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Simple QM question

  1. Sep 7, 2008 #1
    Hi,

    Say if we have two energy eingenfunctions and correpsonding eigenvalues, as specified by:
    [tex] H \psi_1 =E_1 \psi_1 [/tex]
    [tex] H \psi_2 =E_2 \psi_2 [/tex]

    I have written in my notes that since the Hamiltonian operator H is linear, then [tex] \phi=C_1 \psi_1 + C_2 \psi_2 [/tex] must also be a solution. But

    [tex] H \phi =H(C_1 \psi_1 + C_2 \psi_2) =C_1 E_1 \psi_1 +C_2 E_2 \psi_2 =E_1 C_1 \psi_1 + E_2 C_2\psi_2 != E \phi [/tex], where E is some constant. (!= is supposed to represent NOT equal to)

    so how can this combination also be a solution (assuming the eigenfunctions are non degenerate)?

    thanks
     
  2. jcsd
  3. Sep 7, 2008 #2

    George Jones

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    This is true for solutions of the time-dependent Schrodinger equation, but it not, in general, true for solution of the time-independent Schrodinger equation. Consequently, the linear combination of two stationary states is (usually) not a stationary state.
     
  4. Sep 7, 2008 #3

    atyy

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    [tex] H \psi_1 = E \psi_1 [/tex]

    The above equation is linear - this is why degenerate eigenvalues can occur.

    [tex] H \psi_1 =E_1 \psi_1 [/tex]
    [tex] H \psi_2 =E_2 \psi_2 [/tex]

    The above equations are two different linear equations, because [tex]E_1[/tex] and [tex]E_1[/tex] are different, so there is no reason for the sum of their solutions to remain solutions.
     
  5. Sep 7, 2008 #4

    Vanadium 50

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    Note that [tex]C_1 \psi_1+ C_2 \psi_2[/tex] is a solution. However, it is (in general) not an energy eigenvalue. In fact, if one measured the energy, one would find a probability [tex]C_1^2/(C_1^2 + C_2^2)[/tex] that it has energy [tex]E_1[/tex] and [tex]C_2^2/(C_1^2 + C_2^2)[/tex] that it has energy [tex]E_2[/tex].
     
  6. Sep 7, 2008 #5
    ah thank you both! makes perfect sense now...
     
  7. Sep 7, 2008 #6
    I agree with your answer as it pertains to the OP's question, but you touch on a sore point that went unresolved in a recent thread. Is this business of measuring the energy just a hypothetical argument, or is it actually possible to measure the energy of such a state? In the previous discussion we were considering the case of a hydrogen atom in a superposition of states.
     
  8. Sep 7, 2008 #7

    Vanadium 50

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    It's possible. The difficulty is preparing a state like the OP described.
     
  9. Sep 7, 2008 #8
    I guess when I asked "is it possible to measure such a state" I wasn't looking for a yes or no answer; the implication was that I was looking for a description of how you would actually do it.

    As for preparing such states, I would think they are the rule rather than the exception in nature. Any perturbation of a pure energy eigenstate will result in that state going into a superposition with other states mixed in.
     
  10. Sep 7, 2008 #9

    Vanadium 50

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    This is getting pretty far afield of the OP's original question, and I have no idea what your objection is to the idea of measuring energy. I had thought it was a problem with conservation of energy with a mixed state, but I guess it's not. Since there's no short answer, I suggest we end this thread hijack.
     
  11. Sep 7, 2008 #10
    Then why don't you just tell me how it's done?


    Why don't you just ban me from the group altogether?
     
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