# Are eigenfunctions of $E$ also eigenfunctions of $p$?

Given that $$\hat{H}\Psi = \hat{E}\Psi$$ and that $$E=\frac{p^2}{2m}$$

Assuming a non-relativistic, system, does this mean that any eigenfunction of Energy is also an eigenfunction of momentum? Or does it work the other way and every eigenfunction of momentum is an eigenfunction of Energy?

If not, where does this break down? It seems reasonable to me that two particles with the same Kinetic Energy and mass would have the same momentum. Even if the momentum is in different directions I think those would be degenerate states and not strictly count. Although I don't really understand why degenerate states are a problem yet.

The point is that any wavefunction of definite energy sounds like it would also have a definite momentum, or vice versa, and I can't think of a way that doesn't make sense.

• $E$ is not an operator, it shouldn't have a hat. The eigenfunctions of the Hamiltonian operator have eigenvalues of energy – Nihar Karve Nov 11 '20 at 6:21

An eigenfunction of $$\hat p$$ is an eigenfunctions of $$\hat p^2$$ but not necessarily the other way round. An eigenfunction of $$\hat p^2$$ can be a linear combination $$\alpha |p\rangle +\beta |- p\rangle$$ with $$\hat p |\pm p\rangle= \pm p |\pm p\rangle$$ and this is not an eigenfunction $$\hat p$$ unless one of $$\alpha, \beta$$ is zero.

@mikestone's answer is perfect.

More broadly, if two operators commute and both have non-degenerate spectra, an eigenstate of one operator would be an eigenstate of the other and vice-versa. However, when the spectrum of any one of the operators is degenerate, this no longer remains the case.

Let's say $$[\hat{A},\hat{B}]=0$$. This translates to $$\hat{A}\hat{B}=\hat{B}\hat{A}$$. Now, consider an eigenstate of $$\hat{A}$$, say $$\vert a\rangle$$. Then, we can observe that $$\hat{A}\hat{B}\vert a\rangle=\hat{B}\hat{A}\vert a\rangle=\hat{B}a\vert a\rangle=a\hat{B}\vert a\rangle$$. Or, $$\hat{A}\big(\hat B\vert a\rangle \big)=a\big(\hat B\vert a\rangle\big)$$. In other words, $$\hat{B}\vert a\rangle$$ is an eigenstate of $$\hat{A}$$ with the eigenvalue $$a$$. But, by stipulation, $$\vert a\rangle$$ is an eigenstate of $$\hat{A}$$ with the eigenvalue $$a$$. This means that if $$\hat{A}$$ has a non-degenerate spectrum then it has to be the case that $$\hat{B}\vert a\rangle=\lambda\vert a\rangle$$ for some (real) scalar $$\lambda$$. Or, in other words, $$\vert a\rangle$$ is an eigenstate of $$\hat{B}$$.

Notice that this argument relies only on the non-degeneracy of the spectrum of $$\hat{A}$$ but not that of $$\hat{B}$$. This means that what we have found is that if two operators commute then an eigenstate of one of the operators is necessarily an eigenstate of the other operator if the other operator's spectrum is non-degenerate. And it would follow that if both of them have non-degenerate spectra then an eigenstate of one operator would be an eigenstate of the other and vice-versa.

• I don't know much about commutators. Are they calculated based on a specific feature of the system like Potential or do these properties hold for a given operator in any situation? – Disgusting Nov 11 '20 at 20:48

Since you don't specify $$\hat H$$, the general answer is a resounding no. If there is a potential, say:

$$V(x) = -\frac k r$$

then the energy eigenstates are definitely not momentum eigenstates.

Any potential other than $$V(x) = c$$ breaks translational invariance, so that eigenstates of energy cannot have definite momentum.

Note that $$V(x) = -\frac k r$$ is spherically symmetric (it's the hydrogen atom), and in that case, the solution are angular momentum eigenstates.

• So symmetries in the potential define what definite qualities Energy Eigenstates have? – Disgusting Nov 11 '20 at 20:45
• Qualities here meaning definiteness of position/momentum/angular momentum/spin... – Disgusting Nov 11 '20 at 20:46
• So if the potential has a symmetry, that would correspond to the energy eigenstate having some other property of it being specifically defined. – Disgusting Nov 11 '20 at 20:47
• Yes: translation symmetry = momentum eigenstates; rotational symmetry = angular momentum eigenstates; time invariance = energy eigenstates; mirror symmetry = parity eigenstates. In QM, this is expressed as the operator in question commuting with the Hamiltonian, as other answers have pointed out, but think they are all talking about free particle problems ($V(x) = 0$). – JEB Nov 12 '20 at 0:20
• I think the other answers are addressing operators in general and this one is addressing systems in general. Both very useful to me as it means there are multiple ways in which my belief was wrong. Does this mean all of the Noether's theorem correspondances have energy eigenstate matchups? (Apart from Energy/Time obviously because apparently that's complicated.) – Disgusting Nov 12 '20 at 0:33