# Time-independent Schrödinger equation, normalizing

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1. Jun 20, 2017

### WrongMan

1. The problem statement, all variables and given/known data
An electron coming from the left encounters/is trapped the following potential:
-a<x<0; V=0
0<x<a; V=V0
infinity elsewhere
the electron has energy V0
a)Write out the wave function
b)normalize th wave function
2. Relevant equations

3. The attempt at a solution
for -a<x<0
$$Ψ(x)=Acos(kx)+Bsin(kx)$$
$$k^2=\frac{2mV_0}{ħ^2}$$
and for 0<x<a
$$Ψ(x)=Cx+D$$
and 0 elsewhere
i used the sine and cosine because it seemed it would be better for continuity condition in x=0, if you would use exponential form please do explain why.
so this is what my teacher expects for a).
for b)
applying continuity conditions on x=0 i get:
A=D
B=C
and so:$$\int_{-a}^{0}|Ψ(x)|^2=1$$
im a bit confused here, is this the norm or the module? i think its the norm and if so ot might have been worth it to write the wave function in exponential form, so before i transcribe this big integral please clarify this for me.

Furthermore this should look like a particle traped in a box correct? i dont really understand what happens when E=V, i understand the probabiity part, it decays linearly further inside the step, correct?
And what about if E>V0 is it a particle traped in a box, but in the 0-a area the amplitude decreses? And the allowed energy levels for that area start at V0? what about penetration? and when E is smaller what happens?
Thank you!

Edit:would it be easier if i shifted the potential by -a so that it is in the range [0;2a]?

Last edited: Jun 20, 2017
2. Jun 24, 2017

### vela

Staff Emeritus
$B=C$ isn't quite correct. You should also apply continuity conditions for $\psi(x)$ at $x=-a$ and $x=a$.

The normalization requirement is
$$\int_{-\infty}^\infty \lvert \psi(x) \rvert^2\,dx = 1.$$ In this problem, since the wave function vanishes for $|x|>a$, you have
$$\int_{-a}^a \lvert \psi(x) \rvert^2\,dx = 1.$$