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Square well potential

  1. Jul 9, 2015 #1
    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. jcsd
  3. Jul 9, 2015 #2
    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.
     
  4. Jul 9, 2015 #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: Jul 9, 2015
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