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Feynman's Derivation of Maxwell's Equations from Commutator Relations

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S.Daedalus
#1
May31-11, 04:27 AM
P: 211
According to Dyson, Feynman in 1948 related to him a derivation, which, from
1) Newton's: [tex]m\ddot{x}_i=F_i(x,\dot{x},t)[/tex]
2) the commutator relations: [tex][x_i,x_j]=0[/tex][tex]m[x_i,\dot{x}_j]=i\hbar\delta_{ij}[/tex]
deduces:
1) the 'Lorentz force': [tex]F_i(x,\dot{x},t)=E_i(x,t)+\epsilon_{ijk}\dot{x}_j B_k(x,t)[/tex]
2) and the homogenous 'Maxwell equations': [tex]\nabla\cdot\mathbf{B}=0[/tex] [tex]\nabla\times\mathbf{E}+\frac{\partial B}{\partial t}=0[/tex]

Now the derivation is straightforward enough. (And apparently, the inhomogenous equations left underived just provide 'a definition of matter'.) But the question is: what does this mean? I can come up with several interpretations of various strengths:

1) It means nothing; it's just a mathematical oddity.
2) Dyson's view, apparently, is that the proof shows that the only possible fields
that can consistently act on a quantum mechanical particle are gauge fields -- I'm not sure I exactly understand what's meant by that (well, rather, I understand what it means, but I'm not sure I get why the proof implies it).
3) Electrodynamics is somehow 'built in' to quantum mechanics.

Which, if any, of these is right? It just seems odd that all that information ought to be contained in the simple commutation relations -- how are they supposed to 'know' about electromagnetic fields? Moreover, I understand there are various generalizations of the derivation, incorporating explicitly special or general relativity, non-Abelian gauge fields, or even higher dimensions. So... what's it all mean?
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Demystifier
#2
May31-11, 05:06 AM
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It means
4) The derivation above works because the "quantum" commutators involved are essentially the same as the corresponding classical Poisson brackets. So it is really a classical derivation disguised into a quantum one. You can repeat the whole derivation by replacing the commutators with the appropriate Poisson brackets.

One additional comment. The derivation above (either with commutators or Poisson brackets) suggests that electromagnetic forces are the only possible forces on particles. But then, what about gravitational forces? Are they impossible? Of course not. It can be shown that the equations of Einstein gravity in a non-relativistic limit can be approximated by the Maxwell equations above. (Note also that these are only 2 Maxwell equations, not all 4 of them. This truncated set of equations alone does not imply Lorentz invariance.)

For more details see
R. J. Hughes, Am. J. Phys. 60, 301 (1992)
Demystifier
#3
May31-11, 06:32 AM
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Quote Quote by S.Daedalus View Post
2) Dyson's view, apparently, is that the proof shows that the only possible fields
that can consistently act on a quantum mechanical particle are gauge fields -
The inhomogeneous Maxwell equations alone do NOT imply that the fields E and B are gauge fields. In other words, they do not imply that these fields must be derivatives of a potential.

To see this, the most elegant language is the language of differential forms. In this language, the homogeneous Maxwell equations can be written in a compact form
dF=0
which do not imply
F=dA

haushofer
#4
May31-11, 10:09 AM
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Feynman's Derivation of Maxwell's Equations from Commutator Relations

You mean one has to say on top of that something about the topology of spacetime (that it should be simply connected)?
Demystifier
#5
May31-11, 10:44 AM
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Quote Quote by haushofer View Post
You mean one has to say on top of that something about the topology of spacetime (that it should be simply connected)?
Yes. See e.g.
http://en.wikipedia.org/wiki/Closed_...erential_forms
S.Daedalus
#6
May31-11, 12:43 PM
P: 211
Quote Quote by Demystifier View Post
The inhomogeneous Maxwell equations alone do NOT imply that the fields E and B are gauge fields. In other words, they do not imply that these fields must be derivatives of a potential.
Not sure I get this. The way the story is usually told, a source-free field can be written as the rotation of a vector field, [itex]\mathbf{B}=\nabla\times\mathbf{A}[/itex], and that together with the second Maxwell equation above implies [itex]\nabla\times(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t})=0[/itex], which can thus be written as a gradient field, i.e. [itex]\mathbf{E}=-\nabla\Phi-\frac{\partial\mathbf{A}}{\partial t}[/itex]. Is that not right?
Bill_K
#7
May31-11, 02:09 PM
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It means

1.5) There's another assumption being made, namely minimal coupling. A vector gauge field A is the only field that can couple to fermions via P = p + e/c A, and this assumption quickly leads to the homogeneous field equations for the field F that corresponds to A.
Sybren
#8
May31-11, 04:28 PM
P: 47
Here is the article presenting the derivation, further down in the PDF are included the commentaries on the derivation. It's interesting that this thread was started today because I have been working through this derivation during the last couple of days.
Attached Files
File Type: pdf DysonMaxwell041989.pdf (499.5 KB, 73 views)
Demystifier
#9
Jun1-11, 03:56 AM
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Quote Quote by S.Daedalus View Post
Not sure I get this. The way the story is usually told, a source-free field can be written as the rotation of a vector field, [itex]\mathbf{B}=\nabla\times\mathbf{A}[/itex], and that together with the second Maxwell equation above implies [itex]\nabla\times(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t})=0[/itex], which can thus be written as a gradient field, i.e. [itex]\mathbf{E}=-\nabla\Phi-\frac{\partial\mathbf{A}}{\partial t}[/itex]. Is that not right?
This is not really important for the main subject of this thread, but you are right. As haushofer indicated, my remark (which would be better if I didn't make it) is relevant to spaces with nontrivial topologies, which is really not relevant for our discussion here. I apologize for the unnecessary confusion I made.

Anyway, I highly recommend to read the paper I cited in post #2, which is much better than the original Dyson's paper (to which Sybren gave the link).
uw_daryl
#10
Jun2-11, 10:00 AM
P: 1
great source! thanks! -daryl-
lalbatros
#11
Jun4-11, 03:38 PM
P: 1,235
Hello,

I would be interrested to read the same proof based on Poisson brackets instead of commutators.
Would you know a reference for this?

Thanks,

Michel
Demystifier
#12
Jun6-11, 02:16 AM
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Lalbatros, the reference you need is given in post #2.
lalbatros
#13
Jun6-11, 03:49 AM
P: 1,235
Thanks a lot Demystifier.
I found a copy of this paper.
Sybren
#14
Jun6-11, 03:21 PM
P: 47
I also got a copy of the paper now, and in it they mention, just as the first post of this thread does, that the 'Lorentz force' follows from the commutator relations and newton's second law of motion. However, I see from the Dyson paper that the 'Lorentz force' (which I put between quotes because the charge [itex]q[/itex] is missing) is used as a definition. So it is not a result. Only the two maxwell equations are results of the derivation.
unusualname
#15
Jun7-11, 10:47 AM
P: 661
Quote Quote by Sybren View Post
I also got a copy of the paper now, and in it they mention, just as the first post of this thread does, that the 'Lorentz force' follows from the commutator relations and newton's second law of motion. However, I see from the Dyson paper that the 'Lorentz force' (which I put between quotes because the charge [itex]q[/itex] is missing) is used as a definition. So it is not a result. Only the two maxwell equations are results of the derivation.
In the paper Demystifier mentioned by Hughes a fuller derivation is done (do a google search for the paper title and author "on feynman's proof of the maxwell equations hughes")


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