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Number of photon in the electromagnetic wave

  1. Sep 21, 2011 #1
    From EM, the energy of electromagnetic wave in unit volume is [itex] \varepsilon_0 E^2[/itex]. Does that mean the number of photon is [itex]\varepsilon_0 E^2/\hbar\omega [/itex]( [itex]\omega [/itex] is frequency of wave)? In 1-D, [itex]E=E_0cos(\omega t+kx)[/itex], then the number of photon in average is [itex]\frac{1}{(2\pi/\omega)}\int_{0}^{2\pi /\omega}\varepsilon_0 E_0^2cos^2(\omega t+kx)/\hbar\omega dt = \varepsilon_0 E_0^2/\hbar\omega[/itex]. This result looks very weird. Please correct me if something is wrong!
     
  2. jcsd
  3. Sep 22, 2011 #2

    vanhees71

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    Photons are mostly introduced in the most unfortunate way to beginners in quantum physics. The reason is that usually people prefer a historical approach to introduce quantum theory, and the photon has been the first quantum phenomenon ever since Planck discovered the quantum behavior of nature through the quantum nature of thermal electromagnetic radiation. This lead Einstein to his famous 1905 article on a "heuristic point of view" of "light particles", nowadays called photons. One can not emphasize enough that this is an outdated picture of affaires, although it has been a genius invention at the time.

    Nowadays a one-photon state is a well-defined concept in relativistic quantum field theory. It's given by

    [tex]|\psi \rangle=\int \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} \hat{a}^{\dagger}(\vec{p},\lambda) A(\vec{p}) |\Omega \rangle.[/tex]

    Here, [itex]\hat{a}^{\dagger}(\vec{p},\lambda)[/itex] denotes the creation operator for a photon of momentum, [itex]\vec{p}[/itex] and a polarization indicated by [itex]\lambda[/itex] (e.g., the helicity, i.e., left or right circular polarized states). Further [itex]A[/itex] is some square integrable function, and [itex]|\Omega \rangle[/itex] is the vacuum state.

    Such a single-photon state is not so easy to prepare, although nowadays it's common in quantum-optics labs: One uses some crystal to create an entangled pair of photons and measuring one of them to make sure that one is left with a pure one-photon state.

    Often, it's stated, one creates one-photon states by a dimmed laser. This is wrong! Lasers produce something close to a classical em. wave with a small spectral width, i .e., with high spatial and temporatl coherence but never single-photon states (nor any state of determined photon number, or Fock state). Quantum mechanically such a thing is described by a coherent state, which is the superposition of many states of different photon number. For a dimmed laser, where the intensity is such that the energy divided by [itex]\hbar \omega[/itex] is small, e.g., close to 1, the dominating state is in fact the vacuum state superposed by the one-photon state. The photon number in such a state is Poisson distributed with an expectation value (and thus also variance!) of 1. So, what you calculate with [itex]\mathcal{E}/(\hbar \omega)=\epsilon_0 \int \mathrm{d}^3 x E^2/(\hbar \omega)[/itex] is an average photon number for the given (classical approximation of a) coherent state. This is the only physical sense you can make out of this number.

    Of course, a plane wave can never be realized, since this implies and infinite energy. Plane waves are modes, used to provide a Fourier representation of the electromagnetic field, it's not somthing physical!
     
  4. Sep 22, 2011 #3

    clem

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    Your equations are correct. The time average of E^2 just gives a factor of 1/2.
     
  5. Sep 22, 2011 #4

    jtbell

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    Why?
     
  6. Sep 22, 2011 #5

    Dale

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    I just wanted to "second" this excellent description of a coherent state.
     
  7. Sep 22, 2011 #6
    Vanhees, thanks for your fabulous comments.
    clem, I forgot a factor 1/2. Thanks for pointing it out.

     
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