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Spins on electrons versus photons

  1. Feb 11, 2015 #1
    Suppose you measure the spin of an electron with a sensor oriented in the +z direction and find that the spin is up (aligned with the sensor). Now if you immediately measure the spin of the electon with a sensor oriented in the -z direction, you are guaranteed that it will be down (oppositely aligned). It takes a 180-degree rotation of the sensor to guarantee a measurement of opposite spin.

    Now do the same experiment with photons. If the first measurment of photon spin in the +z direction is up, then as I understand it, it will take a 90-degree rotation of the sensor (in the +x direction) to get a guaranteed measurement of down. Is this correct?

    As I understand it (though I could be wrong), photon spin is related to polarization. For a photon traveling in the +y direction, the opposite of polarization in the z direction is polarization in the x direction; hence the 90-degree rotation. Is this related to the fact that electrons are fermions (having spin 1/2), while photons are bosons (having spin 1)?
     
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  3. Feb 11, 2015 #2

    mfb

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    For photons, "up" and "down" are problematic descriptions. Where is the difference between "up" and "down"?
    Rotating the sensor by 90° gives some sort of opposite, yes.
     
  4. Feb 12, 2015 #3
    Thank you for your reply. Looking into the matter further, I find that in many cases, photons are much easier to work with than electrons. The first experimental verification that Bell's Theorem is violated was done on photons, not electrons. Quoting Wikipedia,

    When the polarization of both photons is measured in the same direction, both give the same outcome: perfect correlation. When measured at directions making an angle 45 degrees with one another, the outcomes are completely random (uncorrelated). Measuring at directions at 90 degrees to one another, the two are perfectly anti-correlated. In general, when the polarizers are at an angle θ to one another, the correlation is cos(2θ). So relative to the correlation function for the singlet state of spin half particles, we have a positive rather than a negative cosine function, and angles are halved: the correlation is periodic with period π instead of 2π.

    I know that classically, light can be polarized along planes, but I thought that individual photons had to be circularly polarized, either clockwise or counterclockwise when viewed in their direction of propagation. The up versus down would then refer to their angular momentum vector. Is this correct?

    And I imagine that the difference between the correlation factors: cos(θ) for electrons, cos(2θ) for photons must be directly related to the fact that the former are fermions and the latter are bosons.
     
  5. Feb 12, 2015 #4

    bhobba

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    For how circular polarisation fits into the QM picture check out:
    http://www.if.ufrj.br/~carlos/fismod/seminarios/polarizacaoFoton/LeBellac_QuantumPhysics_3_1.pdf'
    'This result suggests that the Hermitian operator with eigenvectors R and L is associated with the physical property called “circular polarization.” We shall see in Chapter 10 that Jz is the operator representing the physical property called “z component of the photon angular momentum (or spin).”

    Tricky hey - if its circularly polarised its intuitively spinning in the z direction - which is the z component of photon spin - or maybe not that tricky really o0)o0)o0)o0)o0)o0).

    Thanks
    Bill
     
    Last edited: Feb 12, 2015
  6. Feb 13, 2015 #5

    mfb

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    You can have quantum-mechanical superpositions of the two polarizations, which (can) look like linear polarizations.
     
  7. Feb 14, 2015 #6

    vanhees71

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    Photons are massless, and that makes them different from massive quanta. The only way to really understand these issues properly is to go through Wigner's famous analysis on the unitary representations of the proper orthochronous Lorentz group.

    It turns out that massless spin-1 bosons within a local microcausal quantum field theory have to be described by quantized U(1) gauge fields in order to have only a descrete number of spin-like field-degrees of freedom. What comes out of this analysis are precisely photons as can be described also by canonical quantization of the classical Maxwell field in the radiation gauge (I'm talking about free photons here, of course).

    The result is that one way to build the Fock space is to use the momentum single-photon basis. Instead of 3 spin components as in the analogous case of a massive spin-1 boson you find that there are only 2 polarization degrees of freedom. The natural physical choice are those with definite helicity, where helicity is the projection of the total (!) angular momentum of the photon to the direction of its momentum.

    Now you can build single-mode coherent states with definite helicity, and these turn out to describe what's called left or right (helcity -1 or +1 respectively) electromagnetic waves in classical electrodynamics/optics.

    As in classical electrodynamics you can build any other possible photon states by superposition of the Fock states (or in a generalized way also from coherent states). Particularly you can construct any elliptically polarized or also linearly polarized single-photon states. In the latter case two orthogonal states are linear polarizations in two perpendicular directions. Again it's all completely analogous to the classical em.-field case.
     
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