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Quantum Hamiltonian Question

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  1. Sep 16, 2014 #1
    So, I was reading about the exchange interaction, and trying to work out what it referred to, and came across something strange in the treatment of the hydrogen molecule (I think it was on wikipedia):

    The hamiltonian given for the system included a term e2/(4πε0 * Rab) for the repulsion between nuclei a and b, even though the wave equation was only for the motion of the electrons, and the nuclei are treated as fixed.

    In classical physics, you can add a constant to the hamiltonian, and it doesn't change the physical picture of what is being described. It doesn't matter where zero reference energy is, in terms of the forces on the system.

    In wave-mechanics, however, adding a constant to the hamiltonian does shift the frequency of the wavefunction, if not the spatial form of the stationary states. This leads me to wonder about the equivalence of the physical picture for non-stationary states. (time varying states, transient states when the electron cloud is in transition between two energy levels, behavior of free particles, etc.)

    (Thinking back to optics:) If you have something like a beam focusing down to a focus point, then diverging - the minimum spot size for a given initial aperture is very much dependent on the wavelength, and here the wavelength seems to be dependent on this reference energy that (at least in classical mechanics) should be arbitrary.

    So: Do these constant terms added to the hamiltonian, and shifted frequencies change the system physically? (ie, the way the squared magnitude of the wavefunction moves around, Pauli velocities/currents, etc)?
     
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  3. Sep 16, 2014 #2

    Orodruin

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    The short answer to your question is "no".

    If you are using the Schrödinger representation, where the states evolve, it would simply correspond to adding an additional (and equal) time dependent phase to all states and only phase differences are relevant.

    If you are using the Heisenberg representation, where operators evolve, the evolution is given by the commutator of the operator and the Hamiltonian. Since anything commutes with a constant (something proportional to the identity operator), adding a constant does not change the evolution of operators.
     
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