# 2 Level Finite Potential Well

1. Oct 10, 2014

### IainH

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. Oct 15, 2014