# Wave Function problem

#### OhNoYaDidn't

This was in my Introduction to modern physics exam, but i don't quite know what i should do here... My teacher said there was an easy trick. Can you guys help me?

An electron is described by the following wave function
ψ(x)=(ax+b for 0<x<L,
cx+d for L<x<3L,
0 for x<0 V x>3L)

a) Determine the constants a, b and c, and sketch the graph of Ψ.

3. The attempt at a solution
I tried using the boundary conditions, but everything turns out being zero. Guys, please help me, i'm lost here.

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#### Fightfish

The gist of the problem as you pointed out, is to match the boundary conditions between the different regions. However, I don't see why this should turn out to be zero everywhere. Could you show your work that led you to this conclusion?

#### OhNoYaDidn't

Ok, so applying the boundary conditions of the wave functions.
ψ(0-)= ψ(0+)
ψ(L-)= ψ(L+)
ψ(3L-)= ψ(3L+)
i get
b=0
a=-2c
d=-3L
---
Then i use the continuity of the derivative:

and i get
a = 0 = c = b, only d=-3L.

Is this right? I hope i'm not doing any ridiculous mistake.

#### Fightfish

Ok, so applying the boundary conditions of the wave functions.
ψ(0-)= ψ(0+)
ψ(L-)= ψ(L+)
ψ(3L-)= ψ(3L+)
i get
b=0
a=-2c
d=-3L
d = -3Lc. I presume that is a typo since u managed to arrive at a = -2c.

From this you should be able to figure out the shape of the wavefunction. The values of a, c and d (which are related to each other), can be determined through imposing normalisation.

It is okay for the derivative not to be continuous. This can happen when the potential function itself is discontinuous to begin with. In this case since the potential is not given, we assume that the wavefunction given is not the trivial (zero) solution.

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