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Acceleration operator and the electron in a hydrogen atom

  1. May 18, 2013 #1
    I am wondering about acceleration in quantum mechanics. Let's consider spherically symmetric potential [itex]V(r)[/itex]. From the Heisenberg equation of motion, one finds the time derivative of the momentum operator

    [tex]\dot{\hat{p}}=\frac{i}{\hbar}\left[\hat{H},\hat{p}\right] = -\nabla V,[/tex]
    from which we can construct an acceleration operator simply by

    [tex]\hat{a} = -\frac{1}{m} \nabla V .[/tex]
    I then want to apply this to the electron in a hydrogen atom. The expectation value of the acceleration is undoubtedly zero for every state. But the RMS-value could be expected to be non-zero. The calculation of the expectation value

    [tex]\langle \Psi_{nlm} | \hat{a}^2 | \Psi_{nlm} \rangle[/tex]
    for the ground state

    [tex]\Psi_{100}(r,\theta,\phi)=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}[/tex]
    gives a divergent result due to the Coulomb potential. The same evidently happens with all other states of hydrogen as well. I don't know how to interpret this result. Is acceleration not a good observable in quantum mechanics?
  2. jcsd
  3. May 19, 2013 #2


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    You should get a finite value if you take into account that the proton radius is finite. That does not help for positronium, of course.
    Alternatively, you could consider <|a|>.
  4. May 19, 2013 #3
    Thanks mfb!

    I thought about assigning the problem to the Coulomb potential, but as you say, it does not save the case of a "point nucleus" (e.g. positronium).

    You are right, [itex]\langle |\hat{a}| \rangle[/itex] gives a finite value. But it seems strange that the RMS expectation cannot be calculated. I wonder if this reflects some deeper property of the "acceleration operator", or just misuse of the Coulomb potential in ordinary QM, etc.
  5. May 19, 2013 #4
    You would have the same problem if, say, you wanted to calculate the expectation value of 1/|x| for the 1D harmonic oscillator ground state. The expectation value really is infinite. I don't think this indicates any sort of pathology (it does suggest that in practice no apparatus can really measure the quantity 1/|x|, which is sensible: any real apparatus will have a finite position resolution which will cut off the divergent integral).
  6. May 19, 2013 #5
    Is there some other way to differentiate an operator?
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