Heaviside step fn in infinite potential well

[davo789]

By considering the wavefunctions within the potential described below, determine the incident, reflected and transmitted amplitudes of each part of the wavefunction at the step boundary (as necessary).

For x<0, V = infinity.
For 0<x<a, V = 0
For a<x<L, V = V1
For x>L, V = infinity.

Relevant equations
Nil.

Solution ideas

My first hang up is initially trying to determine the wavefunctions for each part of this potential. My guess is that they will not be in exponential form since the particles are not free, but I could be wrong. If this is correct, then I should have 3 wavefunctions (incident, reflected and transmitted) in trig (i.e. sin/cos) form. Something like A.sin(k.x), B.sin(-k.x), C.sin.(l.x) I would guess? Then is it just a case of taking the derivatives at x=a and solving the resulting simultaneous equation?


Thanks v much in advance!

 

[vela]

By "exponential form," I think you mean e^ikx and e^-ikx. That's the same as using sin kx and cos kx. In linear-algebra speak, the pairs are just different bases that span the same space.

You need to consider two cases: E>V1 and E<V1, where E is the energy of the state.

As you noted, you'll need the derivatives to match up at x=a, but you need the wave functions to be continuous there as well, which yields a second equation.

You'll need to take into account the effect of the infinite walls, which reflect anything that hits them back toward the center of the well.

 

[davo789]

Thanks vela! I had forgotten about the reflected wavefn from the RH side. If I can write the wavefns as exponentials, (and let's assume that E<V1 for the minute), then the wavefns involved should be something like this?

Ae^kx = incident wavefn
Be^-kx = reflected wavefn
Ce^lx = transmitted wavefn
De^-lx = reflected transmitted wavefn

Thanks again!

 

[vela]

Not quite. The Schrodinger equation will be

$$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi(x) = (E-V(x))\psi(x)$$

Depending on which section of the well you are in, E-V(x) will have different signs if E<V1, so you'll get different types of solutions. In one section, you'll have E>V(x)=0, so you'll have oscillating solutions. In the other section, E<V(x)=V1, so you'll have exponentially decaying and growing solutions.

 

[davo789]

aah I see! Will work on it tonight. Thanks!

 

[davo789]

...one more thing... would the wavefns I gave in my post before last be appropriate for E>V1?

 

[vela]

If E>V1, then the wave function will be oscillatory in both regions, so you'll have sines and cosines everywhere.

 

[davo789]

aah yes; it's been too long since I've done a problem like this!
Going back to the E<V1 bit, here is what I've come up with.

Asin(kx) - Inc.
Bsin(-kx) - Refl.
Ce^-ikx - Trans.

Does that look at all right?

 

[vela]

No, I think it'll help if you work through the math. In the region 0<x<a, you have V(x)=0, so the wave function satisfies

$$-\frac{\hbar^2}{2m}\psi''(x) = E\psi(x) \Rightarrow \psi''(x) = -\frac{2mE}{\hbar^2}\psi(x)$$

The characteristic equation is $r^2 = -2mE/\hbar^2$, which has roots $r=\pm ik$ where $$k=\sqrt{2mE/\hbar^2}$$. The solution is therefore of the form

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

You could also rewrite this as

$$\psi(x) = A'\sin kx + B'\cos kx$$

The boundary condition $\psi(0)=0$ leads in either case to $\psi(x)=A\sin kx$.

In the other region where V(x)=V₁ and E<V₁, what kind of roots do you get from the characteristic equation? What implication does this have for the solution in that region? What about when E>V₁?

 

[davo789]

OK, so between a<x<L;
(unfortunately latex isn't playing ball so I'll have to do it in 'text' form):

l² = 2m(V1-E)/hbar²

Where V1>E, so l is real and +ve.

The general solns to this are Ce^ilx + De^-ilx

Inserting the boundary conditions (psi = 0 at x=L) implies C = 0, since for large L, e^ilx tends to infinity, so the wavefn for a<x<L = De^-ilx?

Thanks again for your continued help!

 

[vela]

OK, so between a<x<L;
(unfortunately latex isn't playing ball so I'll have to do it in 'text' form):

l² = 2m(V1-E)/hbar²

Where V1>E, so l is real and +ve.

OK up to here.

The general solns to this are Ce^ilx + De^-ilx

This is wrong. Do you see why the i appeared in the solution for the other reason? I didn't just throw it in there arbitrarily.

Inserting the boundary conditions (psi = 0 at x=L) implies C = 0, since for large L, e^ilx tends to infinity, so the wavefn for a<x<L = De^-ilx?

Two problems with your reasoning here. First, e^ilx is oscillatory. Its modulus is equal to 1; it doesn't tend to infinity in any limit. Second, even if the function did diverge for large L, it doesn't matter because in this problem L is fixed and finite. You need to wave function to vanish at a finite value of x.

Thanks again for your continued help!

 

[davo789]

I think I'm slowly getting there...

Using the characteristic equation, r= +/- sqrt[2m(V1-E)]/hbar (no i's)

So now the general soln is: Ce^lx + De^-lx, which needs to = 0 for x = L.
If this is correct so far, then for the wavefn to = 0 for x = L, then C=-D, which means the wavefn is now C(e^lx - e^-lx) or 2Csinh(lx)...maybe that's too big a jump!

 

[vela]

I think I'm slowly getting there...

Using the characteristic equation, r= +/- sqrt[2m(V1-E)]/hbar (no i's)

So now the general soln is: Ce^lx + De^-lx, which needs to = 0 for x = L.

Good to here.

If this is correct so far, then for the wavefn to = 0 for x = L, then C=-D, which means the wavefn is now C(e^lx - e^-lx) or 2Csinh(lx)...maybe that's too big a jump!

You don't get C=-D. When you plug in x=L, you get

$$Ce^{lL}+De^{-lL} = 0 \hspace{0.25in} \Rightarrow \hspace{0.25in} C = -De^{-2lL}$$

With a little algebra, you can eventually express the wave function as $\psi_2(x) = D' \sinh l(L-x)$, which clearly vanishes when x=L.

 

[davo789]

Of course C doesn't equal -D; that'll teach me for doing algebra on a Saturday afternoon! After all that, do I need to worry about the reflected wave (0<x<a) at all, or is it simply packaged up in the A constant prefixing sin(kx)?

 

[vela]

It's packaged up in the A. Writing the wave function in terms of sine obscures the fact it's made up of a wave going to the left and a wave going to the right. If you write sin kx in terms of complex exponentials, it becomes clear.

 

[davo789]

Thanks again!