The absorption of a linearly polarized photon.

[DrDu]

R.Feynman (the Lectures, Quantum Mechanics, Ch. 18) calculates the radiation pattern of an excited atom and the distribution of spin in this radiation at l = 1, ignoring the electron's spin of the excited atom. Can it really be true, these distributions do not depend on the relative orientation of the electron spin and its orbital angular momentum in the initial state of the excited atom, ie on: j = 1/2 or j = 3/2?
Does somebody know an experiment on the distributions?

I just want to discuss the transition starting from l=1 j=3/2 to l=0 j=1/2:
I take as initial excited state the one state with l=1 j=3/2 which is a product state |l=1, m=1 > |m_s=+1/2> (all others with same j correspond to rotated situations).
There are two possible finals states with l=0:
l=0 j=1/2: |l=0, m=0> |m_s=+1/2>
and {l=0, m=0 > |m_s=-1/2>
For light of long wavelength, transition is mainly due to the electric dipole moment d which only couples to the orbital part of the wavefunction hence the transition probability is
<l=1, m=1|\mathbf{d}| l=0, m=0> <m_s=1/2|m_s=\pm 1/2>
The last matrix element is 1 if m_s doen't change and zero else.
A single photon transition to the state with s=-1/2 would require an admixture to the initial state with j=3/2, s=-1/2 and different value of l due to spin-orbit interaction. In case of the 2p ->1s transition in hydrogen, this is completely negigible.

 

[Khrapko]

orbital angular momentum depends on the choice of the origin.

NO! Angular momentum of a rotating wheel does not depend of the position of an observer. Orbital angular momentum of an atom’s electron does not depend on the position of an observer as well. A fortiori, spin angular momentum does not depend on the position of an observer. Angular momentum, L, of an object depends on a displacement, dr, of the position of an observer only if the oject has a linear momentum, P: dL = P x dr.

Spin of electrons is not angular momentum and it doesn't transform as angular momentum under spatial rotations.

Only angular momentum of a wheel, which is a pseudo vector (with an outer orientation), is transformed as a vector under spatial rotations. Orbital angular momentum of an atom’s electron is a characteristic of its Q.M. state, which is a spinor. So, it is transformed not as a vector. All the more spin angular momentum of an electron, which is represented by 1/2-spinor, is not transformed as a vector. Nevertheless, orbital angular momentum of an atom’s electron is added to spin angular momentum of the electron: j = l + s.

 

[Khrapko]

To DrDu #51
It seems to me the matrix element of the transition from l=1, m=1 to l=0, m=0 is <l=0,m=0\mid{\bf d}\mid l=1,m=1>.
But I am interested in the difference between the transitions from m=1, m_s=1/2 and from m=1, m_s=-1/2

 

[DrDu]

You are right, better to say, transition probability is the absolute square of that matrix element, so the order of the bras and kets doesn't matter too much.
The transition probabilities from the j=1/2 state can be worked out in analogy. 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 don't want to determine a and b. However, they can be chosen real and positive.
The transition probability becomes
|a|^2 |&lt;l=1, m=1|\mathbf{d}|l=0,m=0&gt;|^2+|b|^2 |&lt;l=1,m=0|\mathbf{d}|l=0, m=0&gt;|^2 - 2ab Re (&lt;l=1,m=1|\mathbf{d}|l=0,m=0&gt;&lt;l=0,m=0|\mathbf{d}|l=1,m=0&gt;)<br />
There is an analogous state with j=3/2 which can be obtained by interchanging a->b and b-> -a.
As was to be expected, the transitions between states with different spin m_s become mixed up but this is not due to an explicit spin flip. In case of the state with j=3/2, even the mixing interpretation can be avoided by referring m and m_s to a rotated quantization axis.

 

[Khrapko]

I should like to restrict us to the Feynman’s case.
I am interested in the space distributions of spin, angular momentum, and mass in the radiation when an atom makes the transition from l=1, m_l=1, m_s=1/2 or from l=1. m_l=1, m_s=-1/2. Note these radiations have different frequencies (see #34, #36).

 

[DrDu]

I should like to restrict us to the Feynman’s case.
I am interested in the space distributions of spin, angular momentum, and mass in the radiation when an atom makes the transition from l=1, m_l=1, m_s=1/2 or from l=1. m_l=1, m_s=-1/2. Note these radiations have different frequencies (see #34, #36).

The transition from the first state I discussed in #51. When frequencies of the transitions are not the same for the second state, this is due to Spin-Orbit splitting. Then the second state, i.e. l=1. m_l=1, m_s=-1/2 is not an eigenstate but mixes with l=1 m_l=0 and m_s=1/2, which is the situation I discussed in #54. However, in the limit of vanishing Spin Orbit coupling, also the second state is an eigenstate. Then the transition in question does not change spin as the wavefunctions factorize into the spin and the orbital part and the transition is due to the dipole moment operator which only acts on the orbital wavefunction.

 

[hiyok]

Now I have found new reasons for the classical spin tensor. I submitted a paper [1] to PRA. The paper is devoted to an irritative question: how is an angular momentum flux of electromagnetic field distributed in the space?
[1] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files

Hey, Khrapko,
I have discussed your paper with my friends. And I feel that your claim (which, as I understand, is that the expression \vec{J}=\vec{r}\times(E\times B) does not include the spin, let's say, \vec{s}[\tex], which is the polarization of photons.) is not sufficiently justified for the following reason:<br /> <br /> 1. You calculated the spatial distribution of J and s and found that they are spatially separated, and then you claimed an incompatibility between J and s. However, one can express J as a sum of s and another term (which I shall call L), J=s+L. And it is likely that, the distribution of L just makes up for this spatial separation of J and s. <br /> <br /> 2. Actually, the division of J into s and L is well known[<i><b>Quantum Field Theory: from operators to path integrals, Kerson Huang, John Wiley &amp; Sons, 1998</b></i>]. According to this division, s can indeed be related to the polarization of photons and yield results in accord with Feynman&#039;s doing. This means that, L is the orbital angular momentum (the moment of momentum) while J is the total one. Considering Lorentz invariance (invariant under Lorentz group), this must be so. <br /> <br /> The reason why I believe that J has already included the s is because, the above division gives correct results.

 

[Khrapko]

DrDu #51 discussed the transition from l=1 j=3/2 to l=0 m_s=-1/2. It is not interesting.

DrDu #54 stated that |l=1 m=1 s=-1/2> is not an eigenstate,
that the eigenstate is |l=1 j=1/2> = a|l=1 m=1 s=-1/2> - b|l=1 m=0 s=1/2>.
It seems to be wrong because the statement gives rise to a doublet splitting of the term 2P_1/2 just as of ammonia (see Feynman’s Lectures).
Really, 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.
However, as before I am interested in the space distributions of spin, angular momentum, and mass in the radiation when an atom makes the transition from |l=1 m=1, s=-1/2> (#34). And as before I am interested in how can the angular momentum conservation law be satisfied (#32).

Hiyok #57 agrees that the area of the circular polarization is spatially separated from the area where the moment of momentum exists, ie from the area where r\times(E\times B)\ne 0 as is stated in [1].
Hiyok #57 claims that the area where r\times(E\times B)\ne 0 can be expanded to include the circular polarization area by the division \int r\times(E\times B)dV=s+L, but it is strange because a division cannot give rise to an expansion. Unfortunately, the Hiyok’s delusion is a common delusion [2-4]. I explained the matter [5,6].
[1] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files
[2] J. Humblet, Physica, 10, 585 (1943)
[3] J. D. Jackson, Classical Electrodynamics, Problem 7.27
[4] H. C. Ohanian, “What is spin?” Amer. J. Phys. 54, 500-505 (1986).
[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/ftpgetfile.php?module=files&id=9

