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2 Level Finite Potential Well

  1. Oct 10, 2014 #1
    Consider a particle of mass m subject to the following potential function (taking Vo and L
    to be positive):
    V (x) =
    40 Vo if x < 0;
    0 if 0 < x < L/2;
    2 Vo if L/2 < x < L;
    40 Vo if x > L.

    (a) Derive the transcendental equation for energy eigenstates having an energy
    2 Vo < E < 40 Vo.
    To simplify the mathematics and numerics, for the rest of this question take h = 1, L = 1,
    m = 1, and choose (in this system of units) Vo = 1.

    2. Relevant equations


    3. The attempt at a solution
    I've found the solution for the wavefunction in all areas.
    For x<0,

    ## \psi_1(x) = A exp(k_1 x)+B exp(-k_1 x) ##

    Using the condition that ##psi(x)##=0 when x approaches infinity,

    ##\psi_1(x) for x<0 = A exp(k_1 x)##

    For region 2, (0<x<L/2), solution is oscillatory as E>V(x), so

    ## \psi_2(x) = \alpha sin(k_2*x) + \beta cos(k_2*x).##

    Applying boundary conditions: at x=0, ## \psi_1(x)=\psi_2(x)##, and the derivatives of## (\psi_1)/\psi_2 ## must also be equal at x=0.

    ##A exp(0) = \alpha sin(0)+\beta cos(0)
    k_1 A exp(0) = \alpha cos(0) -k_2 \beta sin(0)##
    ##\Rightarrow \beta = A,\alpha = {\frac{k_1}{k_2}}A##

    ## \Rightarrow \psi_2(x) = {\frac{k_1}{k_2}}A sin(k_2 x) + A cos(k_2 x).##

    For region 3, (L/2<x<L), psi is also oscillatory as E>V(x), so

    ##\psi_3(x) = C sin(k_3 x)+D cos(k_3 x)##

    Again, I applied the boundary conditions, where ##\psi_2(x=L/2) = psi_3(x=L/2)##, and the respective derivatives at x=L/2, which produced two equations. Dividing one equation by the other allows the 'A' coefficients to be cancelled.

    ## \psi_2(L/2) = \psi_3(L/2)\Rightarrow {\frac{k_1}{k_2}}A sin({\frac{k_2 L}{2}}) + A cos({\frac{k_2 L}{2}}) = C sin({\frac{k_3 L}{2}}) + D cos({\frac{k_3 L}{2}})##(1)

    Taking derivatives and letting x = L/2;

    ##{\frac{k_1}{2}} A cos({\frac{k_2 L}{2}}) - {\frac{k_2}{2}} A sin({\frac{k_2 L}{2}) = {\frac{k_3}{2}} C cos({\frac{k_3 L}{2}}) +{\frac{k_3}{2}} D sin({\frac{k_3 L}{2}})}## (2)

    Dividing ##{\frac{(1)}{(2)}} \Rightarrow {\frac{{\frac{k_1}{k_2}} A sin({\frac{k_2 L}{2}}) + A cos({\frac{k_2 L}{2}})}{{\frac{k_1}{2}} A cos({\frac{k_2 L}{2}}) - {\frac{k_2}{2}} A sin({\frac{k_2 L}{2})}}} = {\frac{C sin({\frac{k_3 L}{2}}) + D cos({\frac{k_3 L}{2}})}{{\frac{k_3}{2}} C cos({\frac{k_3 L}{2}}) -{\frac{k_3}{2}} D sin({\frac{k_3 L}{2}})}}(3)##

    For region 4, x>L, psi_4(x) is decaying, and applying the boundary condition as in region 1,

    ##\psi_4(x) = G exp(-k_1 x).##

    I let ##psi_3(x=L) = psi_4(x=L)##, and the respective derivatives equal each other when x=L. This produced another two equations.

    ##C sin(k_3 L) + D cos(k_3 L) = G exp(-k_1 L)## (4)

    ##k_3 C cos (k_3 L) - k_3 D sin(k_3 L) = -k_1 G exp(-k_1 L)## (5)

    Dividing (5) by (4) gives:

    ##{\frac{k_3 C cos (k_3 L) - k_3 D sin(k_3 L)}{C sin(k_3 L) + D cos(k_3 L)}} = -k_1## (6)

    Dividing one by the other allows the 'G' coefficients to be removed as above.

    I'm then left with a mess of sines and cosines with the factors C and D. I can't seem to get rid of the C and D coefficients to get the required transcendental energy solution. Most of my efforts has been trying to plug in (6) into (3), to cancel C and D, but I've had no luck so far.

    Am I missing something here? Any help would be much appreciated. I was thinking that perhaps as ## 2 V_0 < E< 40 V_0,## the wave function won't be found below ## E = 2 V_0##, but I don't think this is correct.

    Thanks very much.
     
    Last edited: Oct 10, 2014
  2. jcsd
  3. Oct 15, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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