The absorption of a linearly polarized photon.

[Khrapko]

Linearly polarized light does not carry angular momentum. However, individual photons emitted, say, in the direction of the X-axis, carries a spin angular momentum \hbar in the direction + X or -X. And, when absorbing of the photon, target gets its energy h\nu and its angular momentum \hbar. Is it right?

 

[olgranpappy]

Linearly polarized light does not carry angular momentum. However, individual photons emitted, say, in the direction of the X-axis, carries a spin angular momentum \hbar in the direction + X or -X. And, when absorbing of the photon, target gets its energy h\nu and its angular momentum \hbar. Is it right?

no..

 

[DrDu]

No, the emitted photons in linearly polarized light are in a superposition state of the two helicity eigenstates.

 

[Andy Resnick]

Linearly polarized light does not carry angular momentum. However, individual photons emitted, say, in the direction of the X-axis, carries a spin angular momentum \hbar in the direction + X or -X. And, when absorbing of the photon, target gets its energy h\nu and its angular momentum \hbar. Is it right?

Light can have both spin as well as orbital angular momentum:

http://www.physics.gla.ac.uk/Optics/play/photonOAM/

Photons have spin momentum only, corresponding to $\Delta l = \pm 1$. It's possible to have linearly polarized emission, corresponding to $\Delta m= \pm 1$, but it's an unusual process that typically occurs in the presence of magnetic fields or specially designed materials.

 

[DrDu]

Interesting link, Andy. But it says clearly that also single photons can have an angular orbital momentum.
It is also possible to expand an electromagnetic field into vector spherical harmonics which are angular momentum eigenstates. These can have all values of total angular momentum j starting from 1. This minimal value is what is meant when speaking of the "spin" of the photon. In the narrow sense of the word, the photon doesn't have a spin, as the spin is defined as the angular momentum in the rest frame of the particle, which does not exist for light.
What is l and m ?

I don't think that one can say that the usual emission is circularly polarized. The light emitted from most lamps has no polarization at all, so it can be either viewed as a mixture of circularly polarized photons or linearly polarized photons. Most fluorescent molecules are elongated in one direction and the transition dipole moment also oscillates in that direction which results in the emission of linearly polarized photons. For preferencial circular polarization, one needs either chiral molecules or magnetic fields.

 

[Cthugha]

Light can have both spin as well as orbital angular momentum:

You can also have a conversion from one to the other. See for example
http://people.na.infn.it/~marrucci/oam/index.htm"
and the references on that page.

 

[Khrapko]

Dear DrDu, it is impossible that a photon has no spin (no matter what the state of the photon).
Please consider an analogy: an isotropic emission of particles. The state is a superposition of plane wave in all direction. But an individual particle has concrete momentum.
At the same time you are right concerning vector spherical harmonics which are orbital angular momentum eigenstates. But besides this orbital angular momentum, a dipole radiation can contain spin. http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=7

 

[DrDu]

Dear Krapko, I just wanted to point out that speaking of the spin of a photon is strictly speaking, an abuse of nomenclature. A Photon has helicity.
If I remember correctly, the vector spherical harmonics contain both the spin and the orbital angular momentum.

 

[Khrapko]

DrDu,
(i) Would you please let me know the difference between spin of photon, which it does not have, in your opinion, and helicty, which it has?
(ii) It is strange that the vector spherical harmonics are the same for a photon with its spin (or helisty) 1, for an electron with its spin 1/2, and for a spinless particle, if, as you remember, the vector spherical harmonics contain both the spin and the orbital angular momentum.

 

[Khrapko]

no..

A photon has no spin, i.e. has spin zero, only if the photon's state is an eigenstate corresponding to the zero eigenvalue. But a linear polarization is not an eigenstate.

 

[DrDu]

DrDu,
(i) Would you please let me know the difference between spin of photon, which it does not have, in your opinion, and helicty, which it has?
(ii) It is strange that the vector spherical harmonics are the same for a photon with its spin (or helisty) 1, for an electron with its spin 1/2, and for a spinless particle, if, as you remember, the vector spherical harmonics contain both the spin and the orbital angular momentum.

