# 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?

• Just because the energy has no correction at first order in $\epsilon$ doesn't mean that the energy eigenstates $|\psi\rangle$ have no correction at first order in $\epsilon$. Your textbook probably describes how the corrections to $|\psi\rangle$ are derived. – Michael Seifert Dec 20 '16 at 15:08

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 $E_n^{(1)}$.
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?
• A 1st order perturbation of the eigenvalue always reads $E_n \rightarrow E_n^{(0)} + \epsilon E_n^{(1)}$, where $E_n^{(1)}$ is constant, like $E_n^{(0)}$, not a function of $\varphi$ as in your $\frac{2E_n ma^2}{\hbar^2} \leftarrow \frac{2E_n ma^2}{\hbar^2} +\frac{2\epsilon m a^3 q}{\hbar^2} \cos \varphi$. – udrv Dec 21 '16 at 7:04