Quantized non free EM field


by NeroKid
Tags: field, free, quantized
NeroKid
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#1
Nov9-13, 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
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fzero
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#2
Nov9-13, 11:25 AM
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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
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#3
Nov9-13, 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

fzero
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#4
Nov9-13, 01:58 PM
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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.
NeroKid
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#5
Nov10-13, 12:16 AM
P: 44
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
fzero
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#6
Nov11-13, 12:12 AM
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Quote Quote by NeroKid View Post
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
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
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#7
Nov12-13, 08:08 AM
P: 44
Quote Quote by fzero View Post
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 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


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