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Photon spin - where is the missing state?

  1. Jan 10, 2004 #1
    Photons have spin 1, yet only show two possible spin values along the direction of travel. For a spin 1 particle there should be three possible values, along the direction of motion, away from it and zero. What happens in the wave equation solutions to stop the zero spin state from coming into existence???
  2. jcsd
  3. Jan 10, 2004 #2
    there are a couple of ways to see what happens to the other degrees of freedom of the photon. they all boil down to the fact that the photon is massless.

    for one thing, the classical electromagnetic wave must be transverse, by Maxwells equations. thus, the classical wave has only 2 degrees of freedom, so the corresponding quantum particle better also have 2 degrees of freedom, or else we are in trouble.

    that number 2j+1 is the dimension of the rotation group for a spin-j particle. it counts how many degrees of freedom a spin-j particle has in its rest frame due to spin. however, the photon, being massless, always travels at the speed of light, and therefore doesn t have a rest frame.

    to count the degrees of freedom of a particle, we start with the Poincaré group, and look for its representations. for a massive particle, we can find those by going to the rest frame, and we find 2j+1 degrees of freedom, since in the rest frame, the Poincaré group reduces to the rotation group (the rotation group SO(D-1) is the little group of the Poincaré group ISO(D-1,1) for a massive particle)

    but for a massless particle, we cannot go to a rest frame. instead, we go to a frame where the momentum of the photon is (1,1,0,0,......,0), and here, the symmetry group is SO(D-2). this has one fewer degrees of freedom for a spin-1 particle.

    another way to see it is to look at how we quantize the photon. we start with a vector field, which actually has 4 degrees of freedom. but gauge invariance forces 2 of these degrees of freedom to be unphysical. gauge invariance again relies on the photon being massless.

    in short: the photon loses that degree of freedom because it is massless.
  4. Jan 10, 2004 #3
    Lethe, that's a great explanation - many thanks. I had suspected it must all be down to the zero mass of the photon. Is there any similar impact on spin states for the spin 1/2 neutrino?

    I like the argument around the Poincare group. Is there a spin-1 version of the Dirac spin 1/2 equation? I would really like to work through the algebra and see all this fall out, like the spin states of the electron do for spin 1/2. I would much prefer to see it done that way - the argument around the photon transversality in the classical Maxwell equations is all well and good, but I don't understand how those translate directly via the Correspondence Principle from a vector field to a wave function.......
  5. Jan 11, 2004 #4
    No need for any answer to these questions - found a really great set of lectures on the representations of the poincare group and the resulting wave equations at:

    Thanks for the hints that got me to go an look for this.
    Last edited by a moderator: Apr 20, 2017
  6. Jan 11, 2004 #5


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    And thank YOU for the nifty link. I like to keep sites like this bookmarked - all the basic math on one topic boiled down for handy reference.
  7. Jan 11, 2004 #6
    The pleasure's mine. The URL is interesting - the lectures there are from the 1950's, and are part of a large collection of stuff by Leite Lopes, including some jointly authored with Richard Feynman.
  8. Jan 13, 2004 #7


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    Re: Re: Photon spin - where is the missing state?

    I think I know what you meant by this, but just to be safe, what did you mean?
  9. Jan 14, 2004 #8
    Lethe's argument on gauge invariance got me interested also. Basically, I think the argument here is two fold, firstly gauge invariance of the EM field and the zero-mass of the photon go hand in hand (otherwise the field equations are just not gauge invariant.) Secondly, if the EM field is gauge invariant, then you can show that it is possible to choose a gauge in which not just one but TWO of the four degrees of freedom vanish. This is followed through in detail here:
    http://courses.washington.edu/phys55x/Physics%20557_lec13.htm [Broken]
    Last edited by a moderator: May 1, 2017
  10. Jan 14, 2004 #9


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    Your original question was how can photons be spin-1 but have only two polarizations (The term "helicity" rather than "spin" is used in the case of massless fields. In 4-dimensional spacetime helicity-σ fields have helicity states ±σ). I queried lethe because I think what he posted may create confusion about what role the U(1) symmetry of QED plays in explaining why photons are transversally polarized, which is none at all: Because the electromagnetic potential doesn't transform under lorentz transformations as a true lorentz vector, the U(1)-invariance is required to consistently couple it to charged currents since these are true lorentz vectors, but isn't required to explain the transverse polarization, this following from the representation theory of the lorentz group for massless fields.
    Last edited by a moderator: May 1, 2017
  11. Jan 16, 2004 #10
    Re: Re: Re: Photon spin - where is the missing state?