 

[DrDu]

Of course you get two solutions, but the other one corresponds to j=3/2, m_j=1/2, not j=1/2. The factors a and b are the Clebsch Gordan coefficients. You can calculate them e.g. here:
http://personal.ph.surrey.ac.uk/~phs3ps/cgjava.html
a= sqrt(2/3) and b=sqrt(1/3) for the j=1/2 m_j=1/2 state.
For the j=3/2, m_j=1/2 state, you have to change a->b and b->-a. You may check this in the calculator explicitly.

Btw. your h_12 is due to spin-orbit interactions which are small. If you neglect them, the two states stay degenerate and you can work with the states F and G equivalently.

 

[Khrapko]

DrDu #59
Of course you are right: (J=3/2, M=1/2) = sqrt(2/3) (m=0,s=1/2) + sqrt(1/3) (m=1,s=-1/2).
Are the distributions in the radiation from (J=3/2, M=1/2)-state the superpositions of the distributions in the radiations from (m_l=0)-state and (m_l=1)-state? And the electron has no role in the distributions? Electron can be ignored when calculating?

 

[DrDu]

No, the radiation from the states with specified J is not just the superposition of the one from states with fixed L and M.


What do you mean with "the electron plays no role?".

 

[Khrapko]

What do you mean with "the electron plays no role?".

Sorry, I meant, “the electron’s spin plays no role”
As is known [Jackson], an energy distribution from an electron in the (l=1, m=1)-state is \cos^2\theta+1. Feynman’s Lectures yields this also (see [1])
The distribution from (l=1,m=0)-state is \sin^2\theta.
I think the distribution from the state sqrt(2/3) (l=1,m=0;s=1/2) + sqrt(1/3) (l=1,m=1;s=-1/2), as in #60, is the superposition of the distributions.
If no, how can we calculate the distribution?
[1] http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files

 

[DrDu]

Well, there should be also an interference term, as you don't superpose the energy distributions but the wavefunctions. This is something like 2abFG in your nomenclature. I would have to delve into the derivation of the expressions of Feynman myself to work out these expressions.

 

[Khrapko]

Well, we can observe \cos^2\theta+1 power distribution from (J=3/2, M=3/2) = (l=1,m=1;s=1/2)-state.
But we cannot observe \sin^2\theta power distribution from (l=1,m=0)-state (according to Jackson, 9,9) because (l=1,m=0,s=1/2)-state is not an eigenstate?

I would have to delve into the derivation of the expressions of Feynman myself to work out these expressions.

You don’t need to delve. Feynman did not obtain the result \cos^2\theta+1. Please see (4.4) in http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files

 

[DrDu]

You are asking complicated questions. My analysis up to now treated the field as a classical variable. Hence it is not possible to read of directly the change in the field upon emission.
To treat it quantum mechanically, one presumably has to introduce photon eigenfunctions with given l>0 and m, |l m> so that the electric and magnetic field are something like <x |E|lm> and <x|B|lm> and are products of some kind of vector spherical harmonics and Bessel functions (outgoing Hankel functions to describe a resonance). I can only sketch the reasoning: Before emission we have an excited state of the form aF +bG (both in case of the j=3/2 m_j=1/2 and the j=1/2 m_j=1/2 case) and after emission one with aF'+bG'. Here, F' and G' are obtained from F and G by replacing the excited orbital wavefunction with l and m by the corresponding photon wavefunction with l and m times the (unique) electronic ground state wavefunction with l=0 and m=0. Hence the wavefunction of the photon becomes entangled with the spin wavefunction of the electron.
However, orthogonality of the spin wavefunctions of the electron, will lead to the result that the angular distribution of the radiation emitted is simply |a|^2 (cos^2(theta)+1)+|b|^2 sin^2(theta), so there will be no interference effects in contrast to what I first assumed.

