The absorption of a linearly polarized photon.

[Khrapko]

I am not interested in the energy splitting that really can be neglected. I interested in the space distributions of spin, angular momentum, and mass in the radiation of the excited atoms with different j. Does somebody know experiments on the distributions?

 

[hiyok]

@ Dr Du,

I think it is incorrect when you say a photon does not have angular momentum but only spin. Actually, in some cases when the system has spherical symmetry, it is more convenient to have a photon with definite angular momentum (aside from its spin).

 

[DrDu]

Where did I say this?

 

[hiyok]

Sorry, Dr Du, I made a mistake. It is Andy Resnick who said that. Really sorry for this slip !

hiyok

 

[Khrapko]

I think it is incorrect a photon does not have angular momentum but only spin.

This sentence is strange because spin is an angular momentum.

in some cases when the system has spherical symmetry, it is more convenient to have a photon with definite angular momentum (aside from its spin).

This sentence is strange because a spherical symmetric system is described by Y_0^0=Const spherical harmonic, and has no angular momentum.
By the way, I think spherical harmonics do not describe spin

 

[hiyok]

@Khrapko:

(1) You may infer from the context, that I reserve angular momentum only for orbital angular momentum. While spin, although behaving as angular momentum under rotation, I use it to mean internal degrees of freedom;

(2) No. A spherically symmetric system means the communtation between rotation operator and Hamiltonian. It means the Hamiltonian and the angular momentum can be simutaneously measured with 100% certainty. It means the basis states of the system can be written as |n;l,m>, where n labels energy levels while (l,m) labels angular momentum. Y00 corresponds to only (0,0). But the system can actually take other states with non-zero angular momentum.
Spherical symmetry does not mean the wave function must be a constant.

(3)Of course, spherical harmonics can be used to describe the integer spin (but not half integers). For example, for spin 1, you use Y10,Y11 and Y1-1 as the basis states. And in this case, the angular operator can be represented by 3*3 matrice.

hiyok

 

[Khrapko]

A spherically symmetric system means the communtation between rotation operator and Hamiltonian.

This sentence is strange because angular momentum operators commute with Hamiltonian always independently of a system. Symmetry of a system means that an operator exists that does not change the system. A sphere is spherically symmetric, but a cylinder is not spherically symmetric. Only s-states (Y00) of an electron is spherically symmetric (if we disengage from electron spin). An electron with spin has no spherically symmetric states.

While spin, I use it to mean internal degrees of freedom;
spherical harmonics can be used to describe the integer spin

A question arises: if spherical harmonics can be used to describe spin as well as an orbital angular momentum, why do you name namely spin an internal degree of freedom?

 

[hiyok]

@Khrapko:
(1) As regards what is meant by symmetry and the C.R. between angular operators and Hamiltonian, we don't have to debate here. Because, this is a textbook content. You may consult any textbook in quantum mechanics.

(2)That I name it as an internal degree of freedom is largely my personal preference. Fpr example, I also name isospin as an internal degree of freedom, which can actually be described by Pauli Matrice, the same Pauli matrice that are used to describe spin half. Indeed, such matrice are even used to describe any two level system.

(3)I would say, all the above two points are textbook content.

 

[Khrapko]

As regards what is meant by symmetry and the C.R. between angular operators and Hamiltonian, we don't have to debate here.

We will not have the debate if you agree that “a symmetry operation on a system is an operation that does not change the system” (The Pengium Dictionary of Physics) while the angular momentum operators commute with Hamiltonian always independently of a system because of the symmetry (isotropy) of our space rather than of a system.

I also name isospin as an internal degree of freedom, which can actually be described by Pauli Matrice, the same Pauli matrice that are used to describe spin half.

