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Hamiltonian operator

  1. Aug 20, 2014 #1
    Is there any difference between Hamiltonian operator and E? Or do we describe H as an operation that is performed over (psi) to give us E as a function of (psi)??
     
  2. jcsd
  3. Aug 20, 2014 #2

    stevendaryl

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    Usually, [itex]H[/itex] is an operator, and [itex]E[/itex] is a real number, its eigenvalue. For example, for a free particle,

    [itex]H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}[/itex]

    When applied to [itex]\psi(x) = e^{ikx}[/itex] you get:

    [itex]H \psi = \frac{\hbar^2 k^2}{2m} \psi[/itex]

    So for this particular [itex]\psi[/itex], [itex]E = \frac{\hbar^2 k^2}{2m}[/itex], which is a real number.
     
  4. Aug 20, 2014 #3

    Orodruin

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    Adding to what steven said, any wave function (or more generally, quantum state) does not fulfill the time-independent SE. This only happens for the eigenstates of the Hamiltonian (in fact the time independent SE is just the eigenstate equation for H). The time dependent SE describes how any quantum state evolves, not only the Hamiltonian eigenstates (although if we know the evolution of the eigenstates, i.e., for all possible E in the time independent equation, then we can easily reconstruct the general time evolution).
     
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