# Definition of charge of a free electron

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1. May 29, 2015

### metroplex021

Hi there,
I have a question about the definition of a charge of a free electron.
Let's suppose that QED is the true theory of the interactions of charged particles. Presumably the charge on an (effectively) free electron, then, is the charge on an electron in which the electromagnetic interactions have been turned off -- ie, the coupling constant e is set to zero. But since, as well as being the coupling, e is also the charge on an electron, how can there be any charge on a free electron?!

Sure the answer here is obvious, but any help much appreciated!

2. May 29, 2015

Staff Emeritus
Your question is "if we use magic to turn off the electromagnetic interaction, would there still be an electric charge"? There's no answer to that question - how could we tell?

3. May 29, 2015

### metroplex021

Sure -- but then we talk about 'the charge on a free electron' all the time, and I'm wondering how this is defined. Maybe another way to ask the question is this: given that we take QFT to admit free-field models, are any of these models in which electrons have charge?

Here's why I'm struggling with this. Often it is said that we can identify charge with the invariant of the global U(1) symmetry from the free-particle Dirac Lagrangian. But it strikes me that, if we want to interpret this Lagrangian as that of free electrons, those models need to be understood as the limit of QED in which the interaction is turned off. (Otherwise how else can we see them as describing free versions of the particles that exist at this world?) But given that that's equivalent to setting e to zero, I don't see how that could work without having a conserved charge of zero as well.

4. May 29, 2015

Staff Emeritus
You're reading too much into this. A "free electron" is an abstraction where it is far away from other charges.

5. May 29, 2015

### fzero

The $U(1)$ charge in the interacting theory is just $-1$. The charge that appears in Maxwell's equations carries a factor of the coupling constant along with whatever units we're using. At low energies, the coupling constant doesn't change much, so it can be approximated by the value at the $E=0$ intercept.

If we turn off the EM interaction, we still have a conserved $U(1)$ charge, but it doesn't have a measurable interpretation as electric charge anymore. We can only measure the electric charge on something if we can interact with it.

6. Jun 2, 2015

### metroplex021

Thanks a lot. I appreciate that of course we can't measure an electron in the absence of interactions, but I'm wondering if it even makes theoretical sense to ascribe a charge to a field occuring in a free theory. Of course there will be a global U(1) charge in this case, but it seems to me that if we want to interpret that as electric charge then the value would have to be equal to the value in the interacting theory with the interaction set to zero. Since in the gauge theory the charge has a dual role of both invariant and coupling, it seems that there's no way to do that. (To my mind anyway!)

7. Jun 2, 2015

### fzero

If someone had just handed you the action for QED and asked you to describe the physics, you would use the $U(1)$ charge to define the "electric charge" $q$ of an elementary particle and call $e$ the coupling constant. When you quantized the theory, you'd be happy because the charge defined in this way will be invariant under the renormalization group as long as the theory has no anomaly.

When you went to write down Maxwell's equations

$$\partial^\nu F_{\mu\nu} = - e \bar{\psi} \gamma_\mu\psi ,$$

you would interpret $-1=q_e$ as the charge of a single electron. Then you would describe the macroscopic charge density of a collection of points charges to be

$$\rho(\vec{r}) = \sum_i q_i \delta (\vec{r}-\vec{r}_i),$$

and write Gauss' law as

$$\nabla\cdot \vec{E} = e \rho.$$

(Well, I am really not being honest since we hate having extraneous factors sitting in equations. We would of course absorb the factor of $e$ into the definition of the charge density, but we would not forget that we did so when we compare microscopic formulae to macroscopic ones. )

Historically however, Gauss' law was discovered first, so $\rho$ and the corresponding definition of a point charge were defined in such a way that the coupling constant $e$ was absorbed in the definition of $q$. There was no reason not to do this, since Gauss found the law c. 1835 (Coulomb's law was even known c. 1785) and the elementary electric charge wasn't measured until c 1909. People simply didn't know that $e$ was even a thing back then.

By now there's no great advantage to expend the effort to revise the set of units as long as we're going to use meters and kilograms to describe everything else, so classical mechanics is still taught according to the 19th century conventions. When we do QFT, we use the sensible definition of charge that excludes the value of the coupling constant, not the definition that we would have inherited from the classical, macroscopic physics.