(i)To answer your question I have to delve into group theory. According to Wigner, particles are characterized as representations of the Poincare group. The representations of massless particles like photons and massive particles like electrons differ in that the wavevector k is timelike for massive particles, i.e. it a certain frame of reference (the rest frame in that case) it has the form (mc,0,0,0)^T, while for a massless particle it can at best be brought to the form const.*(1,0,0,-1)^T. The sub-group of the Poincare group which leaves the wave-vector invariant is known as the "little group" of the wavevector. In case of massive particles it is SO(3), the rotation group which leads directly to spin, while in the case of massless particles it is E(2), the group of all translations and rotations in a plane. It has a sub-group U(1), which leads to the classification according to helicity.
(ii)Why do you think that the vector spherical harmonics, which describe e.g. particles with spin 1 also describe particles with spin 1/2 or 0?
For spin 0 we use the simpler (scalar) spherical harmonics, which are prominent as solutions of the angular dependence of the hydrogen wave functions.
For spin 1, there are three different vector spherical harmonics. For light there are only two
allowed due to transversality restriction. Due to this restriction it is no longer possible to separate the angular momentum into a sum of spin and orbital momentum.

 

[Khrapko]

DrDu,
(i) Well, what is a numerical value of photon's helicity?
(ii) Sorry, "the simpler (scalar) spherical harmonics, which are prominent as solutions of the angular dependence of the hydrogen wave functions", is, in reality, the electron's wave function rather than hydrogen's one; and an electron has spin 1/2. So, we are forced to introduce electron's spin besides orbital AM.
Next, consider the decay $\Lambda\to N+\pi$:
$\psi=A_s\chi_+Y_0^0+A_p(\chi_+Y_1^0/\sqrt{3}-\chi_-Y_1^1/\sqrt{2/3}$ where $\chi$ is the Pauli spinor. So, we are forced to introduce $\pi$'s spin besides the orbital AM.
I think, we must introduce photon's spin besides orbital AM. This is even more the case that the radiation patterns of orbital AM and spin are orthogonal (see Difference Between Spin and Orbital Angular Momentum http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=7)
The belief is naive that spin will be automatically introduced by the Jackson's procedure 9.6 "Spherical Wave Solution of the Scalar Wave Equation"

Sorry, I cannot write formulae but I can send attachments

 

[DrDu]

Yes, in case of an electron we have to take spin into account. However, we can consider wavefunctions which are a product of a spin part and the orbital angular momentum part so it is possible to consider spin and OAM separately.
The same holds true for a spin 1 particle. But for a photon, only those solutions are allowed where the field is transversal. These states are superpositions of different product states of spin and orbital angular momentum in general, so it becomes impossible to disentangle spin and orbital angular momentum. This is a well known result:
V. B. Berestetskii, E. M. Lifgarbagez, and L. P. Pitaevskii, Quantum Electrodynamics (Pergamon Press, Oxford, 1982).


Btw. to write latex formulas, you have to begin the formulas with "square bracket" "tex" "square bracket" and end with "square bracket" "/tex" "square bracket" instead of $ signs.

 

[Khrapko]

I know very well the well-known statement about the impossibility. I know you cannot disentangle spin and orbital angular momentum. But why are you sure that you have spin in the frame of Maxwell electrodynamics (quantum electrodynamics brings nothing into Maxwell electrodynamics). Naturally, spin cannot be separated from an angular momentum if the angular momentum does not contain spin. Maxwell angular momentum is moment of momentum, $${\bf L}={\bf r}\times({\bf E}\times{\bf B})$$. But let us defer this problem.
What about the absorption of a linearly polarized photon? I repeat: A photon has no spin, i.e. has spin zero, only if the photon's state is an eigenstate corresponding to the zero eigenvalue. But a linear polarization is not an eigenstate. So, an individual linearly polarized photon does have spin or helicty.

 

[DrDu]

This is only true insofar as S^2 or the square of helicity is different from zero. However,
a linear polarized photon does certainly not have either positive or negative helicity as long as I am not measuring it. That's the same situation as in a double slit experiment where claiming that the particle has passed through only either of the slits leads to contradictions.

