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Photon energy and light quantum hypothesis

  1. Dec 12, 2015 #1
    [Moderator's note: Spun off from https://www.physicsforums.com/threa...city-in-a-medium.843312/page-2#post-5294258.]

    The web page http://plato.stanford.edu/entries/quantum-field-theory/ says:

    "Quantum Field Theory (QFT) is the mathematical and conceptual framework for contemporary elementary particle physics. In a rather informal sense QFT is the extension of quantum mechanics (QM), dealing with particles, over to fields, i.e. systems with an infinite number of degrees of freedom."

    In my understanding from above, quantum mechanics should be included in modern QFT. Quantum mechanics is developed based on Einstein light-quantum hypothesis; for example, Rayleigh scattering, Stokes Raman scattering, and anti-Stokes Raman scattering are all explained based on Einstein light-quantum hypothesis: E = hf. Now I have a question for you: In modern QFT, photon energy is derived or is taken from Einstein light-quantum hypothesis?
     
    Last edited by a moderator: Dec 12, 2015
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  3. Dec 12, 2015 #2

    mfb

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    Standard quantum mechanics is a special case of QFT.
    It is way more general than that.

    The energy/momentum and energy/frequency relations can be calculated from QFT, but the theory was designed to satify those that got observed before...
     
  4. Dec 13, 2015 #3

    bhobba

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    We know a lot more about QM these days. The best modern take on its foundations IMHO is the following:
    http://arxiv.org/pdf/quant-ph/0101012.pdf

    QFT comes from taking a field, breaking it into a large number of blobs, applying QM to those blobs, then taking the blob size to zero. Its exactly analogous to classical field theory.

    Thanks
    Bill
     
  5. Dec 27, 2015 #4
    What I know
    "energy-frequency relation" of a photon in free-space comes from Einstein light-quantum hypothesis, while
    "energy-momentum relation" of a photon in free-space comes from the principle of relativity and Einstein light-quantum hypothesis.
    My question is: How to calculate the "energy-frequency relation" of a photon in free-space only from QFT (without Einstein light-quantum hypothesis used)? Could you please roughly explain the specific process? For example, what postulates or assumptions are needed? Or tell me the locations of specific references. Thanks a lot.
     
  6. Dec 27, 2015 #5

    bhobba

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    Dirac sorted it out. You can find his approach in Chapter 10 of his classic textbook - Principles of QM. A more modern treatment can be found in Chapter 19 of Ballentine

    At the lay level its like the harmonic oscillator:
    https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

    You start with a quantum field which is simply taking a field, breaking it into a large number of blobs, and treating each blob using standard QM. You then let the blob size go to zero. You then do what's called a Fourier transform and find its analogous to the harmonic oscillator with creation and annihilation operators. The creation operator creates a photon which exactly the properties postulated by Einstein.

    Thanks
    Bill
     
    Last edited: Dec 27, 2015
  7. Dec 27, 2015 #6

    vanhees71

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    The equations for the field operators of free electromagnetic field operators in QFT (whose one-particle Fock states define the excitation of this field which we call photon nowadays; it has not too much to do with Einstein's original "heuristic idea") are the same as for the free classical electromagnetic field. Since it is a massless field the field operators in radiation gauge fulfill the wave equation as their equation of motion
    $$\Box A^{\mu}=\left (\frac{1}{c^2} \partial_t^2-\vec{\nabla}^2 \right) A^{\mu}=0$$
    and the gauge-fixing conditions
    $$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
    The decomposition into eigenmodes (together with the information that the electromagnetic field carries no charge-quantum numbers, i.e., it describes strictly neutral quanta (photons)):
    $$\vec{A}(t,\vec{x})=\sum_{\lambda=\pm 1} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{\sqrt{(2 \pi)^3 2 \omega_{\vec{k}}}} \left [\vec{\epsilon}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda) \exp[-\mathrm{i} \omega_{\vec{k}} t+\mathrm{i} \vec{k} \cdot \vec{x}]+\text{h.c.} \right ].$$
    The dispersion relation reads
    $$\omega_{\vec{k}}=c |\vec{k}|$$
    and ##\vec{e}(\vec{k},\lambda)## are two polarization vectors that are perpendicular to ##\vec{k}##. If ##\vec{k}=k \vec{e}_z## you can choose the helicity eigenstates (circular polarized states) ##\vec{\epsilon}(k \vec{e}_z,\lambda=\pm 1)=(\vec{e}_x \pm \mathrm{i} \vec{e}_y)/\sqrt{2}##.

    The dispersion relation tells you that the energy-momentum relation is as if photons were massless particles
    $$E=\hbar \omega_{\vec{k}}=c \hbar |\vec{k}|=c|\vec{p}|.$$
    Note that this is the only thing which photons have in common with point-like particles. The intuitive idea of photons being massless particles, is however almost always misleading!
     
