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Raising and lowering operators homework

  1. Nov 18, 2006 #1
    Having a lot of trouble with this one. I'm given that the Hamiltonian of a certain particle can be expressed by H = A(a+a) + B(aa+), where A and B are constants and a+ and a are the raising and lowering operators, respectively. I'm supposed to find the energies of the stationary states for the particle.

    I'm also given that the operators satisfy the communtation relation [a, a+] = 1. So from this I'm getting [tex]aa_+\psi - a_+a\psi = \psi[/tex]. From the Hamiltonian I have [tex]A(a_+a)\psi + B(aa_+)\psi = E\psi[/tex]. I tried using the first equation to replace one of the terms on the left side in the second, to get [tex]A(aa_+\psi - \psi) + B(aa_+\psi) = E\psi[/tex]... but I don't see how this really helps. I'm stuck on what to do next, so any help would be appreciated :)
     
    Last edited: Nov 18, 2006
  2. jcsd
  3. Nov 19, 2006 #2
    Anybody have a clue on what I could do next? Even an inkling of an idea? Because I'm really lost. Do I need to provide more info? Thanks :)
     
  4. Nov 19, 2006 #3

    nrqed

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    Looks like nobody is helping so here are some thoughts

    Recall that for the h.o. [itex] H = (a_+a_- +{1 \over2}) = a_- a_+ - { 1\over 2}) [/itex](in units of [itex]\hbar \omega [/itex]).

    Then one shows that is [itex] \psi_n [/itex] is a solution with energy E_n, [itex] a_+ \psi_n [/itex] is also a asolution, with energy E_n + 1, and [itex] a_- \psi_n [/itex] is a solution with energy E_n - 1.

    Then one imposes that for the ground state, [itex] a_- \psi_0 = 0 [/itex] and one uses [itex] H \psi_0 = E_0 [/itex] to find E_0 and the ground state wavefunction. Then one uses the raising operator to find all the energies and the excited states.

    Just go through exactly all the same steps. The difference is that in your case, you will get a slightly different relation for the relation between the energy of psi_n and of [itex] a_+ \psi_n [/itex] and a different expression for the ground state energy and so on. But the steps are all the same.

    I hope this helps.

    Patrick
     
  5. Nov 19, 2006 #4

    Hurkyl

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    It would probably help to group like terms. I.E.

    [tex](A + B) a a_+ \psi = (E - A) \psi[/tex]

    [tex]a a_+ \psi = \frac{E - A}{A+B} \psi[/tex]
     
  6. Nov 19, 2006 #5
    Thanks guys. I think I did it in a convoluted way (before I saw your post, Hurkyl), but I ended up getting that a- operating on a state of energy E would give a state of energy E - A - B, and a+ operating on a state of energy E would give a state of energy E + A + B, which seems to make at least a little sense.

    Using this, for the ground state I got E0 = A + B. So that means E1 = 2A + 2B, etc.. so my question is when do I know when to stop? Am I just looking for a general formula, which would just be En = (n+1)(A+B)? Or have I screwed it up.

    Thanks so much for the help. This was a fun problem once I knew what I was doing :)
     
  7. Nov 19, 2006 #6

    nrqed

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    Are you sure your ground state energy is right?
    The hamiltonian you gave in your first post was [itex] A a_+ a_- + B a_- a_+ = (A+B) a_+ a_- + B [/itex], right? Applied on the ground state the lowering operator should give zero so that the ground state energy should be B, no?

    The general formula will indeed be of the simple form [itex] C_1 + C_2 n [/itex] where C1 and C2 will be functions of A and B.

    Patrick
     
  8. Nov 19, 2006 #7

    I'm not sure where the right half of that equation is coming from :(. I got the ground state energy by just saying [tex]a_-\psi_0 = 0[/tex] and [tex][A(a_+a_-) + B(a_-a_+)]\psi_0 = E_0[/tex], and since a- operating on a state of energy E gives a state of energy (E - A - B), I reasoned that a- operating on E0 must give E0 - A - B = 0. So E0 would be A + B. I probably did mess up somewhere though =/
     
  9. Nov 19, 2006 #8

    nrqed

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    I just rewrote [itex] a_- a_+ = a_+ a_- + 1 [/itex] in your original hamiltonian.
    I see what you did. But that's not what we do in the usual harmonic oscillator problem. We assume that [itex] a_- \psi_0 0[/itex] without saying anything about the energy. If we used the same reasoning you used, we would say [itex] E_0 - \hbar \omega =0 [/itex] so the ground state energy would be [itex] \hbar \omega [/itex] but that's not the correct answer. Instead, we assume [itex] a_- \psi_0 = 0 [/itex] without any condition on the energy. We get the energy by applying the hamiltonian.

    Hope this makes sense

    Patrick
     
  10. Nov 19, 2006 #9
    OK, I see what you're saying. But since I don't know the lowering function explicitly here.. how can I calculate [tex]a_-\psi_0[/tex]?
     
  11. Nov 19, 2006 #10

    nrqed

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    we simply *define* this to be zero (so that there is a state of minimum energy). If there was not such a state of lower energy, the particle could jump down forever, and we would have a source of infinite energy.

    So the key point is that we *define* psi_0 by imposing [itex] a_- \psi_0 = 0 [/itex]. *then* you go back to the hamiltonian which you apply on the ground state and that gives you the ground state energy. If you go back to the harmonic oscillator, this is the logic followed.

    Hope this makes sense

    Patrick
     
  12. Nov 19, 2006 #11
    OK, maybe we didn't use that approach in my class because I can't find anything about it in my notes or text. So if we apply the hamiltonian to the ground state.. you get [tex][A(a_+a_-) + B(a_-a_+)]\psi_0 = E_0[/tex]. So since [tex]a_-\psi_0 = 0[/tex], you're just left with [tex]Ba_-(a_+\psi_0) = E_0[/tex].. which is [tex]Ba_-(E_1 + A + B) = E_0[/tex] and then [tex]BE_0 = E_0[/tex], which is clearly not right. Where did I go wrong?
     
  13. Nov 19, 2006 #12

    nrqed

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    :confused: [itex] a_+ \psi_0 [/itex] is *not* [itex] (E_1 +A + B) \psi_0 [/itex]!! All you know is that [itex] a_+ \psi_0 [/itex] gives a state [itex ] \psi_1 [/itex] with an energy E_0 + A + B (I did not check this part of your calculation, I assume it's right).

    By the way, you must be careful about not dropping the wavefunction (which you did in what you wrote).

    No, what you do is to use the commutator to rewrite [itex] a_- a_+ = a_+ a_- +1 [/itex] which *then* you apply to the ground state wavefunction. At this stage of the calculation you certainly don't want to apply the raising operator and get psi_1 all mixed in in your calculation.

    What textbook are you using? You could see what I mean by looking at their derivation of the ground state energy.

    Patrick
     
  14. Nov 19, 2006 #13
    Sigh. Sorry. I feel like an idiot.

    OK. Thanks for spelling that out for me. So I did what you said and got [tex]a_-a_+\psi_0 = \psi_0[/tex]. Applying the Hamiltonian and then using that, you get:

    [tex][A(a_+a_-) + B(a_-a_+)]\psi_0 = E_0\psi_0[/tex]
    [tex][(A(a_+a_-) + B]\psi_0 = E_0\psi_0[/tex]
    [tex]E_0\psi_0 = B\psi_0[/tex]

    So E0 is just B. E1 is A + 2B, E2 is 2A + 3B, etc., and in general it's just En = An + B(n+1). I *think* this is what you were saying earlier.
     
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