electromagnetic spin from Noether theorem and spin photonby paolorossi Tags: electromagnetic, noether, photon, spin, theorem 

#1
Nov2312, 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=F^{αβ}F_{αβ}/4 After some calculation I find a charge J_{αβ} 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? 



#2
Nov2312, 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.




#3
Nov2412, 02:53 AM

P: 987





#4
Nov2412, 06:51 AM

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electromagnetic spin from Noether theorem and spin photon
"
A_{ν,μ}  A_{μ,ν} = F_{μν} F_{μν,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, σ_{μν}. For a given state ψ, the spin is something like ψσ_{μν}ψ. 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 F_{μν} and A_{μ}. 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! 



#5
Nov2412, 10:10 AM

P: 22

hi guys, I'm very stubborn so I quantized the spin
[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? 



#6
Nov2412, 02:47 PM

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#7
Nov2512, 12:33 AM

P: 987

see the page 181 here,the footnote
http://books.google.co.in/books?id=l...uation&f=false There are three matrices S_{1},S_{2},S_{3},now to find out spin character,evaluate S^{2}=S_{1}^{2}+S_{2}^{2}+S_{3}^{2},which comes out 2,now comparing with the usual relation, S^{2}=S(S+1),it implies S(S+1)=2 and hence S=1,as one wants. 



#8
Nov2512, 03:49 AM

P: 22

andrien maybe you are my salvation! I don't have the book, can you get a little more detail? thanks to all 



#9
Nov2512, 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 S_{k} 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 S_{k} 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)? 



#10
Nov2512, 07:18 AM

P: 987

(iS.∇i∂_{0})ψ=0,where the three S satisfy the commutation relation S_{1}S_{2}S_{2}S_{1}=iS_{3} 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. 



#11
Nov2512, 07:49 AM

Sci Advisor
P: 1,136

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 fourvector field which we can write down explicitly as a 3d vector and a timecomponent. \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 ChernPontryagin density [itex]{\cal A}=\partial_\mu C^\mu[/itex], is nonzero 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://physicsquest.org/Book_Chapte...SimonsSpin.pdf Hans 


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