# Square well potential

## Homework Statement

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?

## 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##.

## Answers and Replies

Dr. Courtney
Education Advisor
Gold Member
2020 Award
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.

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##.

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