Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ladder operators homework

  1. Apr 27, 2012 #1
    I have a homework problem which asks me to compute the second and third excited states of the harmonic oscillator. The function we must compute involves taking the ladder operator to the n-power. My question is this: because the ladder operator appears as so, -ip + mwx, and because I am using it as a differential operator, is it ok to move the x around as I see fit? For example, when the ladder operator is squared, I get cross terms which contain x and d/dx. Am I able to move the x outside of the d/dx, in which case it will appear as 2(hbar)mwx(d/dx), or will I have one cross term which reads (hbar)(d/dx)mwx and the other which reads mwx(hbar)(d/dx)? (minus signs ignored)
  2. jcsd
  3. Apr 27, 2012 #2
    You can't take the x outside of the derivative, as the derivative is with respect to x. So the two cross terms are inequivalent.

    In general, the square of an operator means: "Apply the operator once. Then apply it again to the resulting state" and similarly for nth powers. This makes it clear that you can't move the x in the first raising operator to the left of the second raising operator--it's part of what the second raising operator is acting on.
  4. Apr 28, 2012 #3


    User Avatar
    Science Advisor

    You start with [itex]\psi_0[/itex] and the calculate [itex]a^3\,\psi_0[/itex] which means

    [tex]a^3\,\psi_0 = a\left( a\left( a\,\psi_0\right) \right)[/tex]
  5. Apr 28, 2012 #4
    Awesome thanks.
  6. Apr 29, 2012 #5
    Remember though that the square of the ladder operator is not just a+a+ since it is not hermitian. You have to take the hermitian conjugate and multiply with the operator you want to square, so that (a+)2=(a+)+ (a+) = a a+

    And in Quantum Mechanics you are dealing with operators that in general do not commute, so you always have to care about the order they are in.
  7. Apr 29, 2012 #6


    User Avatar
    Science Advisor
    Gold Member

    It is standard for (a)^2 to be defined as (a)(a) rather than taking some hermitian conjugate.
  8. Apr 30, 2012 #7


    User Avatar
    Science Advisor

    Sorry, the dagger is missing: [itex](a^\dagger)^3\,\psi_0[/itex]
  9. Apr 30, 2012 #8


    User Avatar
    Science Advisor

    In this case the wave function ψ3 is generated by the third power of the creation operator w/o any ordering ambiguity, hermitean conjugate or something like that
  10. Apr 30, 2012 #9
    Yes, I see that I expressed myself incorrectly. You are of course correct, if one writes a2 this means aa and not what I wrote. So to answer the OP, ignore my previous post, correct would be: (a+)3=a+a+a+. Unambiguously.

    So by expressing the creation operators in terms of x and p and expanding the parentheses (and keeping the ordering!) and then acting with the result on the ground state wavefunction you will get an expression for the third excited state, corresponding to energy [itex]E_3 = (7/2) \hbar \omega[/itex]

    [On a side note: I guess my source of confusion stems from something like wanting to express p2 in terms of a and a+, in which case you would take the hermitian conjugate of the linear comb. in a and a+ comprising p and multiply by p in terms of a and a+ (that is, not by multiplying p in terms of a and a+ times itself). Wouldn't this be correct at least? (That is at least how my professor did it when I read QM.)

    EDIT: Sorry, above is obviously incorrect since p is hermitian so that it is equal to its herm. conjugate. We have [itex]p \propto i(a-a^{\dagger})[/itex], then [itex]p^{\dagger} \propto -i(a^{\dagger}-a)=+i(a-a^{\dagger})[/itex] ]
    Last edited: Apr 30, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook