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Find <H> for infinite square well

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data

    This problem comes from David Griffiths' quantum mechanics book which I have been going through on my own.

    A particle in the infinite square well has its initial wave function as an even mixture of the first two stationary states

    [tex]\Psi(x,0) = A(\psi_1 (x) + \psi_2 (x)[/tex]

    I am to find the expectation value of H and state how it compares with E1 and E2.

    (There were a few sub questions but I've done ok with them, except that finding the expectation value of x was incredibly tedious.)

    3. The attempt at a solution

    I'm a bit new to this. I'm assuming here that the particle can possess either of these two energies given by

    [tex]E = \frac{n^2 {\pi}^2 \hbar^2}{2m a^2}[/tex]

    which gives

    [tex]E_1 = \frac{{\pi}^2 \hbar^2}{2m a^2}[/tex]
    [tex]E_2 = \frac{2{\pi}^2 \hbar^2}{m a^2}[/tex]

    I'm not sure what the probability that a given particle can assume either energy is. Can I take the question's word that there is an "even mixture" of the two states, so the probability for each state is 0.5?

    If that is the case, then if I use

    [tex]<H> = \int \Psi^{"*"} \hat{H} \Psi dx = E[/tex]. How do I reconcile that with E1 and E2?
     
  2. jcsd
  3. Mar 26, 2009 #2
    You are on the right track. I assume one of the subquestions are that you have to normalize the wave function.

    When you have normalized it, you can find the probability of the wave function collapsing to psi_1 and psi_2, i.e. the probability of measuren E_1 and E_2. When you have these probabilities, you can find the expectation value.

    You could also find it using the integral you wrote, but in order to do that, you would still have to normalize.

    The results should of course give the same.
     
  4. Mar 26, 2009 #3

    CompuChip

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    You can just use the expression you gave to calculate <H>.
    Note that the wave functions are orthogonal and that <H> = Ei when [itex]H \Psi = E_i \Psi[/itex] (i.e. [itex]\Psi[/itex] is an eigenfunction).

    By the way, what did you answer to the other part of the question ("what values might you get and with what probabilities")?
     
  5. Mar 28, 2009 #4
    Thanks guys... but I'm still stuck. Based on the wave function (they share the same constant A) I gather that I have an equal chance of my measurement having E1 or E2. I'm going to literally guess the answer would be the average of the two, ie, 0.5(E1+E2).

    Since expectation value of x corresponds to the average of several measurements of x, and E = H, so expectation of H must be the average of several measurement of E. Is that right?

    CompuChip, I gave E1 and E2 as above and assume that they have an equal chance of occurring (ie 0.5).
     
  6. Mar 28, 2009 #5

    CompuChip

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    For the first one, you don't have to guess. You have calculated A in question a), haven't you? Now (re-)read the text below example 2.2 (in my edition it's at the bottom of page 36): "Loosely speaking, ..." up to equation 2.38
    Actually, you might want to read up to example 2.3.

    You can also do the calculation explicitly, as I suggested. Let me get you started on the calculation:
    [tex]
    \begin{align*} \langle H \rangle
    & {} = \int \Psi^* \hat H \Psi \, dx \\
    & {} = \int (A^* \psi_1^* + A^* \psi_2^*) \hat H (A \psi_1 + A \psi_2) \, dx \\
    & {} = |A|^2 \int (\psi_1^* + \psi_2^*) \hat H (\psi_1 + \psi_2) \, dx \\
    & {} = |A|^2 \left[ \int \psi_1^* \hat H \psi_1 \, dx + \int \psi_1^* \hat H \psi_2 \, dx + \int \psi_2^* \hat H \psi_1 \, dx + \int \psi_2^* \hat H \psi_2 \, dx \right]
    \end{align*}
    [/tex]
    Now you can use that [itex]\hat H \psi_i = E_i \psi_i[/itex], where Ei is a number which you can pull outside the integral, and that the eigenfunctions are orthonormal (equation 2.31).
     
  7. Mar 28, 2009 #6
    Compuchip, unfortunately, I have an old edition without the footnote, and a number of problems are different from other editions.

    Ok... this wasn't so hard! I substituted the stationary state wavefunction into the formula for <H> as you suggested, and due to the orthonormality the two middle integrals are zero, and the two at the ends are E1 and E2, and the value for A when squared gives 0.5(E1 + E2). Checks out! Thanks...
     
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