# Are there non-constant potentials that result in eigenstates of the Hamiltonian that are all plane waves?

It is commonly known that the eigenstates to the Hamiltonian of a constant potential are plane waves, aka

$$V(r) = V_0 \Rightarrow H\psi = n \text{ with } \psi = \exp\left(\frac{ip}{\hbar}x\right)\exp\left(-i\frac{E}{\hbar}t\right)$$

But is there another $$V(r)$$ for which this holds?

I suppose the question can be simplified to: is a Hamiltonian uniquely defined by its eigenstates, because if this is the case then the answer should be no, there are no other hamiltonians with those eigenstates.

Plane waves are eigenstates of the momentum operator. Another observable can have the same eigenstates only if it commutes with the momentum operator (see this answer for more details).

Now, the momentum operator generates translations, so the Hamiltonian commutes with it if it is invariant under translations. The kinetic part of the Hamiltonian is invariant (i.e. it commutes with the momentum operator). The potential in general not. The only potential that commutes with the momentum operator is the flat potential.

In summary: if $$H$$ has to have the same eigenstates as $$p$$, then you must have that $$[H,p]=0$$. This is equivalent to $$[p^2/2m+V,p]=[V,p]=0,$$ which tells you that $$V=\mathrm{const}$$.

• Accepted this answer as it takes a simpler (to me) route that is similarly general. – Robin Dorstijn Jul 25 '20 at 11:46
• It is a simpler answer, but it's not as general. What if you have a Hamiltonian with more complicated eigenstates than simple momentum eigenkets? – catalogue_number Jul 26 '20 at 5:33

No, there are no other potentials with the same eigenstates. (set $$\hbar = m = 1$$ for clarity)

Suppose there is another Hamiltonian $$\hat{H}'$$, such that the eigenstates of $$\hat{H}$$ and $$\hat{H'}$$ are the same, the complete, orthogonal set $$| \psi \rangle$$ (obtained via the spicy spectral theorem). Then, we know that $$[H, H'] = 0$$, since $$\hat{H} \hat{H}' |\psi \rangle = E' \hat{H} |\psi \rangle = E' E |\psi \rangle$$. Neat!

Then we can do this: (Note here that $$V$$ is an analytic function, and by "$$V(\hat{x})$$", I mean $$\sum_{k=0}^\infty \frac{1}{k!} \frac{\partial V(0)}{\partial x} \hat{x}^k$$ $$0=[\hat{H},\hat{H}'] = [\frac{1}{2}\hat{p}^2 + V, \frac{1}{2}\hat{p}^2 + V'] = [\frac{1}{2}\hat{p}^2, V'] + [V, \frac{1}{2}\hat{p}^2]\\ \Rightarrow [\hat{p}^2, V(\hat{x})] = [\hat{p}^2, V'(\hat{x})]$$ That's pretty big, since that translates (after a bit of algebra, involving expanding V in a Taylor series and using $$[x, p] = i$$ ) to $$-2i \frac{\partial V}{\partial x}\hat{p} - \frac{\partial^2 V}{\partial x^2} = -2i \frac{\partial V'}{\partial x}\hat{p} - \frac{\partial^2 V'}{\partial x^2}$$

By substituting in e.g. momentum eigenstates, this condition begins to look like $$\partial_xV'(x) = \partial_xV(x)$$ By simply integrating this here equation, we see that $$V-V' = constant$$.