# Eigenfunctions of a quantum plane rotor?

I'm trying to determine the energetic levels of a system with Hamiltonian $$H=-\frac{h^2}{2m}\frac{\partial^2}{\partial \phi^2}$$ And the border condition $$\psi(0)=\psi(2\pi)$$ The eigenvalues equation is $$-\frac{h^2}{2m}\frac{\partial^2 \psi}{\partial \phi^2}=E\psi$$ And the general solution of this is in the form $$\psi(\phi)= Acos(k\phi)+Bsin(k\phi)$$ where $$k^2=\frac{2mE}{h^2}$$ Imposing that $$\psi(0)=\psi(2\pi)$$ gives us: $$A=Acos(2k\pi)+Bsin(2k\pi)$$ which is true if $$B=0 \land k\in \mathbb{N}$$, but also for some other values of k if $$B\neq 0$$.
The solutions give as a solution of the differential equations $$\psi(\phi)= N e^{\pm ik\phi}$$ Which has only one constant of integration and gives $$k\in \mathbb{N}$$ when we impose the border condition. Is my approach wrong? Am I missing something?

No there are two in the alternative solution as well. You have one $$N$$ for each exponential, one with the plus sign and one with the minus sign.

What you have is $$\psi(\phi) = A \cos (k \phi) + B \sin (k\phi) = N_+ e^{i k \phi} + N_- e^{-i k\phi}.$$

To answer the comment below: In case you redo the calculation instead with the ansatz $$\psi = N_+ e^{ik\phi} + N_- e^{-ik\phi}$$ you get the same relation between the energy and $$k$$. Using the boundary conditions you can easily arrive at

$$N_+ ( 1 - e^{2 \pi i k} ) + N_- ( 1 - e^{-2\pi i k} ) = 0 .$$

This is true if either $$N_+ = N_- = 0$$ or both of the parenthesis are zero (actually only one of the parenthesis has to be zero, but if one is, then the other automatically is in this case). This gives the same possible values for $$k$$ as you had before.

Note also that you are not in general free to set both of $$N_+$$ and $$N_-$$ to any value since usually you would like to have the wave function normalized in some suitable manner. In this case you probably want to have the integration over the circle of $$| \psi |^2$$ to be unity.

• Thank you, I also thought about that, but then I don't get how I could say that $k$ is always an integer. If I put the boundary condition in that I get that k can be an integer, but there are also other possible solution: for example $A=B=2 \land k=\frac{1}{4}$. That is an eigenfunction of the hamiltoniana, so I can find the system in that energy – E Colnaghi Feb 6 at 12:55
• As I wrote above, you are not free to choose the integration constants freely since these are usually set by normalization. Mathematically you are correct that the thing you suggest is a solution. But when you impose normalization you lose the freedom to choose $A$ and $B$ (or $N_\pm$) freely. You must make sure that they are both not zero. – JezuzStardust Feb 6 at 13:09
• the boundary condition $\psi(0)=\psi(2\pi)$ should be read as $\psi(0)=\psi(2\pi n)$ for any integer $n$. We want the function to be periodic for any integer number of rotations. This means that we demand $$A = A\cos(2\pi n k) + B\sin(2\pi n k)$$ for all integer $n$. This can be achieved only for integer $k$. More generally, $\exp(i n x)$ are a complete basis spanning all the functions which are periodic in $2\pi$, so all solutions can be expanded in terms of these solutions – yu-v Feb 6 at 13:15
• @JezuzStardust Thank you, now it is pretty clear: I have two constant of integration and two condition, the one given by the textbook and normalization. In the solutions I saw the resolver firstly calculated the eigenfunctions and immediatly impose them to be normal, so that the subsequent calculation are easier. – E Colnaghi Feb 6 at 18:49