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I Photon number state and Doppler shift

  1. Mar 31, 2017 #1
    If a mode of light is in the single photon state in the reference frame of the emitter, what will the state look like in a reference frame where the wavelength is, say, 5% less or more? How about a state with, say, 5 photons?

    I saw some online discussions and some papers on arXiv (not necessarily peer reviewed) where it was suggested that the photon number is invariant. Unfortunately it was a while ago and I don't remember details. But It seems to me that you can conserve either photon number or energy but not both when you change to a different frame.

    What is a good reference that gives the correct way to find what |N> looks like in any inertial frame?
     
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  3. Mar 31, 2017 #2

    phyzguy

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    Photon number is frame invariant, but energy is not. So if you have a state with N photons and view it from a different frame, it will still have N photons, but the photon energy will be different. Energy is one component of the energy-momentum 4-vector, so it is not frame invariant. The "length" of the energy momentum 4-vector, [itex]p^\mu p_\mu[/itex], which is zero for a photon, is frame invariant.
     
  4. Apr 1, 2017 #3
    Thank you.

    So then, to go from frame A to frame B, what we do is:

    1. Take the energy-momentum four vector ##(p_A, E_A)##
    2. Apply a transformation to get ##(p_B, E_B)##
    3. Take ##N_B = N_A = N##
    4. Calculate the doppler shifted frequency ##\omega_B = f(\omega_A,v)## using the relativistic Doppler shift equation (which I don't remember)
    5. Our state is now ##|N_{\omega_B}>##

    Another quick question -- and this may be a bit silly, but -- if our original state was a superposition of two ##\omega##'s with their corresponding N's, do we follow the above procedure for each ##\omega## independently and then add them back to get the new superpostion in frame B? And I suppose the coefficients (amplitudes) of the original components would carry over to the new frame?
     
    Last edited: Apr 1, 2017
  5. Apr 1, 2017 #4
    Silly, silly me! Steps 1 and 2 are not needed in the procedure, they only explain why it is so.

    But I'd still like to know about the other thing -- superpositions.
     
  6. Apr 1, 2017 #5

    vanhees71

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    Interesting, but not so easy to understand for me! Can you give a Lorentz-invariant photon-number operator?

    The point is that in relativity, if you have the current of some charge (or number) ##j^{\mu}##, then the corresponding charge
    $$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} j^{\mu}$$
    is, in general, not a Lorentz scalar. That's the case if and only if this charge is conserved, i.e., if ##j^{\mu}## is a Noether current and obeys the continuity equation
    $$\partial_{\mu} j^{\mu}=0.$$

    For photons, there is no charge-like (or number-like) conserved quantity, and indeed it's almost impossible to conserve photon number. It's enough to accelerate any electric charge to change the (average) photon number of the state.

    For free photons, however, the (gauge-invariant) photon-energy-momentum tensor ##T_{\text{em}}^{\mu \nu}## obeys the continuity equation,
    $$\partial_{\mu} T_{\text{em}}^{\mu \nu} =0,$$
    and thus for free photons
    $$P_{\text{em}}^{\mu}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} T_{\text{em}}^{0 \mu}$$
    is indeed Lorentz vector as it should be.

    At presence of charges, however, i.e., when the photons are interacting, that's no longer the case, and only the total energy-momentum tensor ##T_{\text{tot}}^{\mu \nu}=T_{\text{em}}^{\mu \nu}+T_{\text{chrg}}^{\mu \nu}## leads to a energy-momentum four-vector, the total energy and momentum of the system consisting of charges and the em. field (aka photons).
     
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