Is it possible that a particle has a state with both well defined energy and well defined position?

Question

Heisenberg uncertainty principle states that a state cannot have both well defined position and well defined momentum. However, is it possible that a particle has well defined position and well defined energy? And also, how would I prove that it exists or doesn't exist?

Answer 1

The general uncertainty relation for quantum-mechanical observables $A$ and $B$, with operators $\hat{A}$ and $\hat{B}$ and uncertainties $\Delta A$ and $\Delta B$, is

$$\Delta A\Delta B\geq|\langle[\hat{A},\hat{B}]\rangle|$$

(to see why this is true, a good explanation is here: How does non-commutativity lead to uncertainty?).

The reason a state cannot have a well-defined position and momentum is because this commutator $[\hat{x},\hat{p}]=i\hbar$ is nonzero, so the product of their uncertainties must always be nonzero. If the commutator of two operators is zero, on the other hand, then the product of their uncertainties can, in fact, be zero, and one can have a state with well-defined values of both $A$ and $B$.

The operator corresponding to energy is the Hamiltonian $\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{x})$. Using some commutator identities and the fact that $[\hat{x},f(\hat{x})]=0$ for any $f$:

\begin{align*} [\hat{x},\hat{H}]&=\frac{1}{2m}[\hat{x},\hat{p}^2]+[\hat{x},V(\hat{x})]\\ &=\frac{1}{2m}\left([\hat{x},\hat{p}]\hat{p}+\hat{p}[\hat{x},\hat{p}]\right)+0\\ &=\frac{i\hbar\hat{p}}{m} \end{align*}

which means the total uncertainty is

$$\Delta x \Delta E\geq\frac{\hbar}{m}\langle p\rangle$$

Therefore, for situations where the expected value of momentum is zero, it is possible to have simultaneous well-defined position and energy. In any other case, it is impossible.

EDIT: After some objections in the comments, I should probably clarify what I mean by "possible."

Consider a particle-in-a-box potential on the segment $[0,a]$. The ground state of this potential is $\psi(x)=\sqrt{\frac{2}{a}}\sin(\frac{\pi x}{a})$. This state has a well-defined energy $E=\frac{\pi^2\hbar^2}{2ma^2}$. Now let $a\to 0$. As we decrease $a$, the wavefunction $\psi$ gets taller and taller, and sharper and sharper, and the energy $E$ gets higher and higher, but the energy still has an uncertainty of zero for any particular $a$. Eventually $\psi$ is going to approach the Dirac delta $\delta(x)$, which has zero uncertainty in position, and the energy $E$ is also still going to have zero uncertainty (though its value diverges). By this limiting process, we can create a state that has well-defined position and energy.

This is what I mean by "possible" - the uncertainty principle does not expressly forbid us from constructing a state like the above with a limiting process as described. Of course, such a state can never be a physical state. Probably the better way of stating this result is, "You can construct a physical state where the product of position uncertainty and energy uncertainty is arbitrarily small; you cannot do the same for position and momentum."

To put my meaning of "possible" even more clearly, the conclusion would read: "No matter how sensitive your measuring devices are, it is always possible to construct a physical state such that your measuring devices will not measure any uncertainty in either position or energy."

Answer 2

I am afraid the answer by @probably_someone doesn't answer the question correctly:

a) First of all, it is of upmost importance to say that the generic Robertson inequality $\left(\Delta A\right)_\psi \left(\Delta B\right)_\psi \geq \frac 12 \left| \left\langle \left[A,B\right]\right\rangle_\psi \right|$ holds iff the (pure) state vector (representative) $\psi\in D_{[A,B]}\subseteq\mathcal H$ on which the expectation values are calculated has finite norm, therefore eigenvectors of the coordinate operator (therefore also eigenvectors in the continuum spectrum of the Hamiltonian) are automatically excluded (see point b) below).

b) Following from a) we have that, if the vector $\psi$ is of finite norm, then $\left(\Delta A\right)_\psi = 0 \Rightarrow \langle A\rangle_\psi \in\sigma_{pp}(A)$. This again excludes the coordinate operator $x$ from discussion, because $\sigma_{pp} (x) = \emptyset$. Therefore, it makes no sense to ask "Is there a state with well defined position (and whatever other observable)?" and invoke the Robertson inequality to extract any meaningful answer.

c) A conclusion to points a) and b) is that, iff there is a vector $\psi\in D_{[x,H]} \subset \mathcal H$ so that $\Delta x$ and $\Delta H$ are well defined, we have that $\Delta x \Delta H \geq 0$ and there may be states of well defined energy (but not well-defined coordinate) with $\Delta H =0$ (true eigenstates of the Hamiltonian operator corresponding to pure point spectral values) but necessarily $\Delta x >0$. This should give NO as the answer to the question in the title.

N.B. The definition I used for "the observable A has a well defined value" is that $\left(\Delta A\right)_{\psi\in\mathcal H} =0$.