Rectangular Potential Barrier Where E=V0

[GrantB]

Homework Statement

V(x) = 0 for x < 0; V0 for 0 < x < a; 0 for x > a

Particle of mass m

E=V0

Asks for general solution and transmission coefficient.

Homework Equations

Time-Independent Schrodinger Equation

The Attempt at a Solution

I've found region I (the left region where x <0) to be Ae^ikx+Be^-ikx.

Middle region to be C+Dx (not sure where this comes from but I read that this is the solution to the middle region when E=V0 so any help with this is appreciated).

Right region where x > a to be Fe^ikx+Ge^-ikx.

My textbook has the right region in an example as just Fe^ikx. It goes through the explanation of limits mattering as x-> infinity or -infinity. I understand this for when E<V0, but the case seems completely different when E=V0.Any help is appreciated.

Thanks.

 

[vela]

Homework Statement

V(x) = 0 for x < 0; V0 for 0 < x < a; 0 for x > a

Particle of mass m

E=V0

Asks for general solution and transmission coefficient.

Homework Equations

Time-Independent Schrodinger Equation

The Attempt at a Solution

I've found region I (the left region where x <0) to be Ae^ikx+Be^-ikx.

Middle region to be C+Dx (not sure where this comes from but I read that this is the solution to the middle region when E=V0 so any help with this is appreciated).

You do the same thing you did to find Ae^ikx+Be^-ikx for region I, except this time the potential is V(x)=E.

Right region where x > a to be Fe^ikx+Ge^-ikx.

My textbook has the right region in an example as just Fe^ikx. It goes through the explanation of limits mattering as x-> infinity or -infinity. I understand this for when E<V0, but the case seems completely different when E=V0.

Do you know what e^ikx and e^-ikx physically represent? There's nothing different between the E<V₀ case and E=V₀ case.

 

[GrantB]

You do the same thing you did to find Ae^ikx+Be^-ikx for region I, except this time the potential is V(x)=E.

Do you know what e^ikx and e^-ikx physically represent? There's nothing different between the E<V₀ case and E=V₀ case.

For region I, the potential is zero. So I just get the Ae^ikx+Be^-ikx answer. My question for this portion was about the limits. I know that if this was a finite square well, and if the potential was >0 for x < 0, then Be^-ikx would have to be removed, because that would mean that the wave function would go to infinity.

What's confusing me is how in my book, it has psi_III=Ce^ikx for when x > a (to the right of the barrier). I'm getting Ce^ikx+De^-ikx. Why do they remove the "D" term? The potential is zero in that region.

No I don't know what e^ikx and e^-ikx physically represent.

As for the cases being different: http://en.wikipedia.org/wiki/Rectangular_potential_barrier#E_.3D_V0

I don't know where they get the C+Dx from.

 

[vela]

For region I, the potential is zero. So I just get the Ae^ikx+Be^-ikx answer. My question for this portion was about the limits. I know that if this was a finite square well, and if the potential was >0 for x < 0, then Be^-ikx would have to be removed, because that would mean that the wave function would go to infinity.

What's confusing me is how in my book, it has psi_III=Ce^ikx for when x > a (to the right of the barrier). I'm getting Ce^ikx+De^-ikx. Why do they remove the "D" term? The potential is zero in that region.

No I don't know what e^ikx and e^-ikx physically represent.

If you apply the momentum operator to one of these functions, you get\hat{p}e^{ikx} = \frac{\hbar}{i}\frac{d}{dx}e^{ikx} = \hbar k\ e^{ikx}so you can see it is an eigenfunction of the momentum operator with eigenvalue p=\hbar k. In other words, it represents a particle with a definite momentum p. The one with the positive sign corresponds to a particle moving to the right, and the other, to a particle moving to the left.

In these types of a problems, you have a particle incident on the barrier from the left. There will be a reflected wave and a transmitted wave. Can you figure out the rest from here?

I don't know where they get the C+Dx from.

You say you find in region I that \psi_\mathrm{I} = Ae^{ikx} + Be^{-ikx}. How did you get that?

 

[GrantB]

I see. I think I understand why region III only has the Ce^ikx. It is because the only direction a particle will be going is to the right. The other regions can have a particle being reflected.

I find the wave functions for each region by using the Schrodinger Equation.

\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi

For region I, V=0, so it becomes

\frac{d^2\psi}{dx^2}=-k²\psi

where k²=2mE/\hbar^2

which produces Ae^ikx+Be^-ikx since it is a second order diff. eq.

When V=V0, that gives

\frac{d^2\psi}{dx^2}=\alpha^2\psi

where \alpha^2=2m(V0-E)/\hbar^2.

If V0=E, \alpha becomes zero. So,

\frac{d^2\psi}{dx^2}=0

Which would give C+Dx? Since it would be a first order polynomial? That makes sense to me.

--

Is using F²/A² the way to find the transmission coefficient? Ratio of transmitted particles to incident particles? (j_trans/j_inc)

 

[vela]

Yes, exactly. That's how you get Cx+D.

And yes, T = |F/A|².

 

[GrantB]

Thanks a lot!