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The absorption of a linearly polarized photon.

  1. Jul 1, 2010 #1
    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?
     
    Last edited: Jul 1, 2010
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  3. Jul 21, 2010 #2

    olgranpappy

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    no..
     
  4. Jul 22, 2010 #3

    DrDu

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    No, the emitted photons in linearly polarized light are in a superposition state of the two helicity eigenstates.
     
  5. Jul 22, 2010 #4

    Andy Resnick

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

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

    Photons have spin momentum only, corresponding to [itex]\Delta l = \pm 1 [/itex]. It's possible to have linearly polarized emission, corresponding to [itex]\Delta m= \pm 1 [/itex], but it's an unusual process that typically occurs in the presence of magnetic fields or specially designed materials.
     
    Last edited by a moderator: May 4, 2017
  6. Jul 22, 2010 #5

    DrDu

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    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.
     
  7. Jul 22, 2010 #6

    Cthugha

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    You can also have a conversion from one to the other. See for example
    http://people.na.infn.it/~marrucci/oam/index.htm" [Broken]
    and the references on that page.
     
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  8. Jul 22, 2010 #7
    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
     
  9. Jul 22, 2010 #8

    DrDu

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    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.
     
  10. Jul 22, 2010 #9
    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.
     
  11. Jul 22, 2010 #10
    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.
     
  12. Jul 23, 2010 #11

    DrDu

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    (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.
     
  13. Jul 23, 2010 #12
    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
     

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    Last edited: Jul 23, 2010
  14. Jul 23, 2010 #13

    DrDu

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    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.
     
  15. Jul 23, 2010 #14
    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, [tex]{\bf L}={\bf r}\times({\bf E}\times{\bf B})[/tex]. 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.
     
  16. Jul 23, 2010 #15

    DrDu

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    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.
     
  17. Jul 23, 2010 #16
    (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.
     
  18. Jul 23, 2010 #17

    DrDu

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

    Good night!
     
  19. Jul 24, 2010 #18
    No!
    Your answers are somewhat vague
    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
    The eigenvalues of s_x are plus minus hbar. Does it mean that helicity = s_x? If so, then
     
  20. Jul 26, 2010 #19

    DrDu

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    (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.
     
  21. Jul 26, 2010 #20
    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?
     
  22. Jul 27, 2010 #21

    DrDu

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    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.
     
  23. Jul 27, 2010 #22
    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?
     
    Last edited: Jul 27, 2010
  24. Jul 28, 2010 #23
    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!
     
  25. Jul 28, 2010 #24

    DrDu

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    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.
     
  26. Jul 28, 2010 #25
    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?
     
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