Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Photon emission by an excited hydrogen atom

  1. May 29, 2013 #1
    Typically (in popular literature) the process of photon emission by an excited atom is considered as an instant event. But actually it is quite likely that it is a continuous process. Such processes are usually described by evolutionary differential equations (ODEs or PDEs). Assume that we consider a hydrohen atom composed by 1 proton p and 1 electron e. Let [itex]\Psi(t,x_p,y_p,z_p,x_e,y_e,z_e )[/itex] be the time dependent wave function of the atom. The time dependent Shrödinger equation with the Coulomb potential can be written for this function. How many auxiliary functions should be added for describing the photon which is absent in the beginning of the process and which is present in the end of it? How many auxilioary terms should be added to the time-dependent Shrödinger equation? It would be best to see the complete system of differential equations.

    The creation/annihilation operators could be used in deriving these differential equtions, but they should not enter the ultimate equations. These should be differential equations for functions, not for operators.
  2. jcsd
  3. May 29, 2013 #2


    User Avatar
    Science Advisor

    :confused: No, on the contrary, the "popular" view is closer to the continuous one. We often get posts from people with little understanding of the difference between classical and quantum mechanics, asking overly mechanistic questions like, "How many cycles are there in the photon? How long does it take for it to be emitted?"

    The wavefunction evolves continuously but it describes only a probability. The measurement process causes the emission to appear instantaneous. For example for a long-lived alpha emitter, the alpha particle does not gradually seep out over billions of years, it comes out all at once, but at an unpredictable time.

    To calculate the evolution of the wavefunction, just add to H the electromagnetic interaction term (e/c)p·A, and make the electric dipole approximation.
  4. May 29, 2013 #3
    Dear Bill_K. Your answer is just a hint. Please, be more specific. Please, answer the following questions:

    1. How many wave functions should be written?

    2. How many arguments are in each wave function, e. g. time + proton coordinates + electron coordinates? Or maybe time + proton coordinates + electron coordinates + photon coordinates?

    3. What is the nature of [itex]A[/itex] in your formula? If it is an operator written in terms of creation/annihilation operators, then explain how to apply it to the wave functions.

    4. How many differential equations should be written?
  5. May 29, 2013 #4


    User Avatar
    Science Advisor

    Ruslan, I'm beginning to suspect that you're coming into this cold? If so, in quantum mechanics as with any subject, it's important to start with the basics first, then tackle the more difficult problems after you have a firm grounding.

    The reason I say this is your mention of the proton. If you've learned about the states of the hydrogen atom, you've seen how the discussion starts out. The first thing you do is change to center-of-mass coordinates. Thus the proton is essentially out of the picture - the description of atomic energy levels involves only the electron wavefunction.

    And we always try to simplify the problem through the judicious use of approximations. For example even though we know the electron is properly described by the Dirac equation, we don't use that fact except when it's actually necessary.

    In the present case, atomic transitions are usually discussed in a semi-classical framework. That is, the electron is quantum but the electromagnetic field is not. Also we make the dipole approximation, which means that the wavelength of the electromagnetic wave is large compared to the size of the atom.

    With all this said, there is only one wavefunction involved - that of the electron. We write down the interaction Hamiltonian Hint between a charged particle and a plane wave. We take the matrix element of Hint between the initial and final electron states, square it, and plug it into the Fermi's Golden Rule for time-dependent perturbation theory, to get the transition rate.
  6. May 30, 2013 #5
    Dear Bill_K, I do not like to discuss the simplified statement of the problem known from every textbook. I would like to discuss the conceptual statement from which various simplifications should follow as special cases.

    This is true if we study stationary states of the pair. But studying the photon emission in a strict model we should consider the center-of-mass coordinates for the trio or if we deal with the center-of-mass coordinates of the pair, we should remember that its state of motion changes as soon as the photon is emitted.

    I would like to see the exact model in the form of differential equations, from which the approximate model is derived as some limit.

    I agree with this approximation and agree to omit spin terms from the Hamiltonian. But otherwise the model should be exact in order to clarify the conceptual statement of the problem.

    The power of a theory not only in describing particular phenomena using simplifications specific to them, but in unifying various phenomena into a single concept. Otherwise the theory is weak or incomplete.

    Perturbation theory has its limits, so obtaining non-perturbative model in the case of such a simple object as the hydrogen atom is important for understanding the concepts and limits of the Quantum Mechanics as a whole.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Photon emission by an excited hydrogen atom
  1. Exciting a photon (Replies: 36)