Can Maxwell's equations be quantized in the presence of a source particle?

[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

 

[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.

 

[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

 

[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.

 

[NeroKid]

I have been reading Franz Gross book, he define the time translation operator as U₀ U_I, with U₀ stands for free particle time translation operator , so under this definition the field O transform as O = U_IO₀ U_I^-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

 

[fzero]

I have been reading Franz Gross book, he define the time translation operator as U₀ U_I, with U₀ stands for free particle time translation operator , so under this definition the field O transform as O = U_IO₀ U_I^-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

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.

If I am misunderstanding your question, please carefully elaborate.

 

[NeroKid]

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.

If I am misunderstanding your question, please carefully elaborate.

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 U_I^-1 A_$\mu$ U_I(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 U_I ^-1 O₀ U_I 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