# Square well potential

1. Jul 9, 2015

### Dazed&Confused

1. The problem statement, all variables and given/known data
Consider a particle in a square well potential:

$$V(x) = \begin{cases} 0, & |x| \leq a \\ V_0, & |x| \geq a \end{cases}$$

We are interested in the bound states i.e. when $E \leq V_0$.

(1) Show that the even solutions have energies that satisfy the transcendental equation $$k \tan ka = \kappa$$ while the odd ones will have energies that satisfy $$k \cot ka = -\kappa$$ where $k$ and $i\kappa$ are real and complex wave numbers inside and outside the well respectively. Note that $k$ and $\kappa$ are related by $$k^2 + \kappa^2 = 2mV_0/\hbar^2$$

(2) In the ($\alpha = ka, \beta = \kappa a$) plane, imagine a circle that obeys the above. The bound states are then given by the intersection of the curve $\alpha = \tan \alpha = \beta$ or $\alpha \cot \alpha = -\beta$ with the circle.

(3) Show that there is always one even solution and that there is no odd solution unless $V_0 \geq \hbar^2 \pi^2 /8ma^2.$ What is $E$ when $V_0$ just meets this requirement?

2. Relevant equations

3. The attempt at a solution

I was able to solve part (1) by applying the boundary conditions on $e^{-\kappa|x|}$ for $|x| \geq a$ and $e^{ikx} + e^{-ikx}$ and $e^{ikx} - e^{-ikx}$.

I 'sort' of understand why the solutions wouldn't always exist for part (3) after plotting the curve $\alpha \cot \alpha$ as it is only positive, and we are looking for solutions that are negative. I'm not quite sure how to find the minimum value of $V_0$.

2. Jul 9, 2015

### Dr. Courtney

The relevant equation (TISE) should be shown.

The ground state is the even solution.

The first odd solution is the 1st excited state, which requires a minimum depth to the well.

3. Jul 9, 2015

### Dazed&Confused

I assumed I should use the relations which came about from the TISE?

I realised I plotted the incorrect graphs. If the root of $\alpha \cot \alpha$ is at $\pi/2$, then that is the minimum value of the radius. Thus for $k$ and $\kappa$ it is $\pi/2a$, which means that $2mV_0/\hbar^2 = \pi^2/4a^2$.

Last edited: Jul 9, 2015