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Square well potential

  • #1

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?

Homework Equations




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

  • #2
Dr. Courtney
Education Advisor
Insights Author
Gold Member
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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
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:

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