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Homework Help: Rewriting an Initial State with Normalized Eigenfunctions

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the Hamiltonian H=0.5p^2+ 0.5x^2, which at t=0 is described by:

    ψ(x,0)= 1/sqrt(8*pi) θ1(x) + 1/sqrt(18pi) θ2(x), where:

    θ1= exp(-x^2/2); θ2=(1-2x^2)*exp(-x^2/2)

    a) Normalize the eigenfunctions and rewrite the initial state in terms of normalized eigenfunctions. Determine the values of energy corresponding to these eigenfunctions.
    2. Relevant equations
    In (a), to normalize, take the integral of the function's conjugate times the function.

    3. The attempt at a solution
    I began by normalizing θ1 and θ2. The first one:
    A^2∫exp(-x^2)dx=1, so A=1/pi^(1/4)

    For θ2, B^2[∫(exp(-x^2/2) -4∫(exp(-x^2/2)*x^2 + 4∫(exp(-x^2/2)*x^4]=1
    So 1=B^2[sqrt(pi)(1 - 4/2 + 4(3/4)] ==> B= 1/(4*pi)^1/4

    Now I have to rewrite the initial state in terms of normalized eigenfunctions. My question is, do I simply divide the first term in ψ by A and the second term by B? Or do I have to divide the entire ψ function by some common value, based on A and B? Also, I am assuming the question does not require to normalize ψ, based on the fact that the coefficients there are not normalized.

    Thank you!
  2. jcsd
  3. Mar 8, 2013 #2


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    I think your first thought is the way to go as long as you don't need to worry about normalizing ψ. When you divide the first term by A, you are also going to replace θ1(x) by its normalized form, right? Similarly for the second term. After doing this, you could go on and normalize ψ with some overall factor.
  4. Mar 8, 2013 #3
    Thanks for your response! I wanted to clarify what you mean by also replacing θ1(x) by its normalized form. Do you mean including the exponential function that is dependent on x? So that:

    ψ(x,0)= 1/sqrt(8*pi) θ1(x) + 1/sqrt(18pi) θ2(x)


    ψ(x,0)=1/sqrt(8*pi)*[(1/(pi)^1/4]*exp(-x^2/2) + 1/sqrt(18pi)*[1/(4*pi)^1/4]*(1-2x^2)*exp(-x^2/2)

    And then to calculate the energy corresponding to these eigenfunctions, I've usually seen this before when H is a matrix and you get the eigenvalues from there. Would you do something like find Hψ here, and find the ratio of the the values of the answer and the initial state?

    Because H=0.5p^2+ 0.5x^2, I would then think I do something like -0.5h^2(d/dx)^2*[ψ(x,0)] + 0.5x^2[ψ(x,0)]?
  5. Mar 8, 2013 #4


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    No, if you do that you will change the state ψ since you have changed the relative sizes of the coefficients of θ1(x) and θ2(x). For convenience, let η1(x), η2(x) be the normalized versions of θ1(x), θ2(x):

    η1(x) = A θ1(x) ##\;\;\;## η2(x) = B θ1(x)

    You want to write ψ in terms of η1(x) and η2(x). Just replace θ1(x) by η1(x)/A and θ2(x) by η2(x)/B in ψ. This will express the non-normalized ψ in terms of η1(x) and η2(x).

    The individual states η1(x) and η2(x) are eigenfunctions of H. So, you can get the energy of each of these states by applying H to each and finding the eigenvalue. However, you could just as well do this for θ1(x) and θ2(x) since the eigenvalues do not depend on whether or not the eigenfunctions are normalized.

    ψ will not be an eigenfunction of H, so the state ψ does not have a definite energy. You can only find an expectation value for the energy in this state.
  6. Mar 8, 2013 #5
    The problem I run into is when I do this I can't get the x^2 terms to cancel (and I thought eigenenergies had to be constant).

    so for the first term, θ1:

    1/(pi)^1/4*(-0.5h^2(d/dx)^2*(exp(-x^2/2)) + 0.5x^2(exp(-x^2/2)))

    =1/(4*pi)^1/4[-h^2*(exp(-x^2/2)(x^2-1) + x^2(exp(-x^2/2))]

    Now, replacing, (exp(-x^2/2)) with θ1:

    1/(pi)^1/4*(-h^2*(θ1)*(x^2-1) + x^2*θ1)

    =1/(pi)^1/4*(θ1)[(x^2(-h^2+1) + -h^2]

    And this is where I get stuck.

    Also, don't the eigenvalues yield different values if I put the normalization constant in there vs. when I don't?
  7. Mar 8, 2013 #6


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    I suspect your hamiltonian is assuming that you are using units where h = 1 and m =1. Thus p = -i d/dx.

    I would not pull the .5 factor out and try to incorporate it into the 4th root factor.

    The eigenvalue of an eigenfunction for a particular linear operator does not depend on whether the function is normalized or not. Suppose ψ is an eigenfunction of the linear operator O with eigenvalue a: Oψ = aψ. Convince yourself that the function cψ is still an eigenfunction of O with the same eigenvalue for arbitrary nonzero c.
  8. Mar 8, 2013 #7
    Yes, makes sense, thank you!
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