# Potential Well

1. May 7, 2004

### gnome

Given:
in region 1, x < 0, U = u0 > 0
in region 2, 0≤x≤L, U = 0
in region 3, L < x, U = u0 (same as in region 1)
total energy E constant everyplace, E > u0, and a particle is moving towards the right beginning at some x<0.

So, in region 1:
$$\frac{d^2\psi_1}{dx^2} = \frac{-2m}{\hbar^2}\left(E-u_0\right)\psi$$
therefore
$$\psi_1(x) = Ae^{ik_1x} + Be^{-ik_1x}\; \textrm{where}\; k_1 = \sqrt{\frac{2m}{\hbar^2}\left(E-u_0\right)}$$

in region 2, U = 0 so
$$\frac{d^2\psi_2}{dx^2} = \frac{-2m}{\hbar^2}E\psi$$
therefore
$$\psi_2(x) = Ce^{ik_2x} + De^{-ik_2x}\; \textrm{where}\; k_2 = \sqrt{\frac{2mE}{\hbar^2}}$$

in region 3, U = u0 again, but here there is no edge to produce a wave propagating towards the left, so
$$\psi_3(x) = Fe^{ik_1x}\; \textrm{where}\; k_1 = \sqrt{\frac{2m}{\hbar^2}\left(E-u_0\right)}$$

Matching conditions give me
at x=0:
$$\psi_1(0) = \psi_2(0) \implies A +B = C+D$$

$$\frac{d\psi_1}{dx}(x=0) = \frac{d\psi_2}{dx}(x=0)\implies k_1A-k_1B=k_2C-k_2D$$

at x=L:
$$\psi_2(L) = \psi_3(L) \implies Ce^{ik_2L} +De^{-ik_2L} = Fe^{ik_1L}$$

$$\frac{d\psi_2}{dx}(x=L) = \frac{d\psi_3}{dx}(x=L)\implies k_2Ce^{ik_2L}-k_2De^{-ik_2L}=k_1Fe^{ik_1L}$$

I am to show that if 2L = the de Broglie wavelength of the particle in region 2 there will be NO reflected wave in region 1. I understand that this means that, since the path distance L represents 1/2 wavelength, waves reflected by the well edge at x=L will interfere destructively with waves reflected by the well edge at x=0.

But I don't see how to rearrange the equations to show that explicitly. I don't even know exactly what I'm looking for. How would equations in exponential form depict destructive interference? Or do I have to switch to equations in terms of sine and cosine?

I notice that since $$k=\frac{2\pi}{\lambda}$$,
$$\lambda_2 = 2L \implies k_2 = \frac{2\pi}{2L} = \frac{pi}{L}$$
$$k_1=\sqrt{\frac{\pi^2}{L^2} - \frac{2mu_0}{\hbar^2}}$$
but I can't see that that leads to anything useful.

Any hints will be appreciated.

2. May 7, 2004

### turin

Off the top of my head, I would say that this amounts to B = 0. I am basing this on the same justification that you used for the form of &psi;3. Does that make sense?

3. May 8, 2004

### gnome

Thanks, turin, but I can't find anything yet that would compel B to be zero. I think I need to show somehow that the waves represented by
$$Be^{-ik_1x} \; \textrm{and} \; De^{-ik_2x}$$
are 180o out of phase.

But I don't see anything that leads to that conclusion.

4. May 8, 2004

### robphy

Maybe it's easier to show your condition implies that |F|=|A|, that is, |F/A|=1, perfect transmission.

5. May 8, 2004

### gnome

Well, I wouldn't have thought that was the case until you wrote it, but now I'm sure you're right. Thanks.

A more careful reading of the statement of the problem (the part that I ignored the first time around) says that this is a "crude model for the Ramsauer-Townsend effect observed in the collisions of slow electrons with noble gas atoms like argon, krypton and xenon. Electrons with just the right energy are diffracted around these atoms as if there were no obstacle in their path (perfect transmission).

... an example of using the exercises to introduce new material. Hmm..., a few days ago I told turin that I thought this practice was just fine.

But (assuming I can show that |F/A| = 1, which I haven't done yet), I'm still puzzled by this. I would have thought that if there is no wave moving towards the left because of destructive interference, there would not be perfect transmission, i.e. something would be lost due to the interference. On the other hand, I don't know what might be lost since we're talking about a particle, and we're told that the energy is constant. Can you help me make sense of this concept?

6. May 9, 2004

### gnome

Well, I seem to be getting turin's answer, after all.

