3
$\begingroup$

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 $\theta$?

$\endgroup$
1
  • 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
    Feb 1, 2022 at 14:15

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.