Photon occupation numbers in relation to the 4-potential

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  • #1
olgerm
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I have some questions about this video. I have watched other videos in this series. Otherwise very nice series, but I think there may be mistake. Isn't the video flawed because it forgot forgot 0'th component of 4-fector ##A## aka ##\varphi## in 3-vector representation, I think it because lorentz gauge conditiond should be ##\partial_t A_t +\partial_x A_x +\partial_y A_y +\partial_z A_z=0## not ##\partial_x A_x +\partial_y A_y +\partial_z A_z=0## like video in 8.05. Also one maxwell equations ##\partial_t^2 A_0=\nabla^2 A_0## aka ##\partial_t^2 \varphi=\nabla^2 \varphi## is missing.
what is the 4-potential field ##A## if i have occupation number 1 for ##n_{1,0,0;2}## (##k=[1,0,0]## and ##\alpha=2##) and other occupation numbers are 0?

How are these photon occupation numbers related to 4-potential field ##A##. Most confusing is that for each wavenumber k they have only 2 integers although for any wave there are more degrees of freedom(2 numbers more). Or quantum description of EM-field has less degrees of freedom?
 

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  • #2
vanhees71
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It depends a bit on the formalism used to quantize the em. field. For the free em. field the most lucid treatment is to fix the gauge completely before quantizing it, which leads to the radiation gauge. The disadvantage is that it is not manifestly relativistically covariant, and that's why other schemes to deal with the quantization of gauge theories have been formulated. Here, I'd always prefer the path-integral approach, starting with the non-covariant Coulomb gauge in Hamiltonian formulation and then showing that it's equivalent to the covariant gauges using the Faddeev-Popov formalism. For details, see my manuscript on QFT:

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

For the free field, the non-covariant quantization in radiation gauge roughly goes as follows. You start from the classical theory in terms of the potentials ##(A^{\mu})=(\Phi,\vec{A})## and completely fix the gauge for the charge-current-free case, ##\rho=0## and ##\vec{j}=0##. The radiation-gauge conditions read
$$\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
Then the decomposition into plane-wave modes reads
$$\vec{A}(t,\vec{x})=\sum_{h=\pm 1} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{q}}{(2 \pi)^{3/2} \sqrt{2 E(\vec{q})}} \left [\hat{a}(\vec{q},h) \vec{\epsilon}_h(\vec{q}) \exp(-\mathrm{i} E(\vec{q}) t + \mathrm{i} \vec{q} \cdot \vec{x}) + \text{h.c.} \right].$$
Here ##E(\vec{q})=|\vec{q}|## is the energy of a massless mode with momentum ##\vec{q}##, ##\vec{\epsilon}(\vec{q},h)## are the transverse polarization vectors with ##\vec{q} \cdot \vec{\epsilon}(\vec{q},h)=0## with helicities ##h \in \{1,-1 \}## (i.e., the left- and right-circular polarization vectors), and ##\hat{a}(\vec{q},h)## is the annihilation operator for a photon with momentum ##\vec{q}## and helicity ##h##, obeying the bosonic commutation relations
$$[\hat{a}(\vec{q},h),\hat{a}^{\dagger}(\vec{q}',h')]=\delta^{(3)}(\vec{q}-\vec{q}') \delta_{hh'}.$$
The Fock basis is the common eigenbasis of the photon-mode occupation-number operators
$$\hat{N}(\vec{q},h)=\hat{a}^{\dagger}(\vec{q},h) \hat{a}(\vec{q},h).$$
The Hamiltonian and momentum operators (renormalized using the normal-ordering prescription) then read
$$\hat{H}=\sum_{h} \int_{\mathbb{R}^3} \mathrm{d}^3 q E(\vec{q}) \hat{N}(\vec{q},h), \quad \hat{\vec{P}} = \sum_{h} \int_{\mathbb{R}^3} \mathrm{d}^3 q \vec{q} \hat{N}(\vec{q},h).$$
 
  • #3
olgerm
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Thanks you for this extensive reply vanhees71. Now I understand they are using radiation gauge instead of lorentz gauge. This video gave me some intuition about kets.
What would be the decomposition into plane-wave modes in lorentz gauge?
 
