Wave function of a free electron

1. Apr 25, 2013

Rorshach

Hello guys, problem is as follows:
X9) A free electron has energy (kinetic) 10 eV and moves along the positive x-axis.
a) Determine the electron's wave function.
b) The electron reaches a potential step,-V0. Determine V0(expressed in eV), so that the probability of reflectance is 25%.
c) What is the reflection coefficient of the potential step instead is + V0 but with the same V0?

2. Relevant equations

ψ(x,t)=Cexp(i(kx-ωt))

3. The attempt at a solution
Equation above should do (with estimating the constant C) with the first paragraph, however I have problems with normalisation of this function, since there are no boundaries for the particle.

2. Apr 25, 2013

TSny

You won't need to determine a value of C. In addition to the incoming wave, you will have waves for reflection and transmission which also have constants. A certain ratio of the constants will determine the reflectance and you can determine the ratio without having to determine the constants themselves.

3. Apr 27, 2013

Rorshach

ok, I think I got it. Please tell me if am wrong:
for a) Psi(x,t)=Aexp(i(kx-(t*hbarred*k^2/2m)))
for b) V0=8*E
for c) reflection coefficient =0

am I right?

Last edited: Apr 27, 2013
4. Apr 27, 2013

TSny

No, I don't think those answers are correct. You need to specify the full wave function in both regions (before and after the step). How did you get your answers for b) and c)?

5. Apr 27, 2013

TSny

I don't get the same answer as you for c)

6. Apr 27, 2013

Rorshach

answer I got for c) came from a rule I read in Griffiths introduction to Quantum Mechanics, that if E<V0, then T=0. But it only came from a rule, I assumed it as true and didn't push further.

7. Apr 27, 2013

TSny

If you square the h-bar in your answer for (a) then I think that would be correct if there is no step potential for this part.

I'm assuming that for part (b) the negative sign in -Vo means that the potential steps down as you move toward positive x.

8. Apr 27, 2013

TSny

T represents the transmission coefficient. You are looking for the reflection coefficient.

9. Apr 27, 2013

Rorshach

Also- specify the wave function in both regions- I thought it only concerned the situation when a prticle moves to the right(positive in x-axis), so I didn't have to include the step in estimating wave function. But if I was to do this- it would only mean that wave function would be divided into regions-
psi={
psi=Aexp(ikx)+Bexp(-ikx) for x<0
psi=Fexp(ilx) for x>0

10. Apr 27, 2013

TSny

That looks good.

11. Apr 27, 2013

Rorshach

ok, since transmission coefficient is equal to 0, and transmission coefficient and reflection coefficient summed up have to be equal to 1, then reflection coefficient is equal to 1, am I right?

Also, I tried to derivate that formula for free particle, but still had normal non squared hbarred, but will try to derivate again.

12. Apr 27, 2013

TSny

That's correct.

Oops, my mistake. The energy has $\hbar^2$, but the wavefunction will not have the $\hbar$ squared. You were correct. Sorry.