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Here is how I saw the Hamiltonian being written in one Quantum Mechanics book:

$$\hat{H} = -\frac{\hbar^2}{2m_\mathrm{e}r^2} \left[\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{\sin \theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \right] - \frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}$$

Well, that's all very nice and complicated, but looking at the last term, and seeing the use of $r$ in there,

$$-\frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}$$

creates a small doubt in me - what is this $r$ there?

In electrostatics, this was simply the distance between the two charged particles, and that would mean that this was the distance between the proton and the electron, but isn't the position of the electron non-deterministic? Shouldn't that thing be a probability distribution of the distance instead? And why isn't it squared?

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1 Answer 1

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What is the $r$ there?

$r$ is the same in the potential energy term of the Hamiltonian as in the kinetic energy term of the Hamiltonian.

$r$ is the distance from electron to the proton.

But isn't the position of the electron non-deterministic?

In the generally accepted Copenhagen interpretation of quantum mechanics, the position of the electron is not deterministic. However, expressing the kinetic and potential energy of the the electron at each position in space does not contradict this interpretation. It is only saying, for each point, if the electron is at this point, the energy is _.

Shouldn't that thing be a probability distribution of the distance instead?

The wave function solutions represent probability amplitude in the Copenhagen interpretation. The square of the absolute value of the wavefunction represents a probability distribution.

Why isn't it squared?

Force would be inversely proportion to $r^2$, but potential is inversely proportional to $r$.

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    $\begingroup$ The Copenhagen Interpretation is not "generally accepted", it's merely the interpretation that most physicists are least uncomfortable with. But that doesn't compromise your answer, +1. $\endgroup$ May 16, 2014 at 23:10
  • $\begingroup$ I agree, maybe "generally presented in introductory texts" would have been better $\endgroup$
    – DavePhD
    May 16, 2014 at 23:46

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