Let us consider the Hamiltonian for the hydrogen atom

$$ \hat{\mathcal{H}}_{\mathrm{H}}=\hat{\mathcal{T}}_{\mathrm{N}}+\mathrm{\hat{\mathcal{T}}}_{\mathrm{e}}+\hat{\mathcal{V}}_{\mathrm{Ne}}=-\dfrac{\hbar^{2}}{2m_{\mathrm{N}}}\nabla_{\mathbf{R}}^{2}-\dfrac{\hbar^{2}}{2m_{\mathrm{N}}}\nabla_{\mathbf{r}}^{2}-\dfrac{1}{4\pi\epsilon_{0}}\dfrac{e^{2}}{\left|\mathbf{R}-\mathbf{r}\right|} $$

The following discussion can be of course generalized for other systems with point-like interactions.

How the point-like nature of term $\hat{\mathcal{V}}_{\mathrm{Ne}}$ can be related to the non-local nature of the wave function describing the system?

In other words, how can a point-like interaction generate a description for the system in which the particles that interacts are described in a non-local way (by a wave function)?

  • $\begingroup$ What do you mean by asking "how"? It evidently does (as it describes the hydrogen atom correctly, modulo relativistic/spin corrections), and it is obtained from the classical description by canonical quantization. $\endgroup$
    – ACuriousMind
    Commented Nov 25, 2014 at 20:49
  • $\begingroup$ I mean: the electrons see each other as point-like particles (via the term $\hat{\mathcal{V}}_{\mathrm{Ne}}$) but we end up with a non-local description of the system (via the wave function). Of course, as you said, it evidently does. But how can this be conceptually justified? $\endgroup$ Commented Nov 25, 2014 at 21:03
  • $\begingroup$ Without interaction you have two independent waves, i.e., non localized particles. Interaction makes the waves interact or differ from, say, plane waves, but they still are waves with uncertain particle positions. Remember, the total solution is a plane wave of the center of mass and the wave function of the relative motion. Hence, each particle is as non localized as the center of mass, at least. $\endgroup$ Commented Nov 25, 2014 at 21:17
  • $\begingroup$ But the interaction $\hat{\mathcal{V}}_{\mathrm{Ne}}$ it's a point-like interaction not a wave-like one. $\endgroup$ Commented Nov 25, 2014 at 21:22
  • $\begingroup$ Yes, it is "point-like" because it does not involve anything else, but those coordinates. It does not contain an integral over those variables. For example, an atomic form-factor contains an integral over these coordinates. $\endgroup$ Commented Nov 25, 2014 at 21:34

2 Answers 2


It's easy: the argument of the interaction potential, although it depends "locally" on $\mathbf{R}$ and $\mathbf{r}$, is uncertain due to the fact that these variables can take and do take any values in a bound state. Nothing fixes the relative and absolute distances in QM, unlike in CM.

In the Heisenberg picture, these variables are operators (matrices), not just functions of time.


This answer is essentially Vladimis'a answer, but is goes a bit deeper into what the wave-function means. When we say that a quantum object is point-like, we don't say that it has a fixed position. A point-like object of a fixed position is described by a delta-Dirac function. It's not the case of the electron, and proton that you ask about.

Now, the wave-function in your case tell us that the electron may be at the position r_1, or r_2, or r_3, etc., any position where the wave-function is not zero. So, let's consider such a position, and let's give it the name r. Again, r can be r_1, r_2, etc. The same for the nucleus, it can be at R_1, R_2, etc., s.t. let's consider for it an arbitrary position where the wave-function says that the nucleus can be, and let's name it R. Well, for the electron at r and the nucleus at R, we write the Hamiltonian that you mentioned.

In other words, the Hamiltonian is valid for whatever positions r, respectively R, the two particles can take. Now it remains to find the above wave-function. For that, we solve the Schrodinger equation containing this Hamiltonian. The wave-function we get will tell us that that r and R cannot be ANY positions in the space. For some R, the value of r will fall in some allowed range.

I hope that it is clear.

Good luck !


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