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Eigenstates of the Hamiltonian

  1. Oct 7, 2009 #1
    When one says that a system is in an eigenstate of the Hamiltonian, what exactly does this mean?
    I mean, if the Hamiltonian is the total energy of the system, then if it is in an eigenstate of the Hamiltonian, is this saying that its energy is a multiple of its total energy???? Obviously this makes no sense. I hope you can see where I'm confused.
     
  2. jcsd
  3. Oct 7, 2009 #2
    The Hamiltonian is not the total energy of the system. It is an operator. -- not a physical quantity. It is, however, referred to as the "energy operator".

    If a system is in an eigenstate of the Hamiltonian, then the energy carries a specific energy, which is precisely the eigenvalue of the Hamiltonian.

    If a system is in a superposition of eigenstates, then it simply doesn't carry a specific energy. Correspondingly, you can make some statements about the probability of measuring a specific outcome (the usual "coefficient squared").

    But anyways, the morale of the story: the Hamiltonian is not equal to the total energy. A system doesn't even have to carry a specific energy, but can be in a superposition.
     
  4. Oct 7, 2009 #3
    The thing that is an multiple of the state vector is actually the rate of change of the state vector. That's why the vector (state function) preserves its shape with the passage of time.
     
  5. Oct 7, 2009 #4
    If you have a list of eigenstates of the Hamiltonian, you essentially have a list of the possible energy levels you might find the particle in. Which one you'll find it in you won't know until you measure it.
     
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