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How do I find this expectation value?

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data
    A hydrogen like ion (with one electron and a nucleus of charge Ze) is in the state
    [tex] ψ = ψ_{2,0,0} - ψ_{2,1,0} [/tex]

    What's the expectation value of \hat{r} (position operator) as a function of Z?
    Assuming origin at nucleus.


    2. Relevant equations

    for Z=1

    [tex] < ψ | \hat{r} | ψ > = -3 \frac{4 π ε_0 \hbar^2}{m e^2} n_z[/tex]


    3. The attempt at a solution

    Using the values for
    [tex] ψ = ψ_{2,0,0} - ψ_{2,1,0} [/tex]

    I got

    [tex] ψ = \frac{1}{4 \sqrt{2 π a_0 ^3}} e^{- \frac{r}{2 a_0}} ( 2 - \frac{r}{a_0} - \frac{r cos(\theta)}{a_0}) [/tex]

    I wouldn't have a clue how to integrate this and I imagine there must be an easier way to find the expectation value.
     
    Last edited: Oct 19, 2013
  2. jcsd
  3. Oct 19, 2013 #2

    vela

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    What does ##n_z## represent? Is this result for the given ##\psi##?


    You're going to have to normalize the wave function before calculating the expectation value.

    Why not? It's an integral you should be able to do if you've taken basic calculus. Once you have the correct integral set up, it may look intimidating, but it's a straightforward to evaluate.

    Initially, keep the integral in terms of ##\psi_{200}## and ##\psi_{210}##. You'll be able to argue that some of the terms will vanish when integrated.
     
  4. Oct 19, 2013 #3
    It's the unit vector pointing in the z-direction. Yes, this is the Z=1 result for the given ψ.


    The wave function is already normalised.


    As for the integral, this is what I was doing initially:

    [tex] < \hat{r} > = \int_{- \infty} ^ {\infty} ψ * \hat{r} ψ dV [/tex]

    [tex] < \hat{r} > = \int_{- \infty} ^ {\infty} ψ * ψ r^3 dr [/tex]

    [tex] < \hat{r} > = \int_{- \infty} ^ {\infty} \frac{1}{32 π a_0 ^3} e^{- \frac{r}{a_0}} ( 2 - \frac{r}{a_0} - \frac{r cos(\theta)}{a_0})^2 r^3 dr [/tex]

    What I've got here, even wolfram alpha can't integrate.

    By keeping the ψ_{2,0,0} and ψ_{2,1,0} separate, do you mean

    [tex] \hat{r} | ψ > = \hat{r} \psi_{200} - \hat{r} \psi_{210} [/tex]

    [tex] < \psi | \hat{r} | ψ > = ( \psi_{200} - \psi_{210} ) ( \hat{r} \psi_{200} - \hat{r} \psi_{210} ) [/tex]

    [tex] < \psi | \hat{r} | ψ > = \psi_{200} \hat{r} \psi_{200} - \psi_{200} \hat{r} \psi_{210} - \psi_{210} \hat{r} \psi_{200} + \psi_{210} \hat{r} \psi_{210} [/tex]

    And then integrate each part separately and add them up?
     
  5. Oct 19, 2013 #4

    vela

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    No, it isn't. The individual wave functions are, but the linear combination isn't.

    The limits and volume element aren't correct. In spherical coordinates, you want to calculate
    $$\int_0^\infty \int_0^\pi \int_0^{2\pi} f(r,\theta,\phi)\,r^2\sin\theta\,d\phi\,d\theta\,dr$$ in general.

    Yes. Some of those terms you can argue will integrate to 0. (If you don't trust your physical intuition, you could also integrate them and verify that they vanish.)
     
  6. Oct 20, 2013 #5
    I think it is the [tex] < \psi_{200} | \hat{r} | \psi_{200} > and < \psi_{210} | \hat{r} | \psi_{210} > [/tex] terms that will vanish, leaving
    [tex] - < \psi_{200} | \hat{r} | \psi_{210} > - < \psi_{210} | \hat{r} | \psi_{200} > [/tex]

    which would give me.. -6 a_0 after integrating ?
     
  7. Oct 20, 2013 #6

    vela

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    Looks good except that you neglected the effect of Z.
     
  8. Oct 20, 2013 #7
    Ah yes, I guess that would just make it

    [tex] \frac{-6 a_0}{Z} [/tex]

    since r is proportional to 1/Z ?
     
  9. Oct 25, 2013 #8

    vela

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    I don't recall the Z dependence offhand, but what you said does sound familiar. It would probably be best if you looked at how the derivation changes if you use ##Ze^2/r## for the potential instead of just ##e^2/r## in the Schrodinger equation.
     
  10. Oct 25, 2013 #9

    DrClaude

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    That's not possible. The expectation value of ##\hat{r}##, or average position, can't be negative.


    It can't be. The expectation value of ##\hat{r}## is scalar, not a vector.

    You just gave above the equation for ##\langle \psi_{n l m} | \hat{r} | \psi_{n l m} \rangle##. Was it zero?

    Again, you can't have a negative value.
     
  11. Oct 25, 2013 #10
    I'm just writing what it says on the sheet. ##\hat{r}## is written as the position operator, and I guess the electron is just along the negative z axis with respect to the nucleus. I don't think my professor would have made a mistake in his question (at least I rather hope not). I kind of see what you mean, but I've had negative expectation values before. Thanks anyway though.
     
  12. Oct 25, 2013 #11

    vela

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    I think your professor meant the observable corresponding to the position vector ##\vec{r}## rather than the spherical coordinate ##r##. At least, that's what I was assuming given what you had written.
     
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