1. $$A +B = C+D$$

2. $$k_1(A-B)=k_2(C-D)$$

3. $$Ce^{ik_2L} +De^{-ik_2L} = Fe^{ik_1L}$$

4. $$k_2Ce^{ik_2L}-k_2De^{-ik_2L}=k_1Fe^{ik_1L}$$

5. $$k_2 = \frac{pi}{L}$$

Substituting (5) into (3) gives:
$$\Rightarrow Ce^{i\frac{\pi}{L}L} +De^{-i\frac{\pi}{L}L} = Fe^{ik_1L}$$

6. $$\Rightarrow C + D = -Fe^{ik_1L}$$ (this is correct, isn't it?)

And then substituting for C + D in (1):
7. $$A + B = -Fe^{ik_1L}$$

Substituting (5) into (4):
8. $$k_2(-C + D) = k_1Fe^{ik_1L} \Rightarrow k_2(C-D) = -k_1Fe^{ik_1L}$$

But from 2:
2. $$k_1(A-B)=k_2(C-D)$$

Therefore
9. $$k_1(A-B)=-k_1Fe^{ik_1L} \Rightarrow(A-B)=-Fe^{ik_1L}$$

Looking at (7) and (9), it appears that B=0.

But if this is correct, I don't see how to show |F/A| = 1. Instead, I seem to have
$$\frac{F}{A} = -e^{-ik_1L}$$
and
$$k_1 = \sqrt{\frac{\pi^2}{L^2} - \frac{2mU}{\hbar^2}}$$

Did I do something wrong? robphy, how did you arrive at |F/A| = 1?

7. May 9, 2004

### robphy

\begin{align*} \left| \frac{F}{A}\right| &= \left|-e^{-ik_1L} \right| \\ &= \sqrt{ ( -e^{-ik_1L})^* ( -e^{-ik_1L}) }\\ &= \sqrt{ ( -e^{ik_1L}) ( -e^{-ik_1L}) }\\ &= \sqrt{ 1 }\\ &= 1 \end{align*}

More generally, without making the assumption for $$k_2$$:

By adding and subtracting k_1*eq(1) with eq(2), you get
A and B in terms of C and D.

By adding and subtracting k_2*eq(3) with eq(4), you get
C and D in terms of F.

I get (with an eye for forming the trig functions from the exponentials)
\begin{align*} A &= F e^{ik_1 L} \left[ \cos(k_2 L) - i \left(\frac{k_1^2+k_2^2}{2k_1k_2}\right) \sin(k_2 L) \right]\\ B &= F e^{ik_1 L} \left[ i \left( \frac{k_2^2-k_1^2}{2k_1k_2}\right) \sin(k_2L) \right] \end{align*}

So, it looks like when $$k_2 L= n\pi$$, that is, $$L=n \frac{2\pi}{2k_2} =n \frac{\lambda_{dB}}{2}$$, we have $$B=0$$ and $$F/A=(-1)^n e^{-ik_1L}$$ (hence, as above, |F/A|=1).

8. May 9, 2004

### gnome

I am not very good with exponentials, and no good at all when it comes to imaginary numbers, so would you please explain the above. I don't understand how that works.

As for the rest, I got as far as
\begin{align*} A &= \frac{(k_1+k_2)C+(k_1-k_2)D}{2k_1}\\ B &= \frac{(k_1-k_2)C +(k_1+k_2)D}{2k_1}\\ C &= \frac{(k_1+k_2)Fe^{iL(k_1-k_2)}}{2k_2}\\ D &= \frac{(k_1-k_2)Fe^{iL(k_1+k_2)}}{2k_2} \end{align*}
(I think)
and I suppose I could combine those into 2 very ugly expressions for A and B but I have no idea how you converted them to trig functions. How do you do that (if it's explainable within the limitations of this forum)?

9. May 9, 2004

### robphy

Given a complex number $$z=a+ib=re^{i\theta}$$ (where $$i^2=-1$$ and $$a$$, $$b$$, $$r$$, $$\theta$$ are real),
the magnitude or norm of z is
$$|z|=\sqrt{z^*z}=\sqrt{a^2+b^2}=\sqrt{r^2}$$
where $$z^*=a-ib=re^{-i\theta}$$ is the complex conjugate of z.
(You certainly must have seen these operations applied to the wavefunction $$\psi$$.)

It's good to recognize that whenever you have $$e^{i\theta}$$, it has norm 1. Graphically, this is a point or vector on the unit circle in the complex plane. In physics, this term is often called the "phase factor".

With your expressions for A,B,C,D,...so far so good.

Note that $$e^{ik_1L}$$ appears as a factor in my final form. So, write $$e^{iL(k_1-k_2)}$$ as $$e^{ik_1L} e^{-ik_2L}$$, and similarly for the factor in your expression for D.

