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Expectation values of QHO in |n> basis

  1. Feb 5, 2012 #1
    Is it possible to express ANY observable A(X,P) in terms of the ladder operators?

    I know how to evaluate expectation values in the |n> basis given the operators in terms of a & a+, but was trying to figure out <1/X^2>. How do you express 1/X^2 in terms of ladder operators? <ψ|(1/X^2)|ψ> can be done in the X basis fine, by evaluating ∫ψ*(1/x^2)ψ dx , but what about in |n> basis?

    i.e
    Given X^2 = (h/2mw)[a^2 + aa' + a'a +a'^2] how do I "invert" this into 1/X^2 , and be able to evaluate it in a state <1|(1/X^2)|1> for example?
    (a' = a-dagger)
     
    Last edited: Feb 5, 2012
  2. jcsd
  3. Feb 6, 2012 #2
    the division process is not defined for operators. unless you mean inversion which gives a rather simple answer, knowing that the lowering and raising operators are each others inverses. If you just want to know what is the inverse of the expectation value of x squared, that isn't too difficult, you just evaluate x squared and take its inverse.
     
  4. Feb 6, 2012 #3
    Yeah I don't mean division, I guess it would be the inverse of X^2? Since <ψ|(X^2)(1/X^2)|ψ> will just be <ψ|ψ>. And yeah I don't just want to be able to work out 1/<X^2> which is easily done, I want <1/X^2>.

    So how do I find the inverse of X^2 = (h/2mw)[a^2 + aa' + a'a +a'^2] ?
    I need some operator in terms of a and a' such that when I have the product of the two, all the terms cancel out to the Identity operator.. how?

    edit*Thinking about it, how would it even be possible to get to say, 1/X, using a = X + iP and a' = X - iP (ignoring the coefficients). So I'm guessing you CANT write any observable using ladder operators? Only if they have powers of X and P that are in the natural numbers including 0.
     
    Last edited: Feb 6, 2012
  5. Feb 6, 2012 #4
    You're kind of missing the point here. You cannot perform a division operation on a vector.
    <ψ|(1/X^2)|ψ> is not mathematically defined. The object 1/<X^2> however makes more mathematical sense.
    Any Hamiltonian with a potential quadratic in position coordinates, and non relativistic kinetic energy, can have raising and lowering operators associated with it. Although I don't think this is what you were asking for. In general you can write any observable that depends on x or p in terms of the familiar lowering and raising operators for the QHO.
    In theory if you write 1/X as X^(-1), then we have X^(-2) = ((h/2mw)^(-2)).([a^2 + aa' + a'a +a'^2]^-2). But then you would have to expand that bracket. No easy task I grant you, but the crucial assumption here is that the X and P are scalars; we are working with a 1 dimensional harmonic oscillator. In that case matrix and scalar inversion are effectively the same operations. If you have trouble with expanding that bracket, try a Binomial expansion, but be careful with the orderings as these are differential equations you are manipulating.
     
  6. Feb 6, 2012 #5
    What do you mean <ψ|(1/X^2)|ψ> isn't defined? Maybe my notation is just bad, what I meant by that is expectation value of (1/X^2), where (1/X^2)|x> = (1/x^2)|x> .
    In general <1/X^2> ≠ 1/<X^2>.
    But yes I see, so I should be able to get the right result by expanding that bracket, looks tedious haha, thanks for your answers.
     
  7. Feb 7, 2012 #6
    yes i was perhaps pressing on a point not quite relevant to your question. any way the bracket needs to be expanded by a multinomial expansion, or a binomial expansion if you prefer that way.
    here's some help on how to do those:
    http://en.wikipedia.org/wiki/Multinomial_theorem
     
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