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 $\theta$?
Best 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.