  8. Dec 27, 2015 #7
    I think this treatment corresponds to harmonic oscillators in quantum mechanics. In other words, Einstein light-quantum hypothesis and de Broglie equation have already been considered in the building of quantum theory; of course, the results from the theory must be consistent with Einstein light-quantum hypothesis and de Broglie equation. Thus I don't think it can be said that E=hf is derived from QFT; otherwise, it is a logic circulation.
     
  9. Dec 27, 2015 #8

    bhobba

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    It is actually.

    But to see why you need to go into the detail of a modern treatment of QM such as Ballentine. From such you see that Planks constant is simply a conversion of units rather than what Einstein postulated which is derived in QFT.

    Thanks
    Bill
     
  10. Dec 27, 2015 #9

    vanhees71

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    Again, it should be stressed, that Einsteins quantum hypothesis was just heuristics, and indeed Einstein titled one of his famous papers of his "annus mirabilis 1905" (although I think that Einsteins true "annus mirabilis" was 1915, where he finished his General Theory of Relativity) "Über einen die Erzeugung und Verwandlung des Lichts betreffenden heuristischen Gesichtspunkt" (which I'd freely translate as "About an heuristic aspect concerning the production and transformation of light"). It is together with the Bohr-Sommerfeld model of atoms and de Broglie's first version of wave-particle dualism, so to say, a precursor theory which is outdated for more than 90 years now by modern quantum theory. Particularly photons have very little in common with a classical-particle picture, and as any quantum in the relativistic realm, cannot be understood in terms of Schrödinger single-particle wave functions as can be to quite an extent for non-relativistic particles in terms of Schrödinger wave mechanics.

    The reason is that in the relativistic realm interactions between quanta alway can lead to the production and destruction of quanta, and there are no conserved positive definite particle numbers. What's conserved are similar quantum numbers like electric charge (which allows to create particle-antiparticle pairs without changing the total charge) or net-baryon number (which allows to create baryon-antibaryon pairs without changing the total baryon number) etc. That's why you need a formalism that is able to describe the production and destruction of particles when it comes to relativistic collision energies (i.e., energies that are larger than the threshold of particle production, and since photons are massless there's no such threshold at all, i.e., the acceleration of any charged particles is always accompanied by the production of photons). Such a formalism now is quantum field theory, and free fields can be decomposed in energy-momentum eigenmodes, and since there's no interaction these modes are decoupled, i.e., you can treat them as an infinite set of uncoupled harmonic oscillators. Now the harmonic oscillator can be solved in quantum mechanics by introducing the annihilation and creation operators of energy eigenmodes, and these are in the QFT precisely the annihilation and creation operators of field quanta. These annihilation and creation operators have as a very convenient side effect the property of leading to the "occupation-number" or Fock basis which automatically take into account the bosonic or fermionic nature of the quanta, i.e., it leads to a basis of properly symmetrized (bosons) or anti-symmetrized (fermions) many-body states.

    The Einstein-de-Broglie rule for free fields, i.e., associating ##\hbar \omega## with the energy and ##\hbar \vec{k}## with the momentum of a single quantum of this field, survives the quantization of the field theory, because for free fields the equations of motion are linear in the fields, and thus the field operators in the non-interacting case obey the same equations of motion as the classical theory. Thus each frequency eigenmode of the quantized field has the same frequency-wavenumber dispersion relation as the classical field and thus, since the Einstein-de-Broglie rule of old quantum mechanics survives the modern way of QT, the energy-momentum relation of single non-interacting particles, i.e., ##E^2 c^2-\vec{p}^2=m^2 c^2## for a quantum with (invariant) mass ##m##. For a photon you have ##m_{\gamma}=0## in the Standard Model of elementary particles, and the empirical upper limit of the photon mass is indeed very small ##m_{\gamma}<10^{-18} \,\text{eV}##.

    For more about the Standard Model, see the marvelous popular-science page by the Particle Data Group:

    http://www.particleadventure.org/
     
  11. Dec 27, 2015 #10
    Since "QFT is the extension of quantum mechanics (QM), dealing with particles, over to fields", namely a particle is treated as a field when describing interactions, is the process of annihilation and creation of particles consistent with the relativity? For example, electron-positron annihilation leads to the creation of two gamma photons. The fields of electron and positron and the fields of photons are different, and the field distributions are, I think, extended infinitively, and the changes of fields must be superluminal to finish the process of annihilation and creation.
     
  12. Dec 27, 2015 #11

    bhobba

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    100% for sure.

    Its cooked up to be like that.

    Thanks
    Bill
     
  13. Dec 29, 2015 #12

    Nugatory

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    Yes, and "consistent with" is too weak a statement. Particles, antiparticles, and annihilation are what you get when you modify classical quantum mechanics to be consistent with relativity, so not only are these phenomena consistent with relativity, they are required by it. Dirac gets the credit for this discovery (he predicted the existence of the positron) although the modern formulation of QFT has moved beyond his original formulation.
     
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