Perturbation Theory for a ring in an Electric Field 
A particle of mass $m$ move on a circular ring  of radius $a$. The only variable of the system is the azimuthal angle, which we will call $\varphi$. The state of the system is described by a wave function $\psi(\varphi)$ that must be periodic,
  $\psi(\varphi + 2\pi) = \psi(\varphi)$ and normalized.
Now assume that the particle has a charge $q$ and that it is placed in a uniform electric field $ε$ in the $x$-direction. We must therefore add to the Hamiltonian the perturbation
  $$\delta H = −q\epsilon a \cos \varphi$$
Calculate the new wave function of the ground state to first order in $ε$. Use this wave function to evaluate the induced electric dipole moment in the $x$-direction: $\langle\psi|q_x|\psi\rangle $. Determine the proportionality constant between the dipole moment and the applied field $ε$. This proportionality constant is called the “polarizability” of the system.

My problem is when I'm trying to estimate the first correction
$$E_1=\langle \psi_0|−q\epsilon a \cos \varphi|\psi_0\rangle.$$
It is coming out zero so I don't understand why the wave function of the ground state will change?
 A: This is a simple exercise in properly applying time-independent perturbation theory to the eigenvalue equation 
$$
\frac{d^2\psi_n}{d\varphi^2}  +\frac{2E_n ma^2}{\hbar^2} \psi_n(\varphi) =-\epsilon\frac{2 m a^3 q}{\hbar^2} (\cos \varphi ) \psi_n(\varphi).
$$
All you have to do is substitute 1st order approximations to the eigenfunction and eigenvalue, something like $\psi_n(\varphi) = \psi_n^{(0)}(\varphi) + \epsilon \psi_n^{(1)}(\varphi)$ and $E_n = E_n^{(0)} + \epsilon E_n^{(1)}$, and then separate terms corresponding to different powers of $\epsilon$. 
If you do this correctly you'll find that the equation for the 1st order correction to the eigenfunction turns out to be 
$$
\frac{d^2\psi_n^{(1)}}{d\varphi^2} + \frac{2E^{(0)}_n ma^2}{\hbar^2} \psi_n^{(1)} =- \frac{2ma^2}{\hbar^2} \left( a q \cos \varphi + E^{(1)}_n\right) \psi_n^{(0)} 
$$
It's not hard to see that even if both $E_n^{(0)}$ and $E_n^{(1)}$ happen to be zero (and I am not saying that they are), the $\psi_n^{(1)}$ correction does not vanish. As a rule, both $\psi_n^{(1)}$ and $E_n^{(1)}$ are determined by the perturbation, but this does not mean that $\psi_n^{(1)}$ is "proportional" to $E_n^{(1)}$.   
A: Let $\epsilon$ point towards positive x. 
Then potential energy according to electrostatic force is proportional to $(a \cos \varphi)$ only. 
Schrödinger's equation states $$
-\frac{\hbar^2}{2m} \nabla^2 \psi -(a \cos \varphi) \epsilon q \psi(\varphi) =E_n \psi.
$$ According to your problem, we choose $\varphi$ to be the (only) canonical variable. And replace Laplacian with cylindrical expression, keeping azimuthal term only, for others all vanish. $$
-\frac{\hbar^2}{2m} \cdot \frac{1}{a^2} \frac{d^2}{d\varphi^2} \psi(\varphi) -(a \cos \varphi) \epsilon q \psi(\varphi) =E_n \psi(\varphi). 
$$ Or, $$
\psi'' +\frac{2E_n ma^2}{\hbar^2} \psi =-\frac{2\epsilon m a^3 q}{\hbar^2} (\cos \varphi ) \psi(\varphi).
$$ The homogeneous part is a sinusoical with frequency $$
\frac{a \sqrt{2E_n m}}{\hbar}
$$ This should equal to $n$, as is the usual "particle in the box", but only now circular. So $$
E_n =\frac{n^2\hbar^2}{2ma^2}.
$$ Resulting unperturbed eigenfunctions $$
\psi_{n,s}^{(0)} (\varphi) =\sin n\varphi \\
\psi_{n,c}^{(0)} (\varphi) =\cos n\varphi
$$ Since we want to perturb $$
\frac{2E_n ma^2}{\hbar^2} \leftarrow \frac{2E_n ma^2}{\hbar^2} +\frac{2\epsilon m a^3 q}{\hbar^2} \cos \varphi
$$ Or, speaking proportionally, $$
E_n \leftarrow E_n +\epsilon a q \cos\varphi
$$ Now, $\sqrt{E_n} \propto n$, and alteration added to $E_n$ has half effect on $n$ by expansion. We have $$
\psi_{n,s}^{(1)} (\varphi) =\sin \left( 1 +\frac{\epsilon a q}{2E_n} \cos\varphi \right) n \varphi \\
\psi_{n,c}^{(1)} (\varphi) =\cos \left( 1 +\frac{\epsilon a q}{2E_n} \cos\varphi \right) n \varphi
$$ For brevity, $$
A :=\frac{\epsilon a q}{2E_n}
$$ Consider $$ \begin{align}
&[\sin^2 (1+A \cos \varphi) n\varphi ] \cdot \cos \varphi \\
=&[\sin n\varphi \cdot \cos (An\varphi \cos \varphi) +\cos n\varphi \cdot \sin (An\varphi \cos \varphi) ]^2 \cdot \cos \varphi \\
\approx& [\sin n\varphi \cdot 1 +\cos n\varphi \cdot An\varphi \cos \varphi]^2 \cdot \cos \varphi \\
\end{align} $$ Where the fact that $\epsilon \ll 1$ is assumed, and only expanded to the 1st order as asked. 
If you take $n=0$, then this is 0. And $\langle \psi_{0,s}^{(0)} \mid \delta H \mid \psi_{0,s}^{(0)} \rangle =0$, and similarly $\langle \psi_{0,c}^{(0)} \mid \delta H \mid \psi_{0,s}^{(0)} \rangle$. Really?
