- #1

doggydan42

- 170

- 18

## Homework Statement

Consider a particle of mass m moving in a one-dimensional double well potential

$$V(x) = -g\delta(x-a)-g\delta(x+a), g > 0$$

This is an attractive potential with ##\delta##-function dips at x=##\pm a##.

In the limit of large ##\lambda##, find a approximate formula for the energy difference ##\Delta E > 0## between the ground state and the first excited state. (Hint: Start by finding approximate expressions for ##\xi_{even}## and ##\xi_{odd}## in the limit of large ##\lambda##.)

Express your answer in terms of lambda for ##\lambda##.

$$\frac{\Delta E}{E_0} = $$

## Homework Equations

$$\kappa \equiv \sqrt{-\frac{2mE}{\hbar^2}}, xi \equiv \kappa a, \lambda \equiv \frac{mag}{\hbar^2}$$

$$\lambda_{even}= \frac{\xi}{e^{-2\xi}+1}, \lambda_{odd}= \frac{\xi}{1-e^{-2\xi}}$$

$$E_0 = \frac{\hbar^2}{2ma^2}$$

$$\frac{E}{E_0} = -\xi^2$$

## The Attempt at a Solution

For large even lambda, ##\lambda = \frac{xi}{2} \rightarrow \xi = 2\lambda##. So,

$$\frac{E_{even}}{E_0}=-\xi^2=-4\lambda^2$$

For odd lambda, ##\xi = \lambda*0 = 0##, so

$$\frac{E_{odd}}{E_0}=-\xi^2=0$$

For the change in energe:

$$\frac{\Delta E}{E_0}=\frac{E_{even or odd}}{E_0}-\frac{E_0}{E_0} = \frac{E_{even or odd}}{E_0}-1$$

Frome here I was not sure how to identify whether to use even or odd E.