Quantum particle moving on the surface of cylinder I have a problem with a spinless particle moving on the lateral surface of a cylinder of radius $r$.
If no Hamiltonian is given, is $H=\frac{p^2}{2m}$ only?
What are the Hamiltonian's eigenfunctions and eigenvalues?
Do I have to write down the Laplacian in cylindrical coordinates?
I've tried as follows:
$H=\frac{p^2}{2m}$ so in cylidrical coordinates the TISE is:
$\frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial z^2}\psi(z,r)+\frac{L_z^2}{r^2}\psi(z,r)=E\psi(z,r)$ since $L_z^2=-\hbar^2\frac{\partial^2 }{\partial \phi^2}$ the Hamiltonian in separable and I can find a solution of the form $\psi(z,r)=F(z)R(r)$ and I don't know what to do next.
 A: Perhaps I'm misunderstanding but it seems the question is simply 'what are the eigenfunctions of a quantum particle confined to the (lateral) surface of a cylinder?'. There is no need to add an infinite potential to the Hamiltonian in order to confine the particle to the surface. For instance, when you solve the problem of a free particle in a 'box' in 2D, one does not add an infinite potential to confine the particle along the 3rd dimension. For simplicity, I will take the radius of the cylinder to be $R=1$. I'll also be assuming that question is about the scenario in which the cylinder is infinitely long.
There is no motion along the radial direction $r$, but there is motion along $z$ and $\phi$. The Hamiltonian given by Salmone is almost correct. It is ($\hbar=1$)
\begin{equation}
 H = -\frac{1}{2m} \frac{\partial^2}{\partial z^2}  -\frac{1}{2m} \frac{\partial^2}{\partial \phi^2},
\end{equation}
which is clearly separable. The eigenfunctions are then given by $\psi(z,\phi) = Z (z) \Phi(\phi)$, where $Z$ and $\Phi$ are the states of a free particle, i.e.,
\begin{align}
Z(z) &= e^{ikz}\\
\Phi(\phi) &= e^{ik\phi},
\end{align}
where I'll let you worry about the correct normalisation of these functions.
For $Z$, the values of $k$ are any real number, while for $\Phi$ we have $\Phi(\phi+2 \pi) = \Phi (\phi)$, which implies that $k = 0, \pm1,\pm2, \dots$.
Then the corresponding eigenvalues for either $Z$ and $\Phi$ are simply $E = k^2/2m$ (where $Z$ and $\Phi$ have different constraints on the possible values of $k$)
Hope this helps.
A: Quantum Particle trapped in a Cylindrical Well
I'll solve this problem more or less 'ab initio''.
Since as $U=0$ inside the cylinder and $+\infty$ outside of it, the TISE is (with $u$ - which saves keystrokes - the wave function):
$$-\frac{\hbar^2}{2m}\nabla^2 =Eu$$
With the $\nabla^2$-operator in cylindrical coordinates:
$$-\frac{\hbar^2}{2m}\left(\frac{u_r}{r}+u_{rr}+\frac{u_{\varphi \varphi}}{r^2}+u_{zz}\right)=Eu$$
The Boundary Conditions, because $U=\infty$ outside of the cylinder ($R_o$ outer radius, $L$ length of cylinder):
$$u(R_o,\varphi, z)=0$$
$$u(r,\varphi,0)=u(r,\varphi,L)=0$$
We assume that:
$$u(r,\varphi,z)=R(r)\Phi(\varphi)Z(z))$$
Now carry out separation of variables:
$$\frac{u_r}{r}+u_{rr}+\frac{u_{\varphi \varphi}}{r^2}+u_{zz}=-\frac{2mE}{\hbar^2}u\tag{1}$$
$$\frac{\Phi Z R'}{r}+\Phi Z R''+\frac{RZ\Phi''}{r^2}+\Phi R Z''=-k^2u$$
$$\frac{R'}{rR}+\frac{R''}{R}+\frac{\Phi''}{r^2\Phi }+\frac{Z''}{Z}=-k^2$$
$$\frac{R'}{rR}+\frac{R''}{R}+\frac{\Phi''}{r^2\Phi}+k^2=\frac{Z''}{Z}=-m^2$$
Here's the first ODE:
$$\frac{Z''}{Z}=-m^2\tag{2}$$
$$\frac{R'}{rR}+\frac{R''}{R}+\frac{\Phi''}{r^2\Phi}+k^2=-m^2$$
$$\frac{rR'}{R}+\frac{r^2R''}{R}+\frac{\Phi''}{\Phi}+k^2r^2=-m^2r^2$$
$$\frac{rR'}{R}+\frac{r^2R''}{R}+k^2r^2+m^2r^2=-\frac{\Phi''}{\Phi}=\ell^2$$
And the second ODE:
$$\frac{\Phi''}{\Phi}=-\ell^2\tag{3}$$
$$\frac{rR'}{R}+\frac{r^2R''}{R}+k^2r^2+m^2r^2=\ell^2$$
Finally:
$$r^2R''+rR'+\left((k^2+m^2)r^2-\ell^2\right)R=0\tag{4}$$

From $(2)$ and the first BC we glean:
$$Z_n(z)=A\sin m_nz$$
$$m_n=\frac{n\pi }{L}\text{ for }n=1,2,3,...$$
$(3)$ looks very similar to $(2)$ but $\varphi$ is an angle, so it has $2\pi$ periodicity. We get:
$$\Phi_{\ell}(\varphi)=B e^{i \ell \varphi}$$
$$\ell=0,1,2,3,...$$
$(4)$ solves to:
$$R(r)=C_1 J_{\ell}\left(r\sqrt{k^2+m^2}\right)+C_2  Y_{\ell}\left(r\sqrt{k^2+m^2}\right)$$
Because:
$$Y_{\ell}\to -\infty\text{ for }r \to 0 \Rightarrow C_2=0$$
$$R(r)=C_1 J_{\ell}\left(r\sqrt{k^2+m^2}\right)$$
So using the first BC:
$$0=J_{\ell}\left(R_o\sqrt{k^2+m^2}\right)$$
If we call the roots of the Bessel function $J_{\ell}$ the values $\rho_{\ell,m}$, then:
$$\rho_{\ell,m}=R_o\sqrt{k^2+m^2}$$
$$\rho_{\ell,m}^2=R_o^2(k^2+m^2)$$
$$k^2=\frac{2mE}{\hbar^2}$$
$$E_{\ell,n}=\hbar^2\frac{\rho_{\ell,m}^2L^2-n^2 \pi^2 R_o^2}{2m R_o^2 L^2}$$
You can find the values of  the $\rho_{\ell,m}$ here.
We get the energy levels. Note that these are not nicely geometrically spaced like for the $\text{1DPB}$ or the hydrogen atom.

If you're interested in probability densities, the solutions to $(2)$, $(3)$ and $(4)$ would need to be normalised on their respective domains (not shown).
