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Expectation value of a product of hermitian operators

  1. Nov 17, 2014 #1

    DrClaude

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    I'm trying to derive something which shouldn't be too complicated, but I get different results when doing things symbolically and with actual operators and wave functions. Some help would be appreciated.

    For the hydrogenic atom, I need to calculate ##\langle \hat{H}\hat{V} \rangle## and ##\langle \hat{V}\hat{H} \rangle##, where ##\hat{H} = p^2/2m + \hat{V}## and ##\hat{V} = -Z\hbar^2/(m a_0 r)## for a state that is an eigenstate of ##\hat{H}##, ##\hat{H}|n\rangle = E_n |n\rangle##. Since both ##\hat{H}## and ##\hat{V}## are hermitian, it follows that
    $$
    \begin{align*}
    \langle n | \hat{V}\hat{H} | n \rangle &= \langle n | \hat{V} E_n | n \rangle \\
    &= E_n \langle n | \hat{V} | n \rangle = - E_n Ze^2 \langle n | \frac{1}{r} | n \rangle \\
    \langle n | \hat{H}\hat{V} | n \rangle &= \left( \langle n | \hat{H} \right) \hat{V} | n \rangle \\
    &= E_n \langle n | \hat{V} | n \rangle = - E_n \frac{Z\hbar^2}{m a_0} \langle n | \frac{1}{r} | n \rangle
    \end{align*}
    $$
    from which I conclude that ##\langle \hat{H}\hat{V} \rangle = \langle \hat{V}\hat{H} \rangle = E_n \frac{Z\hbar^2}{m a_0} \langle 1/r \rangle##.

    But when I try to calculate the expectations values explicitely on the 1s wave function,
    $$
    \psi_{1\mathrm{s}} = \frac{1}{\sqrt{\pi}} \left( \frac{Z}{a_0} \right)^{3/2} e^{-Z r /a_0}
    $$
    I find that ##\langle \hat{H}\hat{V} \rangle = \frac{5}{2} \frac{\hbar^4 Z^4}{m^2 a_0^4}##, while ##\langle \hat{V}\hat{H} \rangle = \frac{1}{2} \frac{\hbar^4 Z^4}{m^2 a_0^4}##. Since for hydrogenic atoms we have
    $$
    \left\langle \frac{1}{r} \right\rangle = \frac{1}{n^2} \frac{Z}{a_0}
    $$
    and
    $$
    E_n = - \frac{\hbar^2 Z^2}{2 m a_0^2} \frac{1}{n^2}
    $$
    I would expect ##E_n \frac{Z\hbar^2}{m a_0} \langle 1/r \rangle = \hbar^4 Z^4 / (2 m^2 a_0^2)##, which is the value I get for ##\langle \hat{V}\hat{H} \rangle##, but not ##\langle \hat{H}\hat{V} \rangle##.

    By the way, I did the integrations with the actual wave function both by hand and with Mathematica, so I know there is no mistake there.

    Can somewhat figure out what I did wrong?
     
  2. jcsd
  3. Nov 17, 2014 #2

    dextercioby

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    Well, let's think formally (up to functional analytical finesse). You get that formally <n|VH|n> - <n|HV|n> = 0 => <n|[V,H]|n> = 0 whatever n. But {|n>} form a subbasis in the space of solutions of the spectral equation for H as a whole (H=T+V) and are orthonormal one to another. One gets that [H,V] =0. But H = T-V and [V,V] = 0 (trivially). One gets that: [T,V] =0 which is nonsense, because T and V don't have the same domain.
     
    Last edited: Nov 17, 2014
  4. Nov 18, 2014 #3

    DrClaude

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    The problem now is that I have tested the symbolic approach with the harmonic oscillator (edit: I should have said that I compared the symbolic and direct approaches), and it works fine. I'm trying to figure out what could be wrong with what you wrote, and I wonder if the fact that <m|HV|n> is not necessarily 0 even if m≠n could affect your proof.
     
  5. Nov 18, 2014 #4

    DrDu

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    Did you take into account that the Laplace operator applied to 1/r gives rise to a Delta function at the origin?
     
  6. Nov 18, 2014 #5

    dextercioby

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    What do you mean by what I bolded?
     
  7. Nov 18, 2014 #6

    DrClaude

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    Taking
    $$
    \begin{align*}
    \hat{V} &= \frac{1}{2} m \omega^2 \hat{x}^2 \\
    \hat{H} &= -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \hat{V}
    \end{align*}
    $$
    using the "symbolic" approach, I find
    $$
    \begin{align*}
    \langle n | \hat{V}\hat{H} | n \rangle &= \langle n | \hat{V} E_n | n \rangle \\
    &= E_n \langle n | \hat{V} | n \rangle = E_n \frac{m \omega^2}{2} \langle n | \hat{x}^2 | n \rangle \\
    \langle n | \hat{H}\hat{V} | n \rangle &= \left( \langle n | \hat{H} \right) \hat{V} | n \rangle \\
    &= E_n \langle n | \hat{V} | n \rangle = E_n \frac{m \omega^2}{2} \langle n | \hat{x}^2 | n \rangle
    \end{align*}
    $$
    Using ##E_n = \left(n+\frac{1}{2} \right) \hbar \omega## and ##\langle \hat{x}^2 \rangle = \left(n+\frac{1}{2} \right) \frac{\hbar}{m \omega}##, I get
    $$
    \langle \hat{H}\hat{V} \rangle = \langle \hat{V}\hat{H} \rangle = \left(n+\frac{1}{2} \right)^2 \frac{\hbar^2 \omega^2}{2}
    $$

    I obtain the same if I calculate the actual integrals ##\langle \hat{H}\hat{V} \rangle## and ##\langle \hat{V}\hat{H} \rangle##.
     
  8. Nov 18, 2014 #7

    DrClaude

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    No. I'll check my calculations again.
     
  9. Nov 18, 2014 #8

    dextercioby

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    I'm sure you see a trick in the fact that HV is not selfadjoint. Then of course my argument above still applies, because, assuming HV is properly defined (and for Schwartz test functions is), then HV=/=VH, because it's more than obvious that T = H-V and V don't commute (because x and p don't commute). So are the formal maneuvers legal? I'll think about it.
     
  10. Nov 19, 2014 #9

    DrClaude

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    Yeah, that was it. Writing it as I did, I just didn't see the ##\nabla^2 (1/r)## in there. Being more careful, it makes a term go away, and I get that ##
    \langle \hat{H}\hat{V} \rangle =
    \langle \hat{V}\hat{H} \rangle## and all is fine. Thank you and Dexter for the help.
     
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