# Quantized non free EM field

1. Nov 9, 2013

### NeroKid

Hello everyone, I have been wondering about the quantization of Maxwell's equation in free space, but now suppose that we have a source particle, fermion for example, now the field equation is a mixture of both fermion field and photon field so my question is whether you can get out of this by treating the EM potential as mixture of both particle like A = free photon field + retarded fermion field and treating fermion field as free field

2. Nov 9, 2013

### fzero

Since the photon is a boson and the fermion is a fermion, you can't mix them by just add their fields together. The resulting object doesn't have well-defined properties under the Lorentz group. Similarly, the resulting object does not have a well-defined transformation under the gauge group, so it wouldn't be possible in any way to obtain a consistent theory of electromagnetism from such an object.

3. Nov 9, 2013

### NeroKid

what if I define the field as (identity fermion) x free photon + indentity photon x retarded fermion field since we can define the product of different space operator in lagrangian , can you tell me or point me to the literature what the formula has violated

4. Nov 9, 2013

### fzero

The kinetic energy term in the fermion Lagrangian has a term $\bar{\psi} \gamma\cdot{\partial}\psi$ that doesn't involve the photon field. The transformation of this operator under a product kind of field redefinition isn't very well defined. It's impossible to transform the electromagnetic field away unless the gauge group is spontaneously broken for some reason.

5. Nov 10, 2013

### NeroKid

I have been reading Franz Gross book, he define the time translation operator as U0 UI, with U0 stands for free particle time translation operator , so under this definition the field O transform as O = UIO0 UI-1 and under the coulomb gauge the the free field V =0, so V in present of interaction should equals 0, but in the books he present it as coulomb self energy so I have been wondering whether we can do the same for other components of field

6. Nov 11, 2013

### fzero

I am not certain, but I think you are confusing two different uses for the letter $V$. In one case, $V$ is the temporal component of the gauge field $A_0= V$. There is a partial gauge fixing in which one takes $A_0=0$, but this is called the temporal or Weyl gauge, and is different from the Coulomb gauge.

The other use of $V$ is as the interaction potential in a general Hamiltonian, $H = H_0 + V$. Then $H_0$ appears in $U_0$, while $V$ appears in $U_I$, if I have your definitions correct. In QED, $V$ is not the same as $A_0$, but involves

$$V \sim e \int d^4 x A_\mu \bar{\psi} \gamma^\mu \psi .$$

There is no gauge in which we can make every component of $A_\mu$ vanish, so it is impossible for the interaction potential to vanish by a gauge choice.

Im pretty sure that I was using Coulomb gauge not Weyl Gauge, what I meant was if other $A_\mu$ you have the transform in the interaction is UI-1 A$\mu$ UI(the field in between is free field) but not for $A_0$ in the presence of photon and electron the author suddenly add the coulomb self energy which, in my understanding of EM field, should be $$A_0 = \int d^3 x \frac{\bar{\psi} \gamma^0 \psi}{r}$$ and the self Halmitonian is $$H_s = \int \int d^3 x_1 d^3 x_2 \frac{[ \bar{\psi} \gamma^0 \psi]_1 [ \bar{\psi} \gamma^0 \psi]_2}{|r_1 - r_2 |}$$ I don't understand why , I thought the field suppose to transform from free field like UI -1 O0 UI how can a 0 field transfom in this way to a non zero field they are both supposed to describe the EM field and in turn the field quanta and I don't want to make the interaction vanish, what I want to ask is that whether we can generalize both $\psi$ and $\vec{A}$ to the global operator for both electron and photon