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(adsbygoogle = window.adsbygoogle || []).push({});

[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?

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# Acceleration operator and the electron in a hydrogen atom

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