# Why do the boundary conditions change the eigenvalues between unitarily equivalent Hamiltonians?

Let's say we have a particle in a circle (let e.g. $$x = x+1$$) with Hamiltonian

$$H = \frac{1}{2} p^2,$$

Let's also set $$\hbar = 1$$. Now the solution of the Schrödinger equation $$H \psi(x) = E \psi(x)$$ is of the form

$$\psi(x) = A \exp ( i \sqrt{2E_n} x ),$$ where $$E_n = 2 \pi^2 n^2$$ from the condition $$\psi(x) = \psi(x + 1)$$.

Now consider the Hamiltonian

$$H_{\theta} = \frac{1}{2} \left ( p - \frac{\theta}{2 \pi} \right )^2,$$ where $$\theta \in [0, 2 \pi)$$.

Now a solution could be written as $$\psi_{\theta} (x) = A \exp \left ( i \left (\sqrt{2E} + \frac{\theta}{2 \pi} \right) x \right ),$$ since this will bring down a factor of $$+ \frac{\theta}{2 \pi}$$ which will cancel the $$- \frac{\theta}{2 \pi}$$ term in the Hamiltonian. Now imposing the boundary conditions we get $$E_{n, \theta} = \frac{1}{2}\left ( 2 \pi n - \frac{\theta}{2 \pi} \right )^2.$$

Observe that imposing the boundary conditions we have that $$\psi(x) = \psi_{\theta}(x)$$.

I can also show that the Hamiltonians are unitarily equivalent (i.e. $$U^\dagger H_{\theta} U = H$$), since

$$U^\dagger \left ( p - \frac{\theta}{2 \pi} \right ) U = p,$$

where $$U = \exp \left (i \frac{\theta}{2 \pi} x \right)$$.

But then I can also write $$U^\dagger H_{\theta} U \psi(x) = E_{n} \psi(x) \implies H_{\theta} U \psi(x) = E_{n} U \psi(x),$$

yielding that $$H$$ and $$H_{\theta}$$ have the same spectrum, which pretty much seems like a contradiction to me.

On one hand, I get that the boundary conditions give different eigenvalues depending on $$\theta$$. On the other hand, I can show that unitarily equivalent Hamiltonians should yield the same eigenvalues.

Question

Why do I get a contradiction here, and how should I fix it?

Can I argue that the unitary equivalence argument fails since $$U \psi(x)$$ does not adhere to the boundary conditions in general? Or do I have to change the boundary conditions depending on $$\theta$$?

• You claim you can show that the operators are unitarily equivalent, but you didn't actually write down the corresponding $U$. Note that the Weyl relations do not hold on all of $L^2(S^1)$ since otherwise a discrete spectrum of momentum would violate the Stone-von Neumann theorem, so if you want to use an exponential of the position operator this probably won't work. Commented Feb 1, 2022 at 14:15

## 1 Answer

For $$\theta$$ not an integer mutiple of $$2\pi$$, the transformation $$U$$ can not be single-valued. In other words, it can not be a legitimate operator in the Hilbert space of periodic, square-integrable functions.

For any unitary in this Hilbert space, it can not change the "magnetic flux" (if you think about $$x$$ as the coordinate on a ring, then $$\theta$$ can be interpreted as the magnetic flux), except for $$\theta\rightarrow \theta+2\pi$$, which can be achieved by $$U=e^{ix}$$. If you use a $$U$$ that is not single-valued (essentially redefining $$\psi(x)\rightarrow e^{i\theta x/2\pi}\psi(x)$$), then you change the boundary condition of $$\psi(x)$$ from periodic to one that is twisted by the phase $$\theta$$. If you take that into account, the spectrum would be the same.