 

[Khrapko]

(i) Sorry, I ask a simple thing: does target get angular momentum \hbar when it absorbs one linearly polarized photon, or not? Yes or no? Can you predict?
(ii) You did not answer, what is a numerical value of photon's helicity? So, your "the square of helicity" has no sense.

 

[DrDu]

(i) no
(ii) The eigenvalues of helicity are plus minus hbar

Good night!

 

[Khrapko]

Good night!

No!
Your answers are somewhat vague

(i) Sorry, I ask a simple thing: does target get angular momentum \hbar when it absorbs one linearly polarized photon, or not? Yes or no? Can you predict?

(i) no

What “no”? You cannot predict? Or you can predict zero?
If you predict s_x=0 for our photon which is emitted in the direction of the axis X, then you do not take into account my reason that a photon has s_x=0 only if the photon's state is an eigenstate with s_x=0 eigenvalue. But a linear polarization is not an eigenstate of a photon.
Your account of the double slit experiment is needless. The target measures s_x of the photon.
Next

(ii) The eigenvalues of helicity are plus minus hbar

The eigenvalues of s_x are plus minus hbar. Does it mean that helicity = s_x? If so, then

your "the square of helicity" has no sense.

 

[DrDu]

(i) Sorry, I ask a simple thing: does target get angular momentum \hbar when it absorbs one linearly polarized photon, or not? Yes or no? Can you predict?
(ii) You did not answer, what is a numerical value of photon's helicity? So, your "the square of helicity" has no sense.

(i) When giving the analogy of the double slit experiment I wanted to make clear that in quantum mechanics, most questions cannot be answered as yes or no. Their answer is really undefined. (As you will jump on me anyhow, I should add that this statement only holds in standard interpretations of QM but not e.g. in nonlocal hidden variables theories which have some other strange consequences.)

Obviously I can use a beam splitter which sends photons of positive helicity upwards and those of negative helicity downwards. Then I will find in the statistical mean 50% photons of each helicity. However this destroys the information about linear polarization. If a molecule emmits a linearly polarized photon and another one absorbs it, I have no information about helicity at all.

(ii) s_x is the spin of the photon with respect to a space fixed axis and does not exist.
Helicity is the projection of angular momentum on the direction of momentum of the photon.
So spin and helicity have the same unit.

 

[Khrapko]

Dear DrDu, I need not your splitter. We have a linearly polarized photon, i.e. we have a photon, which passed through a linear polarizator. A target absorbs this photon. The absorption means that the target gets energy of h\nu, momentum of h\nu/c, and mass of h\nu/c^2. I ask: does the target get angular momentum of +\hbar, or of -\hbar, or zero? You cannot answer this question. So, you admit that the target maybe get zero angular momentum. Yes? Or you can say nothing?

 

[DrDu]

If the target was in an eigenstate of angular momentum before absorption, it will generally end up in a superposition of angular momentum states. That is, its angular momentum has not a sharp value. Nevertheless, it would be wrong to say that its angular momentum hasn't changed.

 

[Khrapko]

The target is not a QM object. The target is a macroscopic object. The absorption is a measurement of spin AM of the photon (in particular). The wavefunction collapse occures when absorbing. I ask, what change of AM of the target occures when absorbing: +\hbar, or -\hbar, or zero?

 

[Khrapko]

Obviously I can use a beam splitter which sends photons of positive helicity upwards and those of negative helicity downwards. Then I will find in the statistical mean 50% photons of each helicity. However this destroys the information about linear polarization.

Well, now replace the splitter with the target. You have to agree that 50% photons will give +\hbar and 50% photons will give -\hbar to the target!

 

[DrDu]

Yes, but only if you really are measuring the angular momentum. E.g. in case of a hydrogen atom, you may measure whether it got excited from 1s to 2p_+ or 2p_-. Then you will find at random one time the one and one time the other and you get an information about angular momentum. But in another experiment you may ask whether it gets excited to p_x or p_y. Depending on the linear polarization, you may find e.g. allways p_x. But then you don't have any information about angular momentum.

