail to see how such an interpretation is possible

• The electric charge is observable, i.e., gauge-invariant. I call BS. It's Krauss after all. Commented Apr 3, 2018 at 15:19
• @AccidentalFourierTransform good point! But maybe a global phase shift can ne interpreted this way?!
– jak
Commented Apr 6, 2018 at 8:21
• Global phase shifts are generated by global symmetries, which are observable, i.e., non-gauge. Commented Apr 6, 2018 at 13:16

That really depends on what you mean by ''Gauge Transformation''. I will shortly give the mathematical description, which is also the one used in (almost any) physics-book (but in a less technical setting) and once we have this, it's clear that ''charge'' $\rightarrow$ -''charge'' cannot be a gauge transformation.

For the general setting, let us model electromagnetism not on Minkowski Space $\mathbb{R}^{1,3}$ but on $\mathbb{R}^{1,3} \times \mathrm{U}(1)$. (In the following I will use a little bit of gauge theory, but try to minimize the technical details as best as possible.)

Since $\psi$ does transform under a gauge transformation, which are a change of phase, it doesn't seem too off to view $\psi$ as a function which depends on $(x_{\mu},e^{i\theta})$. The latter parameter just explicilty encodes the phase at each point in spacetime.

Thus, if you change the phase, which is what we will call a gauge transformation we have to specify the way in which the field changes. The general setting would look like this:

$$\psi (x_{\mu},e^{i \theta} \cdot e^{i\phi(x_{\mu})}) = \Lambda(e^{i \phi (x_{\mu})})\psi (x_{\mu}, e^{i \theta})$$

Where $\Lambda$ is a representation of $\mathrm{U}(1)$ acting on the vector space in which $\psi$ takes it's values (pre second quantisation).

Now let us assume the representation $\Lambda$ be irreducible and one dimensional (by which I mean, it is given by $\Lambda (e^{i \phi (x_{\mu})})\psi = e^{ \lambda (i \phi (x_{\mu}))} \cdot \psi$ for some function $\lambda$.

Then $\lambda$ is given by multiplication with an Integer $e \in \mathbb{Z}$, giving

$$\psi (x_{\mu},e^{i \theta} \cdot e^{i\phi(x_{\mu})}) = e^{i \cdot e \cdot \phi (x_{\mu})}\psi (x_{\mu}, e^{i \theta}).$$

Finally, $e$ is the elementary (electromagnetic) charge of the particle described by $\psi$.

Thus, changing the charge would be a change of representation and not a gauge transformation, since gauge transformations do not touch the representation, hence, they don't touch the charge.

(going from $e$ to $-e$ would be done by passing from $\Lambda$ to $\Lambda$* which is the complex-conjugate representation)

I'd like to add, that the change of representation described above would correspond to a change of particle, for example passing from the $\pi^+$ to the $\pi^-$-meson.