Potential Steps

1. Aug 20, 2006

Bunting

First it asks a few questions about what if it were a classical particle approaching the barrier. Much of this I understand and am OK with. Then we start treating the particle as a quantum thing so its governed by the TI Schrodinger EQ.

So, what it wants me to do which im a bit unsure about is find the "general solution for the eigenfunctions" in the 2 regions. I have searched a few sites I found on google, and some lecture notes - but they dont give me so much of a derivation or an answer, as a hint so im a bit hazy. What I think there looking for is...

phi(sub1)(x) = A(sub1)*e(i*k(sub1)*x) + A'(sub1)*e-(i*k(sub1)*x)
phi(sub2)(x) = A(sub2)*e(i*k(sub2)*x) + A'(sub2)*e-(i*k(sub2)*x)

But I dont really understand why, or how I will achieve a general solution out of this.

Thanks for anything!

2. Aug 20, 2006

Physics Monkey

Hi Bunting,

In your situation the time indepedent Schrodinger equation is a second order linear differential equation. You probably remember from a course on differential equations that such an equation has two linearly independent solutions. Any linear combination of these two solutions with constant coeffecients is also a solution. In other words, knowing two linearly independent solutions to the equation allows you to write down the general solution to the equation. Clearly then the general solution in a region will contain two undetermined coeffecients that you must fix using boundary conditions, etc.

Physically, the solutions may be chosen to consist of left and right moving waves in both regions because E > V(x) everywhere. Of course, other choices are possible, but the plane waves which you have written down are the most physically transparent choice. As I indicated above, you would determine the unknown coeffecients from boundary conditions. For example, if you were interested in the scattering of particles coming from negative infinity, then you would have an incoming wave in the x < 0 region (incident beam), an outgoing wave in the x < 0 region (reflected beam), and an outgoing wave in the x > 0 region (transmitted wave). You would not have an incoming beam in the x > 0 region, hence one of your coefficients would be zero. Other conditions obtain from considering the boundary at x = 0.

Hope this helps.

3. Aug 21, 2006

Bunting

if I do this in the conventional way (rearrange into an equation with a y'' a y' and a y and come up with two roots of the form...

alpha(1,2) = ± m((4*h(bar)^2*[v(x) - E])/2m) / h(bar)^2

I have read some examples which seem to suggest a better way of going about this is to let phi = A*e^ikx - but if the way I did it works then I dont really see a good reason to do this.

So from here Im assuming your saying plug these roots into a complementary function of the form..

y(x) = Ae(alpha(1)*x) + Be(alpha(2)*x)

But then does this give me a complete solution ? I still have A's and B's in there - I have been given no initial conditions to work work specific to phi(x) = whatever or phi'(x) = whatever, and I have a 0 on the RHS now (as I took the E*phi(x) and substracted it from the other side) like I would be used to being given on a problem such as this.

One other thing thats troubling me is the significance of the incidence from the left and not the right - would this simply be a matter of sign or would it need a reversal of the trigonometric properties also ? (cos->sin and whatnot).

Thanks for the reply, I think its getting there :)

Last edited: Aug 21, 2006
4. Aug 21, 2006

Bunting

ok, scratch that last post - ive been re-reading lecture notes and think ive got a greater understanding - though the "eigenfunctions" are still eluding me. I know what they ARE, just not really if I have foundthem...

First I seperated the equation into the regions defined (x>0 and x<0) then ended up with 2 equations...

phi(x) = Ae[i*k(1)*x] + Be[i*k(1)*x] - I dont know why I need i in this one in particular...
phi(x) = Ce[k(2)*x] + De[k(2)*x]

k(1) = sqrt(2*m*E) / h(bar)
k(2) = sqrt(2*m*(V(0) - E)) / h(bar)

so I was wondering how I can "justify" this, and what I could write as the eigenfunctions of this.

5. Aug 21, 2006

Physics Monkey

Hi Bunting,

Knowing what the eigenfunctions are allows you to easily check if you've actually found them. Recall the definition of an eigenfunction of the Hamiltonian with eigenvalue E: H phi(x) = E phi(x). Thus if you have a function phi(x) which you think is an eigenfunction of H, this definition tells you how to check your suspicion. Simply hit the eigenfunction with the Hamiltonian operator and if what you get out is a constant times your original function, then you've got an eigenfunction. The constant is just the eigenvalue. If you don't get out something proportional to your original function, then your original function is not an eigenfunction.