
#1
Nov913, 08:01 AM

P: 44

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
Nov913, 11:25 AM

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PF Gold
P: 2,606

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 welldefined properties under the Lorentz group. Similarly, the resulting object does not have a welldefined 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
Nov913, 12:31 PM

P: 44

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
Nov913, 01:58 PM

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PF Gold
P: 2,606

Quantized non free EM field
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
Nov1013, 12:16 AM

P: 44

I have been reading Franz Gross book, he define the time translation operator as U_{0} U_{I}, with U_{0} stands for free particle time translation operator , so under this definition the field O transform as O = U_{I}O_{0} 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




#6
Nov1113, 12:12 AM

Sci Advisor
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PF Gold
P: 2,606

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. 



#7
Nov1213, 08:08 AM

P: 44




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