# Asymmetric potential well

1. Dec 31, 2012

### bcrowell

Staff Emeritus
Consider a nonrelativistic electron in one dimension, with the potential

$V(x)=\left\{ \begin{matrix} \infty, \quad x<-a \\ -4V_0, \quad -a<x<0 \\ -V_0, \quad 0<x<a \\ \infty, \quad x>a \end{matrix} \right.$

I use this as an example when I teach the Schrodinger equation in a freshman survey course. Say you're looking for a solution with an energy E=0. The momentum in the left half of the well is double what it is on the right, so the wavelength on the left is half of what it is on the right. To match the two pieces of the wavefunction without a kink, you need to have double the amplitude on the right, which corresponds to four times the probability, $P_R/P_L=4$.

Classically, the particle moves twice as fast on the left, so it spends half as much time there. If you peek at it at some randomly chosen moment, you find $P_R/P_L=2$.

Comparing these two results, you get qualitative but not quantitative agreement, and this seems to violate the correspondence principle. But in the classical limit, you can't really put the electron in a pure energy state. It's in an incoherent superposition of some large number of states with $E\approx 0$. The incoherent superposition gives $P_R/P_L=2$, in agreement with the correspondence principle.

This would seem to be a natural model of an asymmetric diatomic molecule, but my intuition tells me that the electron would "want" to spend more time in the deeper potential, which is where it's attracted to. What's going on? The following possibilities are the ones that occur to me. (1) My intuition is just wrong. (2) The result is qualitatively different if there's a classically forbidden region separating the two sides. (3) The result only holds for $E\approx 0$, and for smaller energies it does what I'd have intuitively expected. (4) The result only holds because I chose $V_L/V_R=4$ to be a small integer, and I've been considering simple solutions where the quantum numbers are small integers.

2. Dec 31, 2012

### atyy

I'll guess (3). Do we know the ground state energy?

3. Dec 31, 2012

Staff Emeritus
I'm afraid the answer is (1).

Consider a classical elliptical orbit. The orbiting body spends less time near the body orbited and more time farther away (apogee dwell). The deeper the potential, the faster it's moving and the less time it spends there.

4. Dec 31, 2012

### atyy

But in the ground state of the harmonic oscillator, the wave function is peaked at the deepest part of the well, which means it's most likely to be found there. At higher energies the wave function is peaked at shallower parts of the potential, suggesting the answer is energy dependent. So naively in the OP case I'd expect the ground state to also be peaked at the deepest part of the well. No?

5. Dec 31, 2012

Staff Emeritus
The ground state is as close to non-moving as you can get: it's like having the pendulum at rest. Gaining intuition on how things move by thinking about states at rest is probably not going to work well.

6. Dec 31, 2012

### bcrowell

Staff Emeritus
Classically, I agree with V50, and the system does have the expected behavior in the classical limit. But quantum-mechanically, don't asymmetric diatomic molecules have dipole moments that arise from the tendency of the electron to be more localized in the atom that it's more attracted to?

I'd think the classical limit of the harmonic oscillator would be an incoherent superposition of high-n states in some range of energies. Check out the third figure (the orange one) in this WP article: http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator The high-n states have the classical behavior, but the low-n ones don't.

7. Dec 31, 2012

### atyy

But are options (1) and (3) contradictory? I do think your intuition is just wrong for the energy you specified - in that sense I agree with V50 (1). But perhaps it is right for the ground state (3).

These guys give solutions for a related situation: http://pubs.acs.org/doi/abs/10.1021/ed085p591. The ground state is peaked over the deeper well.

I tried http://fermi.la.asu.edu/Schroedinger/html/node2.html#applet with

V(x) = -4*pulse(x+1,1)-pulse(x-1,1)
xmin=-2.0
xmax-2.0
Vscale=10.

The ground state is peaked over the deeper well.

I think each of the two intuitions correspond to whether the state is "bound" or "free", roughly speaking. In the former case, the intuition that it should spend its time in the deeper well applies. in the latter case, the arguments from classical kinetic energy apply.

Last edited: Dec 31, 2012
8. Jan 1, 2013

### bcrowell

Staff Emeritus
Cool applet!

Yes, I think I agree with #7. Both the ground state and the high-energy states behave qualitatively as we'd expect classically.

I was able to get a reasonable mockup of my original problem (with infinite walls) with this potential:

a=0.2
b=5
max(a*x^4-b,-4*pulse(x+1,1)-pulse(x-1,1))

With "Nodes" set to n=0, I get the ground state, which is peaked in the lower well. n=1 is roughly equal on the two sides. n=2 is strongly peaked on the right. n=3 and higher is about the same on both sides. The n=2 case is the one where both sides are classically allowed, and the energy is comparable to Vo.

9. Jan 1, 2013

Staff Emeritus
I played a little with the applet to try and build intuition.

Starting from atyy's conditions, I made the left well deeper and deeper. As I did this, the wavefunction moved more and more to the left. Classically, a particle that isn't moving sits at the bottom of the potential well.

I then went back to the original potential, and started increasing n. For n large and/or even, the situation looks like it does classically - the particle spends most of its time in the shallower potential. For n small and odd (i.e. 1 for this potential, 1 or 3 for a deeper potential where the 4 becomes a 20) it starts to look like a finite potential well - the wave function is oscilliatory in the deep well and exponential in the shallow well.

10. Jan 1, 2013

### atyy

While the rough intuition for the 2 extremes (ground state and highly excited states) seems ok, it does indeed seem to fail for small, even n like the first excited state. I guess it's too much to ask for intuition to cover everything? After all, it already fails for explaining why the ground state has non-zero expected kinetic energy?

(It's intuitive if we know it's a wave equation, but I don't think that's what was being asked.)

11. Jan 1, 2013