Heisenberg's uncertainty principle for eigenfunctions of one of the observables

Question

I am studying the Heisenberg's uncertainty principle and I am trying to apply it when one of the observables has zero uncertainty. So far I know that: $$ (\Delta A)^2(\Delta B)^2 \geq \frac{1}{4} \lvert\langle[A,B] \rangle \rvert^2$$ Let's say that both $\Delta A$ and $\Delta B$ are equal to zero, the observables have a common set of eigenfunctions and their commutator is equal to zero. But what if the quantum system is in a state which diagonalizes only the operator A? Two examples are: A is momentum, B is position and the state we are working with is a plane wave (which diagonalizes only the momentum) or A is the hamiltonian of a harmonic oscillator, B is position and the state is one of the eigenfunctions of the hamiltonian. How do we deal with these cases in order to mantain the validity of the uncertainty principle? Here is my reasoning and I would like to know if it's correct: in the first case we are dealing with a plane wave which is physically impossible to have (even though I would like to have a rigorous mathematical explanation), in the second case the resulting commutator is the operator p and the expectation value in the aforementioned state is zero.

Answer 1

To make sense of the uncertainty relation as you wrote it, the eigenstates must be by assumption normalizable. The plane waves are not normalizable, and this makes your first example moot. (I should add that another assumption is that your operators must be self-adjoint and the commutator must satisfy some additional conditions but these types of arguments are not needed for your example.)

If you have an eigenstate of $H$, then $\langle x\rangle$ will not depend on $t$ since the probability density will be independent of $t$ and so $\langle p\rangle=0$ by Ehrenfest’s theorem. Of course this is fully compatible with the uncertainty relation.

Answer 2

Plane waves can be dealt with straightforwardly enough as the limit of ever-widening Gaussian wavepackets. For the plane wave case, as you widen the wavepacket in position space, you make it narrower in momentum space. If the wavepackets are Gaussian, meaning $\psi(x) \propto \exp(-(x - X)^2/a^2 + i P x)$, the uncertainty principle between position and momentum is actually saturated, and as we increase the scale factor $a$, $\Delta p$ goes to zero and $\Delta x$ goes to infinity in precisely the right way so that $\Delta x \Delta p = \frac{1}{2}$.

You're right about the second case. The right-hand side of the inequality has an expectation value $\langle \rangle$, and in the case you've given, it's zero.