# Quantum well combination

1. Apr 8, 2014

### thatguy14

1. The problem statement, all variables and given/known data
The question is attached as a picture. Note: if someone would prefer I type it out I can.

2. Relevant equations
Schrodingers equation

3. The attempt at a solution

PART A
I am pretty sure I got the well right. It looks like a finite well inside an infinite well. I have attached a crudely drawn one called "well.png".

PART B
So I know that because the potential well is even (symmetric about 0) we can choose the eigenfunctions to have some definite parity. Then since we are dealing with bound states where the energy levels are non-degenerate. Then can we say that the lowest state should always be of even parity and every alternating state should be even also? and the ones inbetween are odd? Does this apply to the entire well? I'm not sure I understand this

PART C
This is where I lose all confidence in what I know. So there are 2 cases, one where E < V0 and E > V0. For the E<V0 the solutions would look exactly like the finite well but with the added restriction that the probability of the particle being at b must be 0. For E> V0 I am not really sure... It must be bound right? But how does it behave with the drop in potential?

I'll leave it there for now. Any help would be greatly appreciated

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2. Apr 9, 2014

### thatguy14

Can anyone help please? If I am being unclear please let me know

3. Apr 10, 2014

### thatguy14

I did some work on c, hopefully someone can give me a reply to this to make sure I am on the right path.

so I said that the solution is similar to the finite potential well. In region A which is between -b and -a the wavefunction is of the form:

$\psi$ = Aexp(kx) + Bexp(-kx) (I know that there are i's here to indicate its complex but I am skipping a few steps).

Then I used a boundary condition that at x = -b, $\psi$=0. This led to

$\psi$ = 0 =Aexp(-k*b)+Bexp(k*b)
then I solved for B

Bexp(kb) = -Aexp(-kB)
B = -Aexp(-2kb)

Then I plugged this back into the wave function

$\psi$ = Aexp(kx) - Aexp(-2kb)exp(-kx)
$\psi$ = A(exp(kx)-exp(-k(2b+x)))

Is this the correct approach? I know I have lots more to do but I wanted to make sure this was a step in the right direction. This has the behavior that I want, i.e., that it goes to 0 at b