Well, an internal degree of freedom (spin half) is described by Pauli Matrice and must be added to the external degree of freedom, namely, to orbital angular momentum, which is described by spherical harmonics (see #12). The question is: why they do not add the internal degree of freedom (spin one) of a photon to the external degree of freedom of the photon, namely, to orbital angular momentum, which is described by spherical harmonics? I add spin of photon to the orbital angular momentum [1].
[1] 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

 

[hiyok]

(1) I would agree with you if you said 'a symmetry operation is an operation that leaves the Hamiltonian and the dynamical laws of the system invariant, but not its states.'

(2)I never said that the spin of a photon can not be added to its orbital angular momentum. I just emphasized photons posses both spin and orbital angular momentum. Surely, they can be added, since both transform in essentially the same way under spatial rotation. So, we agree here.

 

[Khrapko]

I would agree with you if you said 'a symmetry operation is an operation that leaves the Hamiltonian and the dynamical laws of the system invariant, but not its states.'

I have cited “The Pengium Dictionary of Physics”. Btw, you ignored my note that cylinder has no spherical symmetry, and so p-state of an electron is not spherically symmetric.

I never said that the spin of a photon can not be added to its orbital angular momentum. I just emphasized photons posses both spin and orbital angular momentum. Surely, they can be added, since both transform in essentially the same way under spatial rotation. So, we agree here.

It is excellent! Thank you very much! I struggle with the scientific community during 10 years! I introduced a spin tensor into the electrodynamics in addition to the moment of momentum. Physicists consider electrodynamics spin as a moment of a linear momentum. My papers were rejected more than 500 times (see http://khrapkori.wmsite.ru/). For example [1], Ohanian wrote: “This angular momentum is the spin of the beam” [2]
DrDu wrote (#13): “it becomes impossible to disentangle spin and orbital angular momentum. This is a well known result”
[1] https://www.physicsforums.com/showthread.php?t=418699&highlight=khrapko #13
[2] Ohanian, What is spin? Am. J. Phys. 54 (1986) p. 502

 

[Sir Beaver]

Just to clarify, what hiyok says is that a Hamiltonian which is spherically symmetric will commute with L^2 and Lz. This is not true for any system. Also, in physics, the statements 'Hamiltonian has this symmetry' and 'system has this symmetry' is the same thing. Finally, just because the Hamiltonian has one symmetry, does not mean that the wave function will have this symmetry. For example, the Hamiltonian in a periodic lattice is periodic with R, but not the wave function, which is a Bloch vector.

 

[Khrapko]

It seems to me that

the angular momentum operators commute with Hamiltonian always independently of a system because of the symmetry (isotropy) of our space rather than of a system.

 

[Khrapko]

I never said that the spin of a photon can not be added to its orbital angular momentum. I just emphasized photons posses both spin and orbital angular momentum. Surely, they can be added, since both transform in essentially the same way under spatial rotation.

It is excellent! Thank you very much! I struggle with the scientific community during 10 years! I introduced a spin tensor into the electrodynamics

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

 

[DrDu]

Just a remark: Spin ( in systems where it is strictly defined, like for electrons but not for photons) is, strictly speaking, not angular momentum and it doesn't transform as angular momentum under spatial rotations. E. g. an electron has always spin 1/2 hbar while orbital angular momentum depends on the choice of the origin. Spin is angular momentum in the rest frame of the particle with the particles' position taken as origin.

 

[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
$$|a|^2 |<l=1, m=1|\mathbf{d}|l=0,m=0>|^2+|b|^2 |<l=1,m=0|\mathbf{d}|l=0, m=0>|^2 - 2ab Re (<l=1,m=1|\mathbf{d}|l=0,m=0><l=0,m=0|\mathbf{d}|l=1,m=0>)$$
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, [tex]\vec{s}[\tex], which is the polarization of photons.) is not sufficiently justified for the following reason:

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.

2. Actually, the division of J into s and L is well known[Quantum Field Theory: from operators to path integrals, Kerson Huang, John Wiley & Sons, 1998]. According to this division, s can indeed be related to the polarization of photons and yield results in accord with Feynman'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.

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