 

[Khrapko]

Thank you. Our discussion is helpful.
1) My conclusion is the distributions \cos^2\theta+1 and \sin^2\theta [1-3] concern the radiation of a rotating dipole and of an oscillating dipole rather than an atom’s radiation. So the problem of the separate radiation of orbital angular momentum and spin by a rotating dipole may be beyond the scope of atom physics. However,
2) Do you agree that the power distribution in the radiation from (J=3/2, M=3/2) = (l=1,m=1;s=1/2)-state of an atom is \cos^2\theta+1? Do you know if experimental evidences of the distribution exist?
[1] J. Jackson, Classical Electrodynamics, #9.9
[2] A. Corney, Atomic and Laser Spectroscopy (1979), Fig.2.6
[3] L. D. Landau, E. M. Lifgarbagez, The Classical Theory of Fields, #67, Problem 1

 

[DrDu]

The transition from the J=3/2 M_J=3/2 state to the J=1/2 M_J=1/2 state is, for small spin-orbit coupling and wavelength sufficiently larger than the atomic dimensions, a purely electronic dipolar transition. Electron's spin doesn't change.
The angular power should presumably be calculated from the wavefunction of the photon |photon>=|el, l=1, m=1> as <el,l=1,m=1|S|el, l=1,m=1> where S=c/8 pi (E x B-B x E) is the Poynting vector operator and the electronic wavefunction is somehow related to the vector spherical harmonics, the "el" referring to the polarization resulting from an electronic dipole transition.
I am not completely sure how to evaluate this expression. Presumably S has to be normal ordered. Then E and B only have matrix elements between |photon> and the vacuum |0>. Then one can introduce the vacuum state so that the matrix element become e.g.
<photon|B|0> x<0|E|photon> =-i omega <photon|B |0> x rot <0|B|photon>. I set b=<0|B|photon>. In the second step, I used Ampere-Maxwells law. I get S \propto \nabla |b|^2 -1/2 (b^*\cdot \nabla b+b \cdot \nabla b^*). From what I found, the angular dependence of b is given as \Phi_{lm}=\nabla Y_{lm}=\sqrt{3/8 \pi}\exp(i \phi)(i \hat{\theta}-\cos \theta \hat{\phi})
Putting all together, the radial component of S in deed varies like 1+cos^2(theta). However, there are also non-vanishing angular components.

 

[Khrapko]

the radial component of S in deed varies like 1+cos^2(theta).

Feynman obtained 1+cos^2(theta) simply (see (4.4) in [1]). Besides, Feynman calculated the distribution of spin, cos(theta) (see (4.2) in [1]). Can you obtain the distribution of spin by your method?

there are also non-vanishing angular components.

Maxwell electrodynamics gives the distribution of orbital angular momentum, sin^2(theta) (see (2.9) in [1], or (2.79) in [2]). Can you confirm this result by your method?
[1] Khrapko, R.I. “Spin and moment of momentum are spatially separated” http://khrapkori.wmsite.ru/ftpgetfile.php?id=56&module=files
[2] A. Corney, Atomic and Laser Spectroscopy (1979)

 

[DrDu]

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?

 

[Khrapko]

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]).

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.

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.

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?

You don’t answer.

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/ftpgetfile.php?id=56&module=files
[3] A. Corney, Atomic and Laser Spectroscopy (1979)
[4] Canonical spin tensor is wrong http://khrapkori.wmsite.ru/ftpgetfile.php?id=49&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/ftpgetfile.php?module=files&id=9
[7] Experiment concerning electrodynamics’ nonlocality http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=46

 

[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&gt; -b |l=1, m=0, m_s=1/2&gt;

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&gt; - |l=1, m=0, m_s=1/2&gt; and \psi_2=a_2/b_2|l=1, m=1, m_s=-1/2&gt; - |l=1, m=0, m_s=1/2&gt; 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.