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

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook