# Homework Help: Quantum harmonic oscillator coupled to electric potential

1. Sep 17, 2016

### rock_pepper_scissors

1. The problem statement, all variables and given/known data

In $1+1$-dimensional spacetime, two objects, each with charge $Q$, are fixed and separated by a distance $d$.

(a) A light object of mass $m$ and charge $-q$ is attached to one of the massive objects via a spring of spring constant $k$. Quantise the motion of the light charge and obtain the ground state energies and wavefunctions. What approximations did you use, if any?

(b) The same light object of mass $m$ and charge $-q$ is now attached to both the massive objects via springs of spring constants $k_1$ and and $k_2$. Repeat the analysis in (a) for this new system.

2. Relevant equations

3. The attempt at a solution

(a) To quantise the motion of the light charge, we need to use the potential energy of the system in the Schrodinger equation to obtain the wavefunctions and the energy spectrum of the theory. Therefore, the first objective is to find the potential energy of the theory.

There are two interactions in the theory. The first is due to the spring which connects the object with charge $Q$ and the object with charge $-q$. The potential energy due to the spring is $\frac{1}{2}kx^{2}$.

The second interaction is due to the charges of the objects. Now, Gauss's law in one dimension gives the electric field as

$\int \vec{E}\ \cdot{d\vec{A}}=\frac{-q}{\epsilon_{0}} \implies 2E = -\frac{q}{\epsilon_{0}} \implies E = -\frac{q}{2\epsilon_{0}}$,

since the electric field in integrated over the two endpoints of the Gaussian "surface" on the one spatial dimension. Letting the distance from the equilibrium position of the charge $-q$ to the fixed position of one of the charges $Q$ be $x_{eq}$ and the displacement (from the equilibrium position) of the charge $-q$ in the direction of the other charge $Q$ be $x$, the potential energy due to the object of charge $-q$ is

$-\frac{qQ}{2\epsilon_{0}}[(w_{eq}+x)+(d-x-x_{eq})]=-\frac{qQd}{2\epsilon_{0}}$.

Therefore, the electric potential energy is a constant and can be ignored.

Am I correct in my analysis so far?

2. Sep 17, 2016

### Simon Bridge

Check by superposition... ie sketch the component potentials and add them graphically.

Lets see if I follow:
Arrange coordinates so that charge Q is at the origin and another charge Q is at x=d, let -q (mass m) be attached to the origin.
You want to solve: $\big[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)\big]\psi(x) = E\psi(x)$
...using the known results for the forces of point charges, and for a spring, then the net force on the mobile charge is given by:
$$-\frac{d}{dx}V(x)= k(x-x_0) + \frac{Qq}{4\pi\epsilon_0(d-x)^2} - \frac{Qq}{4\pi\epsilon_0x^2}$$
... where $x_0$ is the un-stretched length of the spring, and I am assuming that the mobile mass m is always between the two fixed charges (so 0<x<d). Notice that the description also works if the mobile mass is to the left or right of the fixed charges - is it free to pass through the position of a fixed charge?

This 1st order DE can be solved to get V(x)... or you could apply the superposition principle to the known component potentials. I have a feeling the point of the exercise is to get you to think about this.

Whichever the way, $x_0$ is not given.
It is possible that they want $x_0=0$ ... or maybe $x_0=d/2$ to center the imposed HO potential.
Either of these make sense considering part (b) would shift the equilibrium position off-center according to the ratio of $k_1$ and $k_2$... though, if $x_0=d/2$ was intended, I'd have thought part (a) would also use two springs. OTOH: if $x_0=0$ were intended, then the wording is still off.

Note: If $x_0=d/2$ and d is quite big, then the electric potential about the equilibrium is very flat, and you can use perturbation theory to modify the HO potential ground state. If you have just covered perturbation theory, this is a likely focus for the exercise.

3. Sep 24, 2016

### rock_pepper_scissors

This is in $1+1$-dimensional space, so Coulomb's law in three dimensions does not apply.

My application of Gauss's law in $1+1$-dimensional spacetime is the correct way to go.

4. Sep 24, 2016

### Simon Bridge

Oh I see, this is some abstract topology course so 1+1D means exactly that and is not supposed to be taken to mean you are constraining motion to 1D of space.
The rest of the commentary still stands.

You could, then, check from the electric potential in 1D.