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  • #4
vanhees71
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You can do everything in Lorenz (not Lorentz, because the first treatment of the em. radiation problem using that gauge is due to the Danish physicist Ludvig Lorenz and not due to the Dutch physicist Hendrik Antoon Lorentz!) gauge. For the free field, however this doesn't fix the gauge completely, and thus you have to use the more comlicated Gupta-Bleuler formalism, which starts from an improper "Hilbert space" where some unphysical states an have negative norm.

The radiation gauge can be used only in the charge-current free case and obeys of course both the Lorenz and the Coulomb gauge, since ##A^0=0## and ##\vec{\nabla} \cdot \vec{A}=0##, which implies ##\partial_{\mu} A^{\mu}=0##.
 
  • #5
olgerm
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The radiation gauge can be used only in the charge-current free case and obeys of course both the Lorenz and the Coulomb gauge, since ##A^0=0## and ##\vec{\nabla} \cdot \vec{A}=0##, which implies ##\partial_{\mu} A^{\mu}=0##.
This is an important aspect that I have not noticed before.

I understand that with Lorenz gauge and empty space (##J=0##) some combinations of occupation numbers are impossible (unphysical, because against maxwell equations), but I am interested, what would be instead of
$$\vec{A}(t,\vec{x})=\sum_{h=\pm 1} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{q}}{(2 \pi)^{3/2} \sqrt{2 E(\vec{q})}} \left [\hat{a}(\vec{q},h) \vec{\epsilon}_h(\vec{q}) \exp(-\mathrm{i} E(\vec{q}) t + \mathrm{i} \vec{q} \cdot \vec{x}) + \text{h.c.} \right].$$
?
Instead of 2 values of helicity there are more of them, but I don't know which.
 
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  • #6
olgerm
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In lorenz gauge I could you use 1 occupation number for every component 4-potential ##A##. then I had 7 (6) degrees of freedom for every wavenumber (momentum), beacause of lorenz gauge condition.
 
  • #7
PeterDonis
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In lorenz gauge I could you use 1 occupation number for every component 4-potential ##A##.
This doesn't make sense; the components of the 4-potential are not photons and the term "occupation number" makes no sense with respect to them.
 
  • #8
vanhees71
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That's what makes gauge theories so subtle, and local gauge symmetry is a very fragile property, i.e., it can easily be violated by all kinds of mathematical approximations. It's a bit too much to put it in a forum posting, but roughly the Gupta-Bleuler formalism, which is a way to use canonical operator quantization heuristics for a manifestly covariant treatment of QED, and it was extended to the even more subtle case of the covariant operator formalism for non-Abelian gauge theories. As I said earlier, the path-integral quantization method is much more lucid and elegant than these operator methods for gauge theories, but one can learn a lot about the structure of quantized gauge theories from the operator methods too.

The Gupta-Bleuler formalism starts with using all four components of the four-vector potential ##A^{\mu}## and the classical Lagrangian
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{\lambda}{2} (\partial_{\mu} A^{\mu})^2.$$
As usual
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}.$$
Note that this only leads to the correct free-field Maxwell equations, if one considers it under the constraint of the Lorenz-gauge condition,
$$\partial_{\mu} A^{\mu}=0,$$
but to quantize it, one forgets about this constraint for the time being and uses the canonical-quantization formalism, i.e., one calculates the canonical momenta of the fields,
$$(\Pi^{\mu})=\left (\frac{\partial \mathcal{L}}{\partial \dot{A}_{\mu}} \right)=-\begin{pmatrix} \lambda \partial_{\mu} A^{\mu} \\ \dot{\vec{A}}+\vec{\nabla} A^0 \end{pmatrix}.$$
Then the quantization is done making the ##A^{\mu}## operators ##\hat{A}^{\mu}## subject to the commutator relations
$$[\hat{A}^{\mu}(t,\vec{x}),\hat{A}^{\nu}(t,\vec{x}')]=[\hat{\Pi}^{\mu}(t,\vec{x}),\hat{\Pi}^{\nu}(t,\vec{x}')]=0, \quad [\hat{A}^{\mu}(t,\vec{x}),\hat{\Pi}^{\nu}(t,\vec{x}')]=\mathrm{i} \eta^{\mu \nu} \delta^{(3)}(\vec{x}-\vec{x}').$$
Now you cannot implement simply the Lorenz-gauge condition, because it would violate these canonical quantization condition. The ingenious idea of Gupta and Bleuler was to go on with the formalism nevertheless, but then you get with the usual realization in terms of mode-decomposition with creation and annihilation operators trouble with the positive definiteness of the scalar product of the "would-be Hilbert space", i.e., the Fock space you construct in this way is not a real Hilbert space, because there are states with negative squared "norm".

