# Finite potential well

1. Dec 10, 2014

### feynwomann

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I've got these solutions to the Schrödinger equation ($-\frac{\hbar} {2m} \frac {d^2} {dx^2} \psi(x) + V(x)*\psi(x)=E*\psi(x)$):
x < -a: $\psi(x)=C_1*e^(k*x)$
-a < x < a: $\psi(x)=A*cos(q*x)+B*sin(q*x)$
x > a: $\psi(x)=C_2*e^(-k*x)$

$q^2=\frac {2m(E+V_0)} {\hbar^2}$ and $k^2=\frac {2mE} {\hbar^2}$

In a previous exercise i showed that in order for the wavefunction to be continuous the following has to be true:
For x = -a: $C_1*e^-(k*a)=A*cos(q*a)-B*sin(q*a)$
For x = a: $C_2*e^-(k*a)=A*cos(q*a)+B*sin(q*a)$
and for the derived wavefunctions:
x = -a: $k*C_1*e^-(k*a)=q*(A*sin(q*a)+B*cos(q*a))$
x = a: $-k*C_2*e^-(k*a)=-q*(A*sin(q*a)-B*cos(q*a))$

I also know that A*B=0. So here's the question:

I'm asked to show that when A = 0, the four equations above kan be reduced to $k=q*tan(q*a). But I'm not even sure where to start. Last edited: Dec 10, 2014 2. Dec 10, 2014 ### stevendaryl Staff Emeritus Sure, you do. When $A=0$, your equations become: 1.$C_1 e^{-ka}= -B sin(qa)$2.$C_2 e^{-ka}=-B sin(qa)$3.$k C_1 e^{-ka}=q Bcos(qa))$4.$-kC_2*e^{-ka}=q Bcos(qa))##
Using equation 1, you know $C_1e^{-ka} = -Bsin(qa)$
So replace $C_1 e^{-ka}$ by $-B sin(qa)$ in equation 3.

It doesn't look like it results in the same formula as you are expecting, so maybe there was a mistake somewhere.

3. Dec 10, 2014

### feynwomann

Oh it totally makes sense now. I forgot that my teacher said there was a mistake in the exercise. It's supposed to be B = 0. I didn't think of dividing the equations (I had actually done the first step, but didn't see how I could take it further). Anyway, thanks for the help!

Last edited: Dec 10, 2014