You have to substitute your expressions for C and D into A and into B.
Do the B one first because it's easier.

The idea is that
$$\left( e^{ik_2L}-e^{-ik_2L}\right) =2i\sin(k_2L)$$ and
$$\left( e^{ik_2L}+e^{-ik_2L}\right)=2\cos(k_2L)$$,
which follow from the Euler Identity $$e^{i\theta}=\cos{\theta}+i\sin{\theta}$$.

When doing the A one, you have to multiply out each coefficient in this factor you'll encounter
$$(k_1+k_2)^2 e^{-ik_2L} - (k_2 -k_1)^2 e^{ik_2L}$$
then look for the expressions above.

10. May 9, 2004

### gnome

Thanks robphy. I'll have to chew on that for a while.

(Are you saying that any complex number $$a+ib$$ can be written in the form $$re^{i\theta}$$ for some r and $$\theta$$? But if this $$\sqrt{a^2+b^2}=\sqrt{r^2}$$ is the relationship among a, b and r, what determines $$\theta$$?)

11. May 9, 2004

Staff Emeritus

You then have $$cos \theta = x/r, sin \theta = y/r$$, from elementary trigonometry. Then if you expand $$x + iy = rcos\theta + irsin\theta$$ in Taylor series, and cancel common terms, you get the Taylor series for $$re^{i\theta}$$.

12. May 9, 2004

### turin

ei&theta; describes the unit circle in the complex plane. The magnitude of a complex number is the radius.

If you don't understand how B = 0 is the same thing as no reflection, then how did you arrive at the particular form of the wave function in region 3? Why doesn't your region 3 wave function have an e-ikx term?

This wave function is an approximation of a stream of particles as a plane wave. The physical interpretation is very non-classical. The wave function is kind of in a momentum eigenstate (as I've always seen scattering problems treated), and most of us are geared to consider position eigenstates as classical rather than momentum eigenstates. One consequence of this treatment is that everything that will happen for the particle is described all at once; the wave function is showing you what will happen when the scattering process reaches a sort of steady state (enough particles have already come and gone to demonstrate the reflections at both boundaries). When you describe the wave function in terms of the complex exponentials, you are describing it in terms of two oppositely directed plane waves, but this probably isn't entirely clear since you have neglected the time dependence (though this is not incorrect, since we are already assuming nonrelativistic steady state). On physical grounds, you only included the right-going plane wave after the obstacle (in region 3), because you approximate an infinite distance past the boundary. In region 1, however, you had to account for both directions, because in steady state, the wave function is composed of both the incident and reflected particles.

13. May 9, 2004

### gnome

turin: It's not that I didn't understand that B=0 means no reflection. I've understood that all along, but when I posted #3 I hadn't yet solved the equations to see that &lambda;2=2L caused B=0. Also, I hadn't really paid attention to the part of the "story" that mentioned perfect transmission. I was focused on trying to find something in the equations that demonstrated destructive interference, rather than perfect transmission.

It still strikes me as very peculiar (Ha Ha; as if there's any aspect of QM that doesn't strike me as peculiar) that destructive interference of two waves should result in perfect transmission. Intuition tells me that interference should result in the loss of something, although I couldn't say what. I guess it's that I'm trying to apply a physical interpretation to something that's not really physical.

14. May 10, 2004

### turin

The interference does result in the loss of something, but only in each of the regions independently, as for a wave. My point was that the plane waves in this model are about as far from classical as you can get. Classically, the particle would not exist in all three regions but one at time as it went through. In the QM model/approximation for the scattering process it does exist in all three regions, which is accomplished by considering a sort of momentum rather than position eigenstate. The fact that you have 5 coefficients means that you have four independent considerations for the 2 transmissions and 2 reflections (and then one arbitrary normalization). This QM scattering model says that the backward travelling wave in region 2 doesn't have anything to do (in the intuitive classical sense) with the backward travelling wave in region 1. It is precisely this kind of independence that makes this problem profoundly nonclassical. Of course, if there is a certain interrelated condition on the components of the wave function in region 2, then this can in some way relate to the wavefunction in the other regions. But the point is how this relation is induced, as wave behavior instead of particle behavior (i.e. boundary conditions for the wave). This is what makes it nonclassical.

Don't be too hasty. It is very much physical in the sense that an experiment will agree to the same order of approximation that you are willing to apply to the problem mathematically. QM is very physical; it's just not very "classical."

I appologize if I'm telling you stuff you already know.

15. May 12, 2004

### subluminous

don't happen to be using 'Modern Physics' by Serway/Moses/Moyer do you gnome?