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Electromagnetic spin from Noether theorem and spin photon

by paolorossi
Tags: electromagnetic, noether, photon, spin, theorem
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paolorossi
#1
Nov23-12, 05:58 PM
P: 22
hi, I try to use the Noether theorem to determinate the angular momentum of the electromagnetic field described by the Lagrangian density

L=-FαβFαβ/4

After some calculation I find a charge Jαβ that is the angular momentum tensor. So the generator of rotations are
[itex](J^{23},J^{31},J^{12}) = \vec{J}[/itex]

and I find

[itex]\vec{J}[/itex] = [itex]\int d^{3}x ( \vec{E}\times \vec{A} + \sum _{k} E^{k} (\vec{x} \times \nabla ) A^{k} )[/itex]

Now I deduce that the field has an intrinsic angular momentum that is

[itex]\vec{S}[/itex] = [itex]\int d^{3}x ( \vec{E}\times \vec{A} ) [/itex]

but from this, once I quantized the field (for example in the Coulomb gauge, with the modified commutation relations) can I deduce something about the spin of the photon?
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Bill_K
#2
Nov23-12, 06:56 PM
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You want to verify that the angular momentum is ħ per photon, so I guess the thing to do is to determine the number of photons. So take a plane wave, write down the energy density and compare it to your expression for the angular momentum density. The ratio should come out Nħω/Nħ = ω.
andrien
#3
Nov24-12, 02:53 AM
P: 1,020
Quote Quote by Bill_K View Post
You want to verify that the angular momentum is ħ per photon, so I guess the thing to do is to determine the number of photons. So take a plane wave, write down the energy density and compare it to your expression for the angular momentum density. The ratio should come out Nħω/Nħ = ω.
That is the case,but only when the light is circularly polarized.If light is plane polarized,then it can be shown it will have half chance to be in +h- state and half to be in -h- state.If one wants to show that electromagnetic field has spin 1 character then it is possible to write maxwell eqn in a form similar to dirac eqn.,matrices used there can be used to show the spin 1 character.

Bill_K
#4
Nov24-12, 06:51 AM
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Electromagnetic spin from Noether theorem and spin photon

"So take a plane wave..." So take a circularly polarized plane wave...
If one wants to show that electromagnetic field has spin 1 character then it is possible to write maxwell eqn in a form similar to dirac eqn.,matrices used there can be used to show the spin 1 character.
In effect he's already done that. Maxwell's Equations can be written as a set of first-order equations. The variables you'll need to do this are Fμν and Aμ. The equations are

Aν,μ - Aμ,ν = Fμν
Fμν,ν = 0

Similar to the Dirac Equation, I suppose, but there are ten independent variables instead of four. From these equations you can extract a set of very singular 10 x 10 matrices. Then what?

In the Dirac case it's not the gamma matrices that are involved in rotations anyway, it's the other ones, σμν. For a given state ψ, the spin is something like ψσμνψ. Well, this just gets us back to where we started, when the OP wrote S = E x A. This expression can be written as a singular 10 x 10 matrix sandwiched between combinations of the field variables Fμν and Aμ.

So given that matrix, what do you need to do to show the field has intrinsic spin 1 (or 1/2)? You find the eigenvectors and eigenvalues of the matrix.

Guess what the eigenvectors are in the Maxwell case? The circularly polarized plane waves. Which was my suggestion!
paolorossi
#5
Nov24-12, 10:10 AM
P: 22
hi guys, I'm very stubborn so I quantized the spin
Quote Quote by paolorossi View Post

[itex]\vec{S}[/itex] = [itex]\int d^{3}x ( \vec{E}\times \vec{A} ) [/itex]
in gauge of coulomb with [itex] A^{0}=0 [/itex] , and , after some calculations and observations on the polarization vectors , I find

[itex]\vec{S} = \int d^{3}k i \vec{k} (a^{+}_{2}(\vec{k}) a_{1}(\vec{k})-a^{+}_{1}(\vec{k}) a_{2}(\vec{k})) = \int d^{3}k \vec{S}_{\vec{k}}[/itex]

so

[itex]\vec{S}_{\vec{k}} = \vec{k} (a^{+}_{1}(\vec{k}) , a^{+}_{2}(\vec{k})) \sigma_{2} (a_{1}(\vec{k}) , a_{2}(\vec{k}))^{t} [/itex]

Is correct to say that [itex]\vec{S}_{\vec{k}} [/itex] the spin of a photon whit a momentum k ? If the answer is yes, someone can help me to understand what it implies about the spin of the photon?
Bill_K
#6
Nov24-12, 02:47 PM
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Is correct to say that S⃗ k⃗ the spin of a photon whit a momentum k ?
No, you have an extra factor of |k|. Correctly stated, the eigenvalues of spin will be 1.
andrien
#7
Nov25-12, 12:33 AM
P: 1,020
see the page 181 here,the footnote
http://books.google.co.in/books?id=l...uation&f=false
There are three matrices S1,S2,S3,now to find out spin character,evaluate S2=S12+S22+S32,which comes out 2,now comparing with the usual relation,
S2=S(S+1),it implies S(S+1)=2 and hence S=1,as one wants.
paolorossi
#8
Nov25-12, 03:49 AM
P: 22
Quote Quote by Bill_K View Post
No, you have an extra factor of |k|. Correctly stated, the eigenvalues of spin will be 1.
sorry, I make an error, the [itex]\vec{k}[/itex] comes from [itex]\vec{\epsilon}_{1}(\vec{k}) \times \vec{\epsilon}_{2}(\vec{k})[/itex] , so it is a versor! So I must write [itex]\vec{k}/k[/itex] in place of [itex]\vec{k}[/itex]...