 

[Khrapko]

Yes, but only if you really are measuring the angular momentum.

Sorry, I cannot understand your thought. So I will ask in a different way.
(i) Let our macroscopic non-QM target absorbs a circularly polarized photon with p_x=h\nu/c. Do you expect a change of \hbar in J_x of the target? If yes, why? Are you really measuring the angular momentum J_x of the target? (p is momentum, J is angular momentum).
(ii) Let our macroscopic non-QM target absorbs a linearly polarized photon with p_x=h\nu/c. Do you expect a change of \hbar in J_x of the target?

 

[Khrapko]

I do not like alluding to authorities, but now I am forced because my reasons are not apprehended. So I cite Feynman “The Lectures” 17-11:
“Macroscopic measurements made on a beam of linearly polarized light show that it carries zero angular momentum, because in a large number of photons there are nearly equal numbers of RHC and LHC photons contributing opposite amount of angular momentum - the average angular momentum is zero.”
This quotation means that each photon of linearly polarized light contributes an amount of angular momentum, i.e +\hbar or -\hbar, but not zero.

 

[Khrapko]

If Feynman and I are right, i.e. an individual linearly polarized photon moving in x-direction has spin angular momentum s_x=+\hbar or s_x=-\hbar, then an important question arises:
How can the angular momentum conservation law be satisfied? https://www.physicsforums.com/showthread.php?t=413566
Really,
R.Feynman (the Lectures, Quantum Mechanics, Ch. 18) considers an excited atom sitting with its angular momentum along +z-axis, i.e. s_z=+\hbar in the initial state. Then the atom emits a photon in x-direction. So, in the finish state, we have a system consists of our atom without angular momentum and a photon radiated in x-axis direction with s_x=+\hbar or s_x=-\hbar. It troubles me.

 

[DrDu]

If the atom is in eigenstates of s_z before and after emission, then s_x is completely undefined. Hence no problem with conservation.

 

[Khrapko]

s_x is completely defined up to sign because p_x is completely defined. Please be attentive,

the atom emits a photon in x-direction.

photon radiated in x-axis direction with s_x=+\hbar or s_x=-\hbar.

 

[Khrapko]

Dear colleagues. I submitted this question, "Is angular momentum conserved when an atom emits an individual photon?", to Am. J. Phys. on 26 Nov 2001. Jan Tobochnik rejected this submission. It seems that his reasoning is interesting, but I do not understand it completely:
“I think your question is wrong. An atom cannot emit a photon with a particular component of spin. It simply emits a photon with spin 1 and then a detector can be lined up to measure the value of the component of the spin along a certain direction. QM tells us that the spin is never perfectly alligned in any direction. There is always an uncertain amount in some other direction. If one measures the spin of the photon and that of the atom, quantum mechanics will always conspire to conserve angular momentum.”

 

[DrDu]

I did not doubt that the spin s_x of the photon is well defined. I said that the spin s_x of an atom in an eigenstate of s_z is undefined (as long as s^2 isn't 0).

 

[Khrapko]

s_z is defined (of \hbar) in the initial state. s_x is defined (of \hbar) after emitting. How can the angular momentum conservation law be satisfied: that is the question

 

[Khrapko]

Dear colleagues, I have a hypothesis.
We can save the angular momentum conservation law if take into account spin of the atom’s electron. Feynman ignores electron’s spin, but we must take it into account. Then the initial orbital angular momentum of the excited atom, l_z=\hbar, is emitted as an orbital angular momentum of the photon, and spin of the photon, s_x=\pm\hbar, is provided by an overturn of electron’s spin. A corollary: after emitting of a photon with defined p_x, the atom will have j_x=\pm\hbar/2.
Analogically, when an electric dipole rotating in x-y-plane emits a photon in z-direction, one electron of the dipole is being overturned. So, in other words, a rotating dipole is being magnetized in the transverse direction [1].
[1] R.I. Khrapko A rotating electric dipole radiates spin and orbital angular momentum http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=8

 

[Khrapko]

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?

 

[DrDu]

In general yes, but in light atoms, spin-orbit coupling is small so that the splitting of the term components with different j can be neglected.