1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Operators on a Harmonic oscillator ground state

  1. May 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the expectation value for a harmonic oscillator in the ground state when operated on by the operator:
    $$AAAA\dagger A\dagger - AA\dagger A A\dagger + A\dagger A A A\dagger)$$

    2. Relevant equations

    $$AA\dagger - A\dagger A = 1$$
    I also know that an unequal number of raising and lowering operators gives a zero expectation value due to orthogonality requirements.

    3. The attempt at a solution
    I guess that the first term in brackets gives a zero expectation value as it leads to a function which is orthogonal to $$\psi_0$$
    If I say n=1 then:
    $$A\dagger\psi_0 = \psi_1$$
    And
    $$A \psi_1 = \psi_0$$

    I've tried taking the third term and saying:
    $$A\dagger A(1+ A\dagger A) = A\dagger A + A\dagger A A\dagger A$$

    Then doing the same thing with the second term to get
    $$ 1+ 2A\dagger A +A\dagger A A\dagger A$$
    Then I subtract this term from the third and I get
    $$-A A\dagger $$

    But this, operating on $$\psi_0$$ seems to give me an expression which results in an infinity when integrated over all space.

    Can someone tell me where I went wrong?
     
    Last edited: May 17, 2013
  2. jcsd
  3. May 17, 2013 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You shouldn't need to do an integral, but I am curious what integral you're trying to evaluate.
     
  4. May 17, 2013 #3
    It's a sandwich integral to get the expectation value:

    $$ \frac{\hbar^4}{4a^4} \int_{-\infty}^{\infty} \psi_0^* (...) \psi_0 dx$$
    Where the dots in brackets are the operator terms I mentioned above.
    Can you see anything wrong with my reasoning above?
     
    Last edited: May 17, 2013
  5. May 17, 2013 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your reasoning looked fine. I'm not sure now why you think that integral diverges.
     
  6. May 17, 2013 #5
    Actually, it doesn't give me infinity, it gives me the wrong answer, using:

    $$\psi_0 = \left ( \frac{1}{\sqrt{\pi}a}\right )^{1/2} e^{-x^2/2a^2}$$
     
  7. May 17, 2013 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You need to show more details if you're having trouble evaluating the integral explicitly. You shouldn't need to, however, if you understand what ##A## and ##A^\dagger## do.
     
  8. May 17, 2013 #7
    Well, I realise that the $$A$$ and $$A\dagger$$ operators convert energy eigenfunctions from the state $$n$$ to $$n\pm0.5$$ where $$E_n = (n + 0.5) \hbar \omega_0$$, but I am specifically told to use the integral above to calculate the expectation value (it's actually the expectation value of $$p_x^4$$).

    So I get the integrand as \psi_0 (-\psi) and it turns out as -1 by my reckoning. I should get 3 in order to get the required result.

    P.s. how do you type Latex on the same line as text?
     
  9. May 17, 2013 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Not exactly. They take the nth state to the (n±1)th state.

    Your mistake occurred somewhere earlier. For one thing, you shouldn't have had any terms with an odd number of annihilation/creation operators.

    Use #'s instead of $'s.
     
  10. May 18, 2013 #9
    Ok, thanks for your help. If I set out what I'm doing in clear stages, maybe you could tell me in which stage my mistake lies:

    a) I note that of the three terms in the operator above, the first has an unequal number of raising/lowering operators, so must give an expectation value of zero...so I remove it from the integrand.

    b) I take the second term ##-AA\dagger A A\dagger## and I use the commutation relation to rewrite it as ##-(1+A\dagger A)(1+A\dagger A)=-(1 + 2A\dagger A + A\dagger A A\dagger A)##

    c) I note that two of these terms end in the lowering operator, which, when acting on the ground state will give zero. So the middle term becomes -1

    d) I take the third term ##A\dagger A A A\dagger## and use the commutation relation to rewrite as ##A\dagger A(1 + A\dagger A)##.

    e) I note that both terms end in the lowering operator, so when acting on the ground state will be zero. I remove the third term entirely.

    f) my integrand is left as I mentioned in my previous post.

    So could you possibly point out which of these stages involves faulty reasoning? Most appreciated, thanks for the help once again.
     
  11. May 18, 2013 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Instead of describing what you did, show us your actual work.
     
  12. May 18, 2013 #11
    Hi again
    Thanks for all your time...I have solved the problem now. Unbelievably, I simply wrote down the original series of operators wrongly, and it all works out fine now.
    Thanks again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted