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## The absorption of a linearly polarized photon.

Well, Spin density is proportional to ExA. Using E=-dA/dt and Ampere Maxwell you should be able to express this as something proportional to rot b* x rot b. Is there an easy expression for spin flux?

As I understand, your method for finding the power distribution in the atom radiation, $$1+\cos^2\theta$$, is exactly the method used in the frame of the standard electrodynamics [1] because you use Poynting vector, $$E\times B$$, that is a component of Maxwell tensor, Ampere-Maxwell’s law, and vector spherical harmonics, though I should like to see your calculations in details. In contrast, Feynman obtained $$1+\cos^2\theta$$ simply and exclusively by quantum mechanics (see (4.4) in [2]).

 Quote by DrDu there are also non-vanishing angular components.
Maxwell electrodynamics gives the distribution of orbital angular momentum, sin^2(theta) (see (2.9) in [2], or (2.79) in [3]). Can you confirm this result? I ask this for the second time.

 Quote by DrDu Spin density is proportional to ExA.
Unfortunately, you share a common alogism. You use Maxwell energy-momentum tensor, not the canonical energy-momentum tensor, and in the same time you use a component of the canonical spin tensor, $$E\times A$$, which is annihilated by the Belinfante-Rosenfeld procedure.
As is well known, the canonical energy-momentum tensor $$T_c^{ik}$$ is coupled with the canonical spin tensor $$\Upsilon_c^{jik}$$. So the total angular momentum density is $$J_c^{jik}=r^{[j}T_c^{i]k}+\Upsilon_c^{jik}$$. But the construction contradicts experiments [4].
As is well known, Maxwell energy-momentum tensor $$T^{ik}$$ is not coupled with a spin tensor, Maxwell energy-momentum tensor is coupled with zero spin tensor. So, the construction $$r^{[j}T^{i]k}$$ is an orbital angular momentum.

 Quote by Khrapko Do you know if experimental evidences of the distribution $$\cos^2\theta+1$$ in the radiation from (J=3/2, M=3/2) atom exist?

 Quote by DrDu Is there an easy expression for spin flux?
YES, an easy expression for spin flux is well known since 2001. And what is more an experiment for a confirmation of the expression is suggested [2,7]. Unfortunately, the submissions [2,4,7] were rejected without reviewing. Gordon W.F. Drake assessed my paper [2] as “too pedagogical for the Physical Review.” Sonja Grondalski wrote only, “Your manuscript has been considered. We regret to inform you that we have concluded that it is not suitable for publication in Physical Review Letters.”

