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.


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 $\theta$?

  • 2
    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Commented Feb 1, 2022 at 14:15

1 Answer 1


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.


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