# (Hard QM) Consider the double well potential with delta func

1. Mar 30, 2016

### NucEngMajor

1. The problem statement, all variables and given/known data
Consider the double well potential with two wells separated by delta function in middle.

V(x) = V0 > 0 for x<-a and x>a
0 for -a<x<0 and 0<x<a
αδ for x =0
1. Find Bound state energies
2. Find odd solns and their eigenvalue equation. Give solns in graphical form
3. Repeat 2. for even
4. Plot two lowest order even and odd soln
2. Relevant equations
S.E

3. The attempt at a solution
I wrote down BC's noting discontinuity at delta and obtained 6 equations (think they are right?). Prof said to use the fact that the determinant of system must vanish to get energy eigenvalue equation. I don't know how to go about this. I know the other way would be incredibly long and tedious.
Aexp(-la) = Csin(-ka)+Dcos(-ka)
lAexp(-la) = Ckcos(-ka)-Dksin(-ka)
Gexp(-la) = Esin(ka) + Fcos(ka)
-lGexp(-la) = Ekcos(ka) - Fksin(ka)
D=F
k(E-C)=2mαF/h^2 (h is "h bar")

2. Mar 31, 2016

### vela

Staff Emeritus
You have a system of six linear equations with six unknowns (A, C, D, E, F, and G). In matrix form, you'll have $M\vec{x} = 0$ where $\vec{x} = (A, C, D, E, F, G)^T$. If the determinant of $M$ is non-zero, the only solution is the trivial solution $\vec{x}=0$, which isn't what you want, so you need $\det(M)=0$. So start by figuring out what $M$ is.

3. Apr 1, 2016

### NucEngMajor

How does one go about constructing M from this? Not looking for answer, but I quite honestly don't know how to start....

4. Apr 1, 2016

### vela

Staff Emeritus
The same way you convert, for example, the equations
\begin{align*}
x' &= x \cos\theta - y\sin\theta \\
y' &= x \sin\theta + y \cos\theta
\end{align*}
to
$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin \theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

5. Apr 1, 2016

### TSny

NucEngMajor,

You can greatly reduce the number of unknowns by using the fact that your wavefunctions are either even or odd. If you know the wavefunction for x ≥ 0, you automatically know the wavefunction for x < 0. Then, if you want to use the matrix approach, your matrix will be just 3 x 3 instead of 6 x 6.

Last edited: Apr 1, 2016
6. Apr 10, 2016

### NucEngMajor

Ended up using algebra and getting a sensible answer. Thanks