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Expectation Values - Quantum

  1. Feb 4, 2014 #1
    1. The problem statement, all variables and given/known data
    [itex]\Psi (x) = C e^{i k_{0} x} e^{\frac{-x^{2}}{2 a^{2}}}[/itex]

    Find [itex]\left\langle x \right\rangle, \left\langle x^{2} \right\rangle, \left\langle p \right\rangle, \left\langle p^{2} \right\rangle.[/itex]


    2. Relevant equations
    Operators make a "psi-sandwich":
    [itex]\left\langle x \right\rangle = \int_{- \infty}^{\infty} \Psi^{*} \left( x)\right) \Psi[/itex]
    [itex]\left\langle x^{2} \right\rangle = \int_{- \infty}^{\infty} \Psi^{*} \left( x^{2}\right)\Psi[/itex]
    [itex]\left\langle p \right\rangle = \int_{- \infty}^{\infty} \Psi^{*} \left( \frac{\hbar}{i}\frac{\partial}{\partial x}\right)\Psi[/itex]
    [itex]\left\langle p^{2} \right\rangle = \int_{- \infty}^{\infty} \Psi^{*} \left( \frac{\hbar}{i} \frac{\partial}{\partial x}\right)^{2}\Psi[/itex]


    3. The attempt at a solution
    I found <x> to be zero, because of an odd-integrand. (Multiplying psi by psi* in the integrand to simplify). I did the same thing for <x2>, except the integrand was not odd this time because of the x^2 in there. Using a special Gaussian integral formula listed in my book, I was also easily able to find <x2> = x2 / 2.

    Since <p> is just m*(d<x> / dt), <p> is also zero in this case.

    For <p^2>, I take the second derivative of psi, and multiply this result by psi*.

    But when finalizing all the derivatives and simplification in the integrand for <p2>, there is always an i left over (I don't think there should be an i in the momentum..), and the integral is not easily doable (where the professor has said all integrals needed will be listed in the back cover of our book).

    So I must conclude that my approach is fundamentally wrong, even though I'm using the equations I have been given.

    I have found in a separate part of the question the constant [itex]C = \sqrt{\frac{1}{\sqrt{\pi} a}}[/itex]
     
    Last edited: Feb 4, 2014
  2. jcsd
  3. Feb 4, 2014 #2

    vanhees71

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    2016 Award

    The expectation value of an observable is given by
    [tex]\langle A \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi^*(\vec{x}) \hat{A} \psi(\vec{x}),[/tex]
    where [itex]\hat{A}[/itex] is the operator representing the observable in position representation. Your equations given under Sect. 2 are all correct. However, you where somewhat too quickly arguing ;-)).

    The idea to use Ehrenfest's theorem to get the momentum-expectation value is correct in principle, but you forgot that you haven't calculated the expectation value of the position as a function of time but only at a single instant of time. To evaluate expectation values as a function of time you must plug in the solution of the time dependent Schrödinger equation. Then you'd get the correct result, but that's much more complicated than just evaluating the expectation values as given in Sect. 2!

    All other trouble must be simple mistakes in taking the derivatives. Of course, the expectation value of the momentum must be real, because [itex]\hat{p}=-\mathrm{i} \hbar \partial_x[/itex] is an essentially self-adjoint operator on the Hilbert space of square integrable position-wave functions!
     
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