# Electronic component of the Hamiltonian operator and uncertainty principle

This question has to do with the concept of uncertainty principle. The Hamiltonian operator has the electronic component that takes the inverse of the distance between any two electrons. My question is: doesn't this violate the uncertainty principle which says it is impossible to pin down the exact location of a quantum object like electrons. Am I missing something here? And if we are indeed talking about the position vectors of electrons and nucleus, then the construct of Hamiltonian operator is very classical?

I fail to see the connection. The Hamiltonian operator has a well-defined relationship in terms of the momentum and coordinate operators (and also spin) and as of such it has a Hilbert space spectrum of its own. With respect to this we can consider measurements of energies of atomic/molecular levels and discuss whether the expectation values of energy obey certain inequalities or not. So we have "uncertainty principles" involving coordinates and "uncertainty principles" involving Hamiltonians or both, but the fact that $\Delta x \neq 0 \nRightarrow \Delta H~\mbox{is not defined}$.

This "uncertainty principle" issue is nothing more than a myth, it is an intentionally confusing textbook interpretation to some (not so simple) mathematical relationship. Its connection to real life experimental setups is not at all exact, as one is taught/given the impression in textbooks. The Ensemble (Ballentine) interpretation of Quantum Mechanics is even drastic, for example the measurement of $\Delta A$ for a quantum observable which is part of the "uncertainty principle" would require an infinite set of quantum systems identically prepared on which an infinity of observers would try to measure the same quantum observable at the same time and with all the infinity of values they obtain, one (of them) would draw a statistical analysis which would give this quadratic deviation from the mean. No connection to real life here. The naïve interpretation "$\Delta x \Delta p \geq \frac \hbar2$ means that one cannot measure momentum and coordinate precisely" is again a crooked fairytale still, unfortunately, present in the literature...

Now, let's go back to your last question: indeed, atomic/molecular physics in the absence of spin is formulated from a classical Hamiltonian in terms of classical coordinates and momenta (and classical Coulomb interaction) which is "quantized", i.e. brought in the mathematical formalism of Quantum Mechanics [this "quantization" is also a myth, for textbooks typically instruct the reader that any classical system is very easily brought in the quantum domain, all we need to do is use commutators instead of Poisson brackets]. The Xs and Ps become operators subject to these "uncertainty relationships", but they are part of the overall Hamiltonian so that any individual property ("uncertainty") of them is not automatically a property of the Hamiltonian as a whole [think of the harmonic oscillator in 1D. Coordinate and momenta have pure continuous spectrum, the Hamiltonian has a pure point one].

I think we need to break this into a couple issues / areas.

First knowing position with arbitrarily high precision is allowed and does not violate the uncertainty principle. Violation of the uncertainty principle would be knowing the position exactly and still having some knowledge of momentum. The above started mathematically:

$$\Delta x \Delta p => h/2\pi$$

2nd, all of the equations of quantum mechanics involve calculations over precise locations. When you have a probability distribution for the location of an election, you calculate its average position by integrating the probability over all space - that is, for each possible exact position, you multiple that location by the probability of finding the election there. You're always using precise locations in these calculations - the uncertainty manifests in the fact that you need to consider multiple possible locations.