It turns out that you can still get a consistent quantum theory if you impose the Lorenz-gauge constraint in the weak sense, i.e., you assume that only those states in the naively constructed Fock space are physical states, for which
$$\partial_{\mu} \hat{A}_{\mu}^{(+)} |\Psi \rangle=0,$$
where ##\hat{A}_{\mu}^{(+)}## is the part of the mode decomposition with positive frequencies (involving annihilation operators).

This eliminates the unphysical states of the naive canonical formalism, i.e., the "time-like and spatially longitudinal" modes, leaving you with only the physical spatially transverse modes.

Also it turns out that for interacting fields (e.g., by coupling them in the usual minimal-substitution way to a Dirac field, which then describes the most simple version of QED, describing the electromagnetism of photons, electrons, and positrons) that order by order in perturbation theory everything is consistent, i.e., gauge invariance of the original theory implies Ward-Takahashi identities for the Green's functions (and the proper-vertex functions) which guarantee that the unphysical degrees of freedom are non-interacting.
 
  • #9
olgerm
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Then in lorenz gauge just 1 number (##n_0,n_2,n_2,n_3##) for every component of 4-potential ##A##, and number photons (occupation number) is some function of these numbers. I guess i might be ##-n_0^2+n_1^2+n_2^2+n_3^2##, because it is same in all frames of reference (aka invariant).
 
  • #10
PeterDonis
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Then in lorenz gauge just 1 number ##(n_0,n_2,n_2,n_3)## for every component of 4-potential ##A##,
This is true in any gauge, not just Lorentz gauge; it's just another way of saying that the 4-potential is a 4-vector.

number photons (occupation number) is some function of these numbers
To the extent "occupation number" makes sense at all for the states you are talking about, it's the expectation value of the photon number operator. This is not just a "function" of the components of the 4-potential; it's an integral, which you can look up in any textbook on QM.
 
  • #11
olgerm
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This is true in any gauge, not just Lorentz gauge; it's just another way of saying that the 4-potential is a 4-vector.
These numbers are not values of ##A(X)## in given location ##\vec{X}##, but amplitude of ##A(\vec{k})##, with given wavenumber k (and momentum).
These numbers are fourier transform of four-potential.
 
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  • #12
PeterDonis
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These numbers are fourier transform of four-potential.
Which is also a 4-vector. That doesn't change anything I said.
 
  • #13
vanhees71
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This is true in any gauge, not just Lorentz gauge; it's just another way of saying that the 4-potential is a 4-vector.



To the extent "occupation number" makes sense at all for the states you are talking about, it's the expectation value of the photon number operator. This is not just a "function" of the components of the 4-potential; it's an integral, which you can look up in any textbook on QM.
No, it doesn't make physical sense! The point is that massless vector fields are necessarily gauge fields, and of the four field-degrees of freedom in the case of free em. fields only the two spatially transverse modes are physical, while the temporal and spatially longitudinal modes are un-physical degrees of freedom.
 
  • #14
PeterDonis
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No, it doesn't make physical sense!
What doesn't? The expectation value of the photon number operator? That would seem to be a perfectly well-defined quantity.
 
  • #15
vanhees71
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As I said one has to be very careful when calculating expectation values concerning their physical interpretation. Only averages over gauge-invariant quantities make physical sense. So first of all you have to define the "photon number operator" in a gauge-invariant way and then properly calculate the expectation value. First of all, I don't know, how to define a gauge-invariant operator of photon number nor can in the Gupta-Bleuler formalism naive "counting" of the four "number operators" you might build from the ##\hat{a}(\vec{k},\lambda)## make physical sense. Note that out of the four naive polarization-degrees of freedom only the two spatially transverse are physical quantities. As I said, the quantization of gauge theories are a subtle issue.
 