    basically i was thinking of Gupta-Bleuler quantization. there are D degrees of freedom, one for each component of the vector potential, but the states in the Hilbert space created by some of these fields are orthogonal to every physically observable quantity by the physical state condition, which is just the requirement of gauge invariance.
  12. Jan 16, 2004 #11
    this is news to me

    this is also very surprising to me. can you explain further? how exactly does the vector potential transform under change of coordinates, if not as a Lorentz vector?

    certainly we can explain the the polarization of the photon in terms of its massless representation of the Poincaré group. in fact, i think i outlined that argument above. but i do not agree that this has nothing to do with gauge invariance. i believe that any massless representation must necessarily have some gauge degree of freedom. for example, consider the massless spin-2 particle: you discover that it must have the same gauge symmetry that you find in GR, hence the statement that the existence of a massless spin-2 particle in a theory implies that that theory contains GR (a statement that is often heard in string theory)

    but i admit that i don t understand this as well as i would like, so if you can set me straight, i would appreciate it.
  13. Jan 16, 2004 #12
    i think that the same arguments do not apply to the neutrino, but right now i am not sure why not.

    i will think about it (or maybe someone else will tell us)

    yes of course. the massive version of the spin-1 equation is called the Proca equation, and the massless version is called Maxwell's equations (done properly, it is really only one equation though)
    ok, maybe i will post some calculations.
  14. Jan 19, 2004 #13


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    One common technique used in constructing the representations of the lorentz group begins with the canonical re-expression of the standard commutation relations among it's generators in terms of two decoupled spin 3-vectors Ai and Bi satisfying the commutation relations

    [Ai, Aj] = iεijkAk
    [Bi, Bj] = iεijkBk
    [Ai, Bj] = 0.

    We find matrices satisfying these in the same way that we find matrices representing the spins of a pair of uncoupled particles - as a direct sum. That is, we label the rows and columns of these matrices with a pair of integers and/or half-integers a, b, running over the values

    a = -A, -A+1, ⋅⋅⋅, A
    b = - B, -B+1, ⋅⋅⋅, B

    and take

    (A)a'b',ab = δbb' (J(A))a'a
    (B)a'b',ab = δaa' (J(B))b'b

    where J(A) and J(B) are just the standard matrices for spins A and B. The reps are then labelled by (A, B) and are (2A+1)(2B+1)-dimensional. The generators of the rotation group may then be represented by the hermitian matrices

    J = A + B.

    The usual rules of vector addition show that (A, B)-fields have components that rotate like spin-j objects with

    j = A+B, A+B-1, ⋅⋅⋅, |A-B|.

    For example, (½, ½)-fields have j = 1 and j = 0 components corresponding to the spatial and temporal parts of a 4-vector. It turns out that massless (A. B)-fields can be formed only from the annihilation and creation operators for massless particles of helicity ±σ in which

    σ = B - A.

    Since massless lorentz 4-vectors transform in the (½, ½), they can only describe helicity zero. However, the electromagnetic potential aμ(x) is a massless field of helicity ±1 and thus can't be a true lorentz vector, despite the lorentz index. In fact it transforms under general lorentz transformations Λ as

    U(Λ)aμ(x)U-1(Λ) = Λνμaν(Λx) + ∂μΩ(x,Λ)

    where U(Λ) is a unitary rep of the lorentz group and Ω(x,Λ) is a linear combination of annihilation and creation operators.

    As I mentioned, we can use fields like aμ(x) as ingredients in lorentz-invariant physical theories if the couplings of aμ(x) are not only formally lorentz-invariant (that is, invariant under formal lorentz transformations under which aμ(x) → Λμν aν), but also invariant under the "gauge" transformations aμ(x) → aμ(x) + ∂μΩ. This is achieved by taking the couplings of aμ to be of the form aμjμ, where jμ is a conserved 4-vector current, i.e. it satisfies ∂μjμ = 0.
    Last edited: Jan 19, 2004
  15. Mar 19, 2009 #14
    This is generally true, but strictly it should be "the photon loses that degree of freedom because it is massless and we impose Lorentz invariance".

    Phonons are also massless bosonic (quasi)particles, but they have three spin degrees of freedom.
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