The absorption of a linearly polarized photon.

[DrDu]

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.

 

[Khrapko]

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?

 

[DrDu]

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-de-france/1986-87/1986-87.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.

 

[DrDu]

Just found on arxiv an article that might interest you:

http://arxiv.org/ftp/arxiv/papers/1010/1010.1056.pdf

 

[Khrapko]

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/ftpgetfile.php?id=57&module=files
[5] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfile.php?module=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/ftpgetfile.php?id=53&module=files

 

[Khrapko]

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.p...read/thread/6a35d56e15ad1b69/5711f1c2b0ce22dd
[3] http://khrapkori.wmsite.ru/files/struggle-with-arxiv-10?catoffset=10
[4] R. I. Khrapko, Angular momentum of the electromagnetic field, EJP/202604/PAP/ (June 30, 2005) http://khrapkori.wmsite.ru/ftpgetfile.php?id=57&module=files

 

[Khrapko]

Dear DrDu, you wrote in #54:

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.

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

 

[DrDu]

Of course you can. You should first identify the relevant hamiltonian.

 

[Khrapko]

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.

 

[DrDu]

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.

 

[Khrapko]

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?

 

[DrDu]

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.

 

[Khrapko]

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!

 

[DrDu]

Your post is strange.

Well, maybe because I don't exactly understand yours!
The hamiltonian has to interchange with J and also L^2 and S^2. This limits it's general form but of course does not determine the absolute strength of the interaction.
But apparently it fixes the ratio (h22-h11)/h12.