[1] J. Jackson, Classical Electrodynamics, #9.9
[2] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfil...6&module=files
[3] A. Corney, Atomic and Laser Spectroscopy (1979)
[4] Canonical spin tensor is wrong http://khrapkori.wmsite.ru/ftpgetfil...9&module=files
[5] R.I. Khrapko. True energy-momentum tensors are unique. Electrodynamics spin tensor is not zero. - http://arXiv.org/abs/physics/0102084
[6] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfil...ule=files&id=9
[7] Experiment concerning electrodynamics’ nonlocality http://khrapkori.wmsite.ru/ftpgetfil...le=files&id=46
 Recognitions: Science Advisor This weekend I managed to take a quick look on Feynman's calculation. He calculates the probability amplitude to find circularly polarized plane wave photons, i. e., helicity eigenstates. For this kind of photons, electric field is $$\langle \pm, k_z|\mathbf{E}|0 \rangle=E_0 (1, \pm i, 0)^T \exp(ik_z z)$$ (don't ask me, to which helicity the plus and minus refers) for k in z-direction. Other directions can be obtained by rotating first around the z-axis, which introduces only a phase factorn and then around the y-axis ( the rotation matrix has row vectors something like $$(\cos(\theta), 0, -\sin(\theta), (0, 0, 0), (\sin(\theta), 0, \cos(\theta))$$), the relevant vector becomes (up to the phase factor) $$(\cos(\theta), \pm i, \sin(\theta))^T$$. The transition is due to the $$\mathbf{E}\cdot \mathbf{d}$$ coupling in the hamiltonian in the electric dipole approximation. The transition dipole moment has to be calculated with the J=3/2, M_J=3/2 L=1 M_L=1 state and the ground state with J=1/2 M_J=1/2, L=0, M_L=0. As we saw already, spin doesn't change as the operator d only acts on the orbital part. Hence the matrix elements can easily be evaluated from the spherical harmonics. The corresponding matrix element is $$\langle 3/2, 3/2| \mathbf{d}|1/2, 1/2\rangle=d_0 (1, i, 0)^T$$ (again it could be -i instead of i). The scalar product of the electric field and transition dipole moment is then proportional to $$1 \pm \cos(\theta)$$ depending on the polarization.
 Thank you for the attention, but, sorry, I have understood nothing in your post. Feynman does not use E. What is the power T? How can be obtained the scalar product of the electric field and transition dipole moment? Can you recommend textbooks where the notations are used?
 Recognitions: Science Advisor It was not my intention to repeat Feynmans consideration word by word but to make contact with what we discussed before. I wanted to point out especially that Feynman seems to consider circular polarized plane waves, that is states of sharp helicity in his analysis. T means "transposed", i.e. consider the column vector instead of the row vector. The coupling to the electromagnetic field reads different in different gauges. In atomic and molecular physics it is very convenient to start from the multipolar gauge where the coupling is given in terms of a series of couplings to the different electric and magnetic multipole moments of the atom or molecule. The first and most important term is the coupling of the electric field to the electric dipole moment. This can be found in many books e.g. in the book by Craigh and Thirunamachandran, Molecular quantum electrodynamics, Dover publ. or also here, if you know some french: http://www.phys.ens.fr/cours/college...87/1986-87.pdf or in the books of Claude Cohen Tannoudji. In the simplest cases one can argue that in the Coulomb gauge, the coupling of some electrons i to the electromagnetic field has the form $$\sum_i e\mathbf{A}\cdot \mathbf{p_i}$$. Now $$\mathbf{E}=-d\mathbf{A}/dt$$ and $$\mathbf{p_i}=d\mathbf{r_i}/dt$$ hence $$\sum_i e \mathbf{A}\cdot \mathbf{p}_i$$ and $$\mathbf{E}\cdot \mathbf{d}$$ with $$\mathbf{d}=e \sum_i \mathbf{r_i}$$ coincide up to a total time derivative which can always be added to the Lagrangian.
 Recognitions: Science Advisor Just found on arxiv an article that might interest you: http://arxiv.org/ftp/arxiv/papers/1010/1010.1056.pdf
 I know A. M. Stewart very well since 2005. I criticized his articles [1,2,3] in 2005, but James Dimond rejected my submittion [4] to Eur. J. Ph.: “From the confused and inaccurate Abstract right through to the end, this paper contains a large number of errors and misunderstandings. I know that similar papers by Khrapko have been rejected by a number of other journals. The lack of understanding shown by the author is beyond any easy solution; no simple re-writing of the paper can make it sensible. I strongly recommend that the paper be rejected as unsuitable for publication.” Then I criticized Stewart’s articles in my publication [5], but Stewart ignores criticism and continues to publish his mistakes. Unfortunately, L. Allen, M. J. Padgett [6] ignore my criticism in [5] as well. [1] A. M. Stewart, Angular momentum of the electromagnetic field: the plane wave paradox resolved, European Journal of Physics 26, 635-641 (2005). arXiv:physics/0504082v3 [2] A. M. Stewart, Angular momentum of light, Journal of Modern Optics 52(8), 1145-1154 (2005). arXiv:physics/0504078v2 [3] A. M. Stewart, Equivalence of two mathematical forms for the bound angular momentum of the electromagnetic field, Journal of Modern Optics 52(18), 2695-2698 (2005). arXiv:physics/0602157v3 [4] R. I. Khrapko, Angular momentum of the electromagnetic field, EJP/202604/PAP/ (June 30, 2005) http://khrapkori.wmsite.ru/ftpgetfil...7&module=files [5] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfil...ule=files&id=9 [6] L. Allen, M. J. Padgett, “Response to Question #79. Does a plane wave carry spin angular momentum?” Am. J. Phys. 70, 567 (2002) http://khrapkori.wmsite.ru/ftpgetfil...3&module=files
 I am sure that the disgraceful report (see #75) was written by A.M. Stewart rather than by J. Dimond, editor, because my submission [4] concerned Stewat’s paper [1]. However, since editors support anonymity of reviewers, they are responsible for reports. I think it is the situation, which was discussed at [2]. Juan R. Gonzalez-Alvarez wrote on 9 Apr 14:08 “Biased reviewers want to maintain their names anonymous to accept the papers of their friends/colleagues and for rejecting the papers of the people working in rival theories or showing the mistakes contained in referee's work (when as author)...” Note, arXiv publishs A.M. Stewart, but I am blacklisted [3]. [1] A. M. Stewart, Angular momentum of the electromagnetic field: the plane wave paradox resolved, European Journal of Physics 26, 635-641 (2005). arXiv:physics/0504082v3 [2] Peer-Review Under Review http://groups.google.ru/group/sci.ph...11f1c2b0ce22dd [3] http://khrapkori.wmsite.ru/files/str...0?catoffset=10 [4] R. I. Khrapko, Angular momentum of the electromagnetic field, EJP/202604/PAP/ (June 30, 2005) http://khrapkori.wmsite.ru/ftpgetfil...7&module=files

