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Step Potentials

  • Thread starter Fourier mn
  • Start date
  • #1
Fourier mn

Homework Statement


A particle of mass m and E<V0 approaches from the left and encounters a potential barrier, assume the amplitude of the incoming particle is A.
Write the time independent wavefunction for x>0 and x>0 at t=0 E is definite.
Find the reflection coefficient of the particle.

Homework Equations


R=(abs(Ar)2/abs(Ai)2)


The Attempt at a Solution


for x<0 where the potential is 0, [tex]\Psi(x)[/tex]=Aexp(-[tex]\alpha[/tex]x)
and where x>0 where the potential is >E, [tex]\Psi(x)[/tex]=Acos(k0x)
is it correct? does both of the amplitudes are the same?
 

Answers and Replies

  • #2
197
4
For x<0, you have oscillating wavefunction, and for x>0 you have a decaying wavefunction. Is this right? Is it a step up or a step down? In either case, when doing these reflection/absorption problems, I think that it is convenient to use complex exponentials rather than sin and cos.
The idea in these problems is to write the general solution in the different regions, then match the wave functions and their derivatives at the boundaries. This will give you equations that let you solve for the unknown coefficients. Does that help?
 
  • #3
Fourier mn
it's a step up. I agree that it would be easier if I could express both as exponentials, but how do i convert sin/cos to an exponential?
 
  • #4
197
4
Solve the V=0 schrodinger equation
[tex]{d^2x\over{dx^2}}+{2mE\over {\hbar^2}}=0[/tex]
just as thought it is a second order DE.
You probably learned that when you get complex conjugate roots of the characteristic equation,
then you write the solution as sin + cos. But you don't have to do this. You can write your solution as exponentials just like you would if you got real roots. I'll give you an example in a minute.
 
  • #5
197
4
Solve
[tex]{{d^2y}\over{dx^2}}+1 = 0[/tex].
You find that the roots are +i and -i. Right? You were probably taught in your DE class to write
[tex]y = A\cos(x) + B\sin(x)[/tex], because they want you to get a "real" solution. But you can instead write
[tex]y = C e^{+ix}+D e^{-ix}[/tex].
Notice that this is the same solution that you would have written down if you thought that the roots +i and -i were actually real.
Take my word for it. You can cast the second solution into the form of the first solution. If you want to try it, you'll need
[tex]e^{\pm ix}= \cos x \pm i\sin x[/tex]
 
  • #6
Fourier mn
I understand that, but my book uses exponentials only when the V=0, and not when V>0 and E in this case.
 
  • #7
Fourier mn
and then I can set one of the exponentials to zero, and i only have one. so when I take the derivatives to check continuity it'll be easier.
 
  • #8
197
4
That's fine too. The procedure is the same either way:
1. Write the general solution in each region.
2. Match the function and the derivatives at the boundaries.
3. Solve the resulting system for the coefficients (in terms of the one you're given, ie A).
 
  • #9
197
4
Right. Don't use sin and cos. The complex exponentials offer the interpretation of a right/left travelling wave.
 
  • #10
Fourier mn
last question, are all of the amplitudes the same? I mean in the different regions
 
  • #11
197
4
No. Use a different coefficient in front of every term. Use the coefficient A in front of the term representing the initial "incoming" wave.
 
  • #12
Fourier mn
Ok, thank you very much. I'll go wrestle with the questions, I'll come back if I'll have any question. thank you.
 
  • #13
197
4
Cheers!
 
  • #14
Fourier mn
my step potential looks like-
_________
|
|
---------
for tunneling and passing the barrier there are only two wave equations for each?
 
  • #15
Fourier mn
*************___________
************|
************|
----------------
 

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