  • #16
olgerm
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using fourier transform ##A_{k}(k)## of 4-potential: are these

##\forall_{\vec{p}}(-A_{\vec{p}}^2(0)+\sum_{i=1}^D(A_{\vec{p}}^2(i))= N_{footon}(\vec{p})))##
##\forall_{i_1}(\forall_{\vec{p}}((\vec{A}_{\vec{p}}(\vec{p},i_1)\neq 0) \to (-\vec{p}^2(0)+\sum_{i_2=1}^D(\vec{p}^2(i_2))=0)))## (this is maxwell equation for fourier transform of ##A##)
##\forall_{\vec{p}}(\sum_{i=0}^D(\vec{p}(i)*\widetilde{A}_{\vec{p}}(\vec{p},i))=0)## (this is lorenz gauge condition.)

relations correct?
are there more important relations?

  • ##N_{footon}(\vec{p})## is number of photons with momentum ##p##
since using units where ##\hbar=1## and wavenumber and momentum are related by ##p=\hbar*k## argument of fourier transform is ##p## not ##k##.
 
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  • #17
olgerm
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Can I check that photon angular momentum is 1, as it should be, by applying some operator?
 
  • #18
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are these relations correct?
In which textbook do they appear? And if in none, why are you making things up?
 
  • #19
olgerm
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In which textbook do they appear? And if in none, why are you making things up?
1. one is photon number operator that was discussed abow.
2. one is maxwell equation for fourier transform apllyied.
maxwell equation for for potential vector field in vacum is ##-\frac{\partial^2 A(\vec{X};i_1)}{\partial x^2(0)}+\sum_{i_2=1}^D(\frac{\partial^2 A(\vec{X};i_1)}{\partial x^2(i_2)})=0##
applying fourier transform gives the 2. eqation, because fourier transform of deriative of function is frequenzi times fourier transform of the function. And fourier transform konficents cancel out.
3. one is lorenz gauge condition with fourier transform apllyied.
lorenz gauge condition for potential vector is
##\forall_i(\sum_{i=0}^D(\frac{\partial (A(i,\vec{X}))}{\partial \vec{x}(i)})=0)##
applying fourier transform gives the 3. eqation, because fourier transform of deriative of function is frequenzi times fourier transform of the function.
 
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  • #20
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You didn't answer my questions:

In which textbook do they appear? And if in none, why are you making things up?
Especially when it comes to the first equation. Besides:

photon number operator that was discussed abow.
You didn't draw correct conclusions from that discussion. Read @vanhees71 posts again.
 
  • #21
olgerm
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  • #22
vanhees71
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What do you mean bye "Fourier transformation of the number operator"? That doesn't make any sense to me.
 
  • #23
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I apllyied fourier transformation to number operator
And I asked you twice what textbook are you using, where are you getting this from? I know the answer, which is that you're making this up yourself. But that is against PF rules. It appears that you have no proper knowledge of QED, so why aren't you learning this topic from textbooks? You are not even reading responses in your own thread because you would notice that it was said that it's not an easy task to construct operators you are trying to construct.
 
  • #24
olgerm
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I defined new quantity ##A_p##. It is fourier transform of 4-potential. It is 4-vector. I wrote number operator, maxwell equations and lorenz gauge condition for this new quantity.
 
  • #25
jim mcnamara
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@olgerm I am trying to help. Here is a ridiculous example, but it shows your logic:
[ridiculous example]
##2 + 2 = 4##
I just created a new way to prove the Theory of General Relativity.
[/ridiculous example]
The first statement is mathematically correct. The use of second statement makes no sense at all.
So.

If you do not want the thread closed, please explain how you got your "new" result.
We do not want the mathematics. We need you to show why it has meaning.
To do that please cite a Physics reference which explains why you can make your claim.
As it stands what you have does not make sense. No meaning for us readers.
 

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