andrien maybe you are my salvation! I don't have the book, can you get a little more detail? thanks to all
paolorossi
#9
Nov25-12, 04:28 AM
P: 22
summarizing, I find from invariance of e.m. Lagrangian , with the help of Noether th, that the e.m. field has an intrinsic angular momentum. Quantizing in Coulomb gauge, I express this spin of e.m. field as

[itex]\vec{S} = \int d^{3}k i \vec{k}/k (a^{+}_{2}(\vec{k}) a_{1}(\vec{k})-a^{+}_{1}(\vec{k}) a_{2}(\vec{k})) = \int d^{3}k \vec{S}_{\vec{k}}[/itex]

with

[itex]\vec{S}_{\vec{k}} = i \vec{k}/k (a^{+}_{2}(\vec{k}) a_{1}(\vec{k})-a^{+}_{1}(\vec{k}) a_{2}(\vec{k})) [/itex]

so a state with a linearly polarized photon [itex]\left| \vec{k} , s \right\rangle[/itex] (where s=1,2 is the index of polarization) isn't eigenstate of S, but a state

[itex]\left| \vec{k} , 1 \right\rangle +i \left| \vec{k} , 2 \right\rangle[/itex] is eigenstate of Sk with eigenvalue [itex] \vec{k}/k[/itex]

[itex]\left| \vec{k} , 1 \right\rangle -i \left| \vec{k} , 2 \right\rangle[/itex] is eigenstate of Sk with eigenvalue [itex]-\vec{k}/k[/itex]

in pratice they are state of circular polarization... What does this mean? Only states with circularly polarized photon (that have a definite momentum) have a definite spin? And there are only spin states +1 and -1 (in direction k)?
andrien
#10
Nov25-12, 07:18 AM
P: 1,020
Quote Quote by paolorossi View Post
sorry, I make an error, the [itex]\vec{k}[/itex] comes from [itex]\vec{\epsilon}_{1}(\vec{k}) \times \vec{\epsilon}_{2}(\vec{k})[/itex] , so it is a versor! So I must write [itex]\vec{k}/k[/itex] in place of [itex]\vec{k}[/itex]...

andrien maybe you are my salvation! I don't have the book, can you get a little more detail? thanks to all
what do you mean by salvation.By the way,maxwell eqn in empty space can be written in a form like this
(-iS.∇-i∂0)ψ=0,where the three S satisfy the commutation relation
S1S2-S2S1=iS3 and similarly for cyclical order.These three S compared to weyl eqn give appearance of σ matrices.So,these are represented as some spin describing property.I have shown before that they do describe a spin 1 character of EM field.I have said before that light with circular polarization only can be used to predict definite spin.
And there are only spin states +1 and -1 (in direction k)?
Do you want a zero state also?
Hans de Vries
#11
Nov25-12, 07:49 AM
Sci Advisor
P: 1,135
The term [itex]\vec{E}\times\vec{A}[/itex] is not sufficient to represent the electromagnetic spin
because it does not transform as part of an axial vector.

The complete expression for the electromagnetic spin four vector is the
Chern Simons current [itex]{\cal C}^\mu[/itex].

\begin{equation}
{\cal C}^\mu ~~=~~ \epsilon_o\,\tfrac12\varepsilon^{\,\mu\nu\alpha \beta} F_{\alpha\beta}A_\nu ~~=~~ \epsilon_o\,\varepsilon^{\,\mu\alpha\beta\gamma} A_\alpha\partial_\beta A_\gamma
\end{equation}

Which is expressed in matrix form as:

\begin{equation}\label{eq:EM_spin_density2}
{\cal C}^\mu ~~=~~
\mbox{ $\left(
\begin{array}{c c c c}
~ 0 &-\tfrac1c\,H_x &-\tfrac1c\,H_y &-\tfrac1c\,H_z \\
\tfrac1c\,H_x & ~~~ 0 & \ \ ~~D_z & ~-D_y \\
\tfrac1c\,H_y & ~-D_z & ~~~ 0 & \ \ ~~D_x \\
\tfrac1c\,H_z & \ \ ~~D_y & ~-D_x & ~~~ 0
\end{array}
\right) \left(
\begin{array}{c}
\ \ A_0 \\
-A_x \\
-A_y \\
-A_z
\end{array}
\right)$}
\end{equation}

This is a four-vector field which we can write down explicitly as a
3d vector and a time-component.
\begin{equation}
\qquad \vec{\cal C}\ =\
D \times \vec{A}\ +\ \tfrac1c~H~A^o\
,\qquad
{\cal C}^o \ =\ \tfrac1c~H\cdot\vec{A}\ \quad
\end{equation}

So you see that the term [itex]H A^o[/itex] needs to be included. The effect is
that circular polarized light still has a spin of [itex]\pm \hbar[/itex] but linear polarized
light now correctly gets a spin 0.


The Chern Simons current arises in the axial anomaly of the electron
which was discovered around 1969 byS. L. Adler, John S. Bell and
R. Yackiw. It was found that the axial current [itex]J^\mu_A[/itex] of the electron
(its spin) is not conserved independently.

In order to conserve the spin and to keep electromagnetism as a
local gauge theory, it is required by quantum perturbation theory that:

\begin{equation}
\partial_\mu j^\mu_A ~~=~~ - \frac{\alpha}{2\pi}\,\partial_\mu C^\mu
\end{equation}

The rightmost term of the equation, the Chern-Pontryagin density
[itex]{\cal A}=\partial_\mu C^\mu[/itex], is non-zero outside the electron's wave function where
the charge/current density is zero.

See also this chapter from my book where the electromagnetic spin
if worked out for a number of practical cases:

http://physics-quest.org/Book_Chapte...SimonsSpin.pdf


Hans


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