# Mass of Dirac Electron increased by Electromagnetic field?

1. Dec 23, 2011

### johne1618

The Dirac electron in the Higgs vacuum field $v$ and an electromagnetic field with vector potential $A_\mu$ is described by the following equation:

$i \gamma^\mu \partial_\mu \psi = g v \psi + e \gamma_\mu A^\mu \psi$

where $g$ is the coupling constant to the Higgs field and $e$ is the coupling constant to the electromagnetic field.

Let us assume that we are in the rest frame of the electron so that:

$\partial_x=\partial_y=\partial_z=0$

Let us also assume that there is only an electrostatic potential $A_0=\phi$ so that:

$A_x = A_y = A_z = 0$

So the simplified Dirac equation is now:

$i \gamma^0 \partial_t \psi = g v \psi + e \gamma_0 \phi \psi$

Let us choose the Weyl or Chiral basis so that:

$\gamma^0 = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}$

where $I$ is the $2\times2$ unit matrix.

In this representation:

$\psi=\begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$

where $\psi_L$ and $\psi_R$ are left-handed and right-handed two-component Weyl spinors.

Subtituting into the simplified Dirac equation above we get:

$i \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} \begin{pmatrix} \partial \psi_L / \partial t \\ \partial \psi_R / \partial t \end{pmatrix} = g v \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} + e \phi \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$

This equation separates into two equations of two-component Weyl spinors:

$i \partial \psi_R / \partial t = g v \psi_L + e \phi \psi_R$

$i \partial \psi_L / \partial t = g v \psi_R + e \phi \psi_L$

Now let us add these two equations together to obtain:

$i \frac{\partial}{\partial t} (\psi_L + \psi_R) = (g v + e \phi)(\psi_L + \psi_R)$

My question is this:

Does the state $\psi_L + \psi_R$ describe an electron with an effective mass given by $gv + e \phi$?

Does the presence of an electrostatic field increase the electron's mass over and above the mass induced by the Higgs vacuum field alone?

Last edited: Dec 23, 2011
2. Dec 23, 2011

Staff Emeritus
Why do you think the left hand side is a mass? Isn't it an energy?

3. Dec 23, 2011

### blechman

You're trying to go the rest frame in a position-dependent potential (ignoring space derivatives). That doesn't sound strictly kosher. Translation invariance is broken by the presence of the E field, you cannot just ignore the derivatives... Or, in other words, your "rest frame" is not an inertial frame (there's a force!).

4. Dec 23, 2011

### johne1618

Well I'm generalizing from Leonard Susskind's video lecture 6: New Revolutions in Physics:

He uses this reasoning without an electromagnetic potential to explain how interaction with the Higgs field gives the electron mass. I have just incorporated an electromagnetic field into his argument.

5. Dec 23, 2011

### blechman

but the point is that the Higgs vev introduces a CONSTANT, while the potential is not a constant! It makes all the difference.

6. Dec 23, 2011

### johne1618

But what happens if the electron is placed inside a charged hollow insulator at potential $\phi$?

The electron will be at a constant potential but with zero field so it won't feel any force.

7. Dec 23, 2011

### blechman

In that case it might be possible to think of the electron inside the cavity as having an additional mass when compared to the electron outside the cavity, but you must be careful. In particular, when you "measure" the mass of the electron inside the cavity you have to cross the walls where the field is not zero, and that might cause trouble. So I'm not sure what kind of observable effect it might have, if any.

I haven't thought much about this kind of problem. Perhaps it should be placed in the "Quantum Mechanics" section for people to think about it there...

8. Dec 27, 2011

### johne1618

Prof Susskind has kindly answered my question in the following attachment. Adding a constant potential is equivalent to an overall phase shift and so does not effect the electron's equation of motion.

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Last edited: Dec 27, 2011