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Quantum Harmonic oscillator

  1. Sep 18, 2011 #1
    Consider the usual 1D quantum harmonic oscillator with the typical hamiltonian in P and X and with the usual ladder operators defined.
    i) I have to prove that given a generic wave function [itex] \psi [/itex], [itex] \partial_t < \psi (t) |a| \psi (t)> [/itex] is proportional to [itex]< \psi (t) | a | \psi (t) >[/itex] and determine their ratio.

    Here I tried to express [itex] \psi(t) [/itex] as an infinite sum of the eigenstates with time evolution operator applied to it but I think there must definitely be some less clumsy way.

    ii) Construct an operator [itex] Q_k [/itex] such that it only allows transitions between the states [itex] |n> [/itex] and [itex] |n \pm k > [/itex] ([itex] |n> [/itex] being the nth eigenstate of the N operator).

    In this question I really did not get why the answer couldn't just be [itex] a [/itex] or [itex] a^_\dagger [/itex].

    Thanks in advance
  2. jcsd
  3. Sep 18, 2011 #2

    Ken G

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    Gold Member

    In (i), I think you want to use the time-dependent Schroedinger equation to replace the time derivative, after noticing that the time derivative can be applied to the bra or the ket, so you get two terms there. The action of H means the the two terms always differ by the energy difference between a level and the next higher level, so that stays fixed even as n varies. That's why it ends up a proportionality, no matter how many n's are involved in the eigenvalue expansion.

    For part (ii), a single raising or lowering operator only connects to either +1 or -1, not to both +1 and -1, let alone both +k and -k. It seems like you need a net raising by k, or a net lowering by k, in whatever operator you use. Is it as simple as ak+(adag)k?
  4. Sep 18, 2011 #3
    Of course this is what I meant, sorry. So I turn the question back to you. Could the answer be so trivial?
  5. Sep 18, 2011 #4


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    yeah sure, that is an operator that connects those states. If they just want an operator in particular then that answer the question. Here's something to think about, though. For example, suppose k=1. Then your operator is

    a + a^\dagger

    but how about the operator

    a + a^\dagger + a^2a^\dagger + a^5 (a^\dagger)^6

    these also work, right? And in general can have arbitrary complex coefficients in front of each term.
  6. Sep 19, 2011 #5
    For the sake of completeness. I have to say there was one last question and it asked to prove that this operator [itex] Q_k [/itex] commutes with the parity operator whenever k is even. How should I go about proving this even supposing my [itex] Q_k [/itex] is simply [itex] a^k[/itex] or [itex](a^\dagger)^k [/itex]. And would it be true?
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