# Finding the energy eigenfunctions of infinite well with potential step

1. Oct 3, 2014

### deedsy

1. The problem statement, all variables and given/known data
The potential for a particle mass m moving in one dimension is:
V(x) = infinity for x < 0
= 0 for 0< x <L
= V for L< x <2L
= infinity for x > 2L

Assume the energy of the particle is in the range 0 < E < V

Find the energy eigenfunctions and the equation that determines the energy eigenvalues. Don't worry about normalizing the eigenfunctions.

2. Relevant equations

3. The attempt at a solution
I believe I've found the equation for the energy eigenvalues....
in the region 0 < x < L the wave equation has the form:
$\psi_1(x) = e^{ikx} + Be^{-ikx}$ where $k = \sqrt{2mE}/\hbar$
for L < x < 2L, the wave equation is:
$\psi_2(x) = Ce^{-qx} + De^{qx}$ where $q = \sqrt{2m(E-V)}/\hbar$

applying the boundary condition $\psi_1(x=0) = 0$,
I find B = -1 and the wave equation can be rewritten and simplified as $\psi_1(x) = A' sin(kx)$

applying the boundary condition $\psi_2(x=2L) = 0$,
I find $D = -C e^{-4qL}$ and the wave equation can be simplified to $\psi_2(x) = C' sinh[q(x-2L)]$

Now, equating these two equations at x = 0 and their derivatives at x=L, and then dividing the two results gave me a solution i can use to solve for E, the energy eigenstates: $q tan(kL) = -k tanh(qL)$

But now, I'm not sure how to go about finding the energy eigenfunctions... For the plain infinite well, it was easy because you just had one equation, sin(something) = 0 , so something = n*pi. Then you just rewrote the equation with n in it...
I'm thinking this involves another application of the boundary conditions, but I don't see how quantized energy functions will arise from equating my solutions at x = L...

2. Oct 4, 2014

### vela

Staff Emeritus
$\psi_1$ and $\psi_2$ are not wave equations; they're solutions to the wave equation.

Since E<V and you want q to be real, it should be $q = \sqrt{\frac{2m(V-E)}{\hbar^2}}.$

You've found them for the most part. You have
$$\psi(x)=\begin{cases} A' \sin kx & 0 \le x \le L \\ C' \sinh q(x-2L) & L < x \le 2L \end{cases},$$ where $k$ and $q$ satisfy the relationship you found above. You also have $k^2+q^2 = k_0^2$ where $k_0^2 = \frac{2mV}{\hbar^2}.$ You can't solve for $k$ and $q$ exactly. You should write C' in terms of A'.

3. Oct 4, 2014

### deedsy

thanks vela - Just to make sure I understand what is going on in this problem: so to find the different energy eigenfunction equations for the different energy states, I solve my energy eigenvalue equation for E (my book uses a plotting and intersection method for the infinite square well), the number of intersections = my number of different every states? So the energy eigenfunction equations would all be the same except substituting in the different E values for each one? And these energy eigenfunction equations take two different forms over the two regions - before and in the potential well?

Other than that, once I clean up my two equations writing C' in terms of A' and relating k and q, that should just be the answer.

Last edited: Oct 4, 2014
4. Oct 4, 2014

### vela

Staff Emeritus
That's correct.

5. Oct 4, 2014

### deedsy

great- thanks