Infinite potential barrier question.

[xcrunner2414]

Homework Statement

I'm asked to calculate the reflection and transmission coefficients for scattering from a potential energy barrier. The potential energy barrier is,

$$V(x)=\frac{\hbar^2}{2m} \frac{\lambda}{b} \delta(x)$$

where $$\delta (x)$$ is the Dirac delta function.

Homework Equations

See above.

The Attempt at a Solution

Wouldn't the reflection coefficient be 1, and the transmission coefficient be 0? It seems like it's just an infinite potential well problem, except there is only one wall at x=0, so the wavefunction will be zero on the side of the barrier that is opposite the incident wavefunction. But that seems almost too simple. But then again, I'm dealing with an infinite potential barrier, so the idea that not even subatomic particles governed by quantum mechanics can penetrate it makes sense to me...

The actual solution, then, looks like this: Assuming region I is the side of the incident wavefunction, and region II is the other side, then solving the time-independent Schrodinger equation gives,

$$\psi_I (x) = Ae^{ikx} + Be^{-ikx}$$

$$\psi_{II} (x) = Ce^{ikx}$$

where $$k = \sqrt{2mE/\hbar^2}$$. Since $$V(0)\rightarrow \infty$$, this means $$\psi(0) = 0$$, so from the very first boundary condition:

$$\psi_I (x) |_{x=0} = \psi_{II} (x) |_{x=0} \Rightarrow A+B=C=0 \Rightarrow A=-B \Rightarrow R = \frac{j_{ref}}{j_{inc}} = \frac{{|B|}^2}{{|A|}^2} = 1$$

Right?

 

[ideasrule]

Wouldn't the reflection coefficient be 1, and the transmission coefficient be 0?

One would think so, but it actually isn't. Think about it this way: although the potential barrier reaches infinity at x=0, it's never at infinity for a finite distance, because the total integral of the delta function must be 1.

where $$k = \sqrt{2mE/\hbar^2}$$. Since $$V(0)\rightarrow \infty$$, this means $$\psi(0) = 0$$, so from the very first boundary condition:

$$\psi_I (x) |_{x=0} = \psi_{II} (x) |_{x=0} \Rightarrow A+B=C=0 \Rightarrow A=-B \Rightarrow R = \frac{j_{ref}}{j_{inc}} = \frac{{|B|}^2}{{|A|}^2} = 1$$

Right?

Actually, this isn't true: the wavefunction doesn't have to be 0 at x=0.

Try starting off by finding the derivatives of the wavefunction on the 0+ and 0- sides of x=0. Note that the derivative doesn't have to be continuous, but the wavefunction does, since the derivative is never infinite.

Then, integrate Schrodinger's equation and see if you can relate the integral to the derivatives you just found.

 

[xcrunner2414]

Actually, this isn't true: the wavefunction doesn't have to be 0 at x=0.

Try starting off by finding the derivatives of the wavefunction on the 0+ and 0- sides of x=0. Note that the derivative doesn't have to be continuous, but the wavefunction does, since the derivative is never infinite.

Then, integrate Schrodinger's equation and see if you can relate the integral to the derivatives you just found.

- So... dirac delta functions don't go to infinity?

 

[xcrunner2414]

Ok, I'm totally stuck.

Schrodinger's Equation for this problem is,

$$\frac{d^2 \psi (x)}{dx^2} =[ \frac{-2mE}{\hbar^2} + \frac{\lambda}{b}\delta (x) ] \psi (x)$$

and integrating gives,

$$\frac{d\psi}{dx} |_{-\infty}^{0^-} + \frac{d\psi}{dx} |_{0^+}^{\infty} = \frac{-2mE}{\hbar^2} \int_{-\infty}^{\infty} \psi(x) dx + \frac{\lambda}{b} \psi (0)$$

but the wavefunction is always in the form $$Ae^{\pm ikx}$$, the integral of which doesn't converge with infinite bounds... I feel like I'm forgetting something simple...

 

[eys_physics]

Use that

$$\delta (x)= \int_{-\infty}^{\infty}1\cdot e^{2\pi i x \xi}d\xi$$

 

[Matterwave]

Don't worry about normalizing the wave-functions here. Because these are not bound states, they require the formation of wave-packets to normalize; however, that's too difficult for this problem. Just use plane waves and then match boundary conditions and just neglect normalization.

 

[xcrunner2414]

Don't worry about normalizing the wave-functions here. Because these are not bound states, they require the formation of wave-packets to normalize; however, that's too difficult for this problem. Just use plane waves and then match boundary conditions and just neglect normalization.

- Yea, I know the general approach. It's essentially the same as a finite potential barrier. The left has incident and reflected plane waves, i.e. $$\psi_{left}=Ae^{ikx} + Be^{-ikx}$$ and on the right of the barrier there is only a transmitted plane wave, with coefficient C. Apparently the potential does not go to infinity, so the wavefunction at x=0 is just some value $$\psi (0)$$. Correct me if I'm wrong, but the first boundary condition would give $$A+B=C=\psi(0)$$ and the second boundary condition would come from integrating Schrodinger's Equation, as I showed in my previous post. I just wasn't sure what to do with,

$$\frac{-2mE}{\hbar^2} \int_{-\infty}^{\infty} \psi(x) dx$$

I let that equal zero and I got transmission and reflection coefficients that added to 1:

$$R=\frac{1}{1+(\frac{2kb}{\lambda})^2}$$

$$T=\frac{1}{1+(\frac{\lambda}{2kb})^2}$$

But, again, I'm not sure if I should have let that integral be zero...

 

[ideasrule]

Apparently the potential does not go to infinity, so the wavefunction at x=0 is just some value $$\psi (0)$$.

No, the potential does go to infinity at x=0. However, because it does not stay at infinity for any finite length, the wavefunction doesn't have to be 0 at x=0.

I let that equal zero and I got transmission and reflection coefficients that added to 1:

$$R=\frac{1}{1+(\frac{2kb}{\lambda})^2}$$

$$T=\frac{1}{1+(\frac{\lambda}{2kb})^2}$$

But, again, I'm not sure if I should have let that integral be zero...

You got the right answer (assuming you didn't mess up the algebra), but for the wrong reasons. You should have integrated the equation from -epsilon to epsilon, where epsilon is an arbitrarily small number. That way, the term you were worried about would indeed be 0.

 

[xcrunner2414]

No, the potential does go to infinity at x=0. However, because it does not stay at infinity for any finite length, the wavefunction doesn't have to be 0 at x=0.

- I see. Neither my teacher nor my textbook explained this (directly). Thank you.

You got the right answer (assuming you didn't mess up the algebra), but for the wrong reasons. You should have integrated the equation from -epsilon to epsilon, where epsilon is an arbitrarily small number. That way, the term you were worried about would indeed be 0.

- Ahhh... now I get it. I didn't really understand why I had to integrate Schrodinger's equation, other than the fact that it seemed to work. Is that the general approach to getting the boundary condition on the derivatives?

 

[vela]

A side note: If you've already solved the finite potential barrier problem, try taking the limit of those results as the width of the barrier goes to 0 while the total area under the potential remains constant. You should recover the result you found for the delta function potential.