Dear DrDu, you wrote in #54:
 Quote by DrDu The wavefunction with j=1/2 and l=1 is of the form $$a |l=1, m=1, m_s=-1/2> -b |l=1, m=0, m_s=1/2>$$
I should like to obtain the ratio a/b by the use of the way, which was started in #58, though we know a/b=sqrt{2} for (J=1/2, M=1/2), and a/b=-1/sqrt{2} for (J=3/2, M=1/2) from http://personal.ph.surrey.ac.uk/~phs3ps/cgjava.html.
 Quote by Khrapko Denote for short |l=1 m=1 s=-1/2> = F, |l=1 m=0 s=1/2> = G. Then $$\hat H F=h_{11}F+h_{12}G,\qquad \hat H G=h_{21}F+h_{22}G$$ $$\hat H (aF-bG)=ah_{11}F+ah_{12}G-bh_{21}F-bh_{22}G=E(aF-bG)$$ $$ah_{11}F-EaF-bh_{21}F=0,\qquad ah_{12}G-bh_{22}G+EbG=0$$ $$(h_{11}-E)(-h_{22}+E)+h_{12}h_{21}=0$$ has two solutions.
These solutions are .
$$E_{1,2}=(h_{11}+h_{22})/2\pm\sqrt{(h_{11}-h_{22})^2/4+h_{12}h_{21}}$$.
Then
$$a/b=h_{21}/(h_{11}-E)=(h_{22}-E)/h_{12}$$.
So,
$$a_1/b_1=(h_{22}-h_{11})/2h_{12}-\sqrt{((h_{11}-h_{22})/2h_{12})^2+1}$$.
How can we obtain $$a/b=\sqrt{2}$$?
Thank you
 Recognitions: Science Advisor Of course you can. You should first identify the relevant hamiltonian.
 Oh! The Hamiltonian depends on the energy of spin-orbital interaction. Do you mean the energy depends on the Clebsch-Gordan coefficients? It is strange because the coefficients are the expansion coefficients of total angular momentum eigenstates in an uncoupled tensor product basis; they are not connected with the energy of spin-orbital interaction.
 Recognitions: Science Advisor Well, when you take spin orbit interaction into account, then states with different total angular momentum will have different energies. So if you take H to equal the spin orbit part of the hamiltonian, then the clebsch gordan coefficients should result from diagonalizing the hamiltonian.
 I have diagonalized the Hamiltonian. The diagonal is $$E_1,E_2$$ from #77. Respective eigenfunctions of the Hamiltonian are $$\psi_1=a_1/b_1|l=1, m=1, m_s=-1/2> - |l=1, m=0, m_s=1/2>$$ and $$\psi_2=a_2/b_2|l=1, m=1, m_s=-1/2> - |l=1, m=0, m_s=1/2>$$ from #77 or from #54. The Clebsch-Gordan coefficients give $$a_1/b_1=\sqrt{2}$$ and $$a_2/b_2=-1/\sqrt{2}$$ where $$a_1/b_1$$ is the function of $$h_{ij}$$ (see #77). Do you assert that Clebsch-Gordan coefficients contain information about the energy of spin-orbital interaction, i.e. about $$h_{ij}$$, which are in $$a/b$$? Can Clebsch-Gordan coefficients help to identify the relevant hamiltonian?
 Recognitions: Science Advisor Well, in a sense they contain information about the hamiltonian, but this information you already have once you know the symmety ( in this case the rotational symmetry) of the hamiltonian.
 Your post is strange. The energy of spin-orbital interaction, $$h_{12}$$, depends on $$e,h,c,\mu, etc.$$, but Clebsch-Gordan coefficients are the expansion coefficients of total angular momentum eigenstates in an uncoupled tensor product basis; they are not connected with the energy of spin-orbital interaction. So, they cannot determine the energy of spin-orbital interaction, $$h_{12}$$. The equation $$a_1/b_1=(h_{22}-h_{11})/2h_{12}-\sqrt{((h_{11}-h_{22})/2h_{12})^2+1}=\sqrt{2}$$ (#77) is a puzzle!

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