# Two (almost) independent infinite square wells

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1. Jul 20, 2015

### treynolds147

1. The problem statement, all variables and given/known data
Consider a one-dimensional, nonrelativistic particle of mass m which can move in the three regions defined by points A, B, C, and D. The potential from A to B is zero; the potential from B to C is (10/m)(h/ΔL)2; and the potential from C to D is (1/10m)(h/ΔL)2. The distance from A to B is ΔL; the distance from B to C is 10ΔL; and the distance from C to D is chosen such that the ground state has the same probability between A and B as between C and D. Exploit the fact that this potential is very close to that of two independent infinite square wells to estimate the distance from C to D.

2. Relevant equations
For an infinite square well with zero potential, En = h2n2/8ma2 (a being the width of the well), and ψn=√(2/a) sin(nπx/a)

3. The attempt at a solution
It's been quite a while since I've needed to use my physics, so I'm pretty rusty, and I'm trying to start brushing up on it again before I start grad school. This one's kind of thrown me for a bit of a loop, maybe because the wording seems weird to me. From what I'm understanding, the probability that E1,AB is measured is equal to that the probability that E1,CD is measured. Based on this I've been trying to equate the expansion coefficient for the ground state ∫ψn*ψ dx for a particle in region AB and for a particle in region CD, but this approach has proven fruitless so far. Any thoughts on how to proceed, or am I completely misunderstanding this one?

2. Jul 21, 2015

### fzero

This seems to be a pretty tricky problem for a brushing up. To start out with, the problem wants you to compare the probability to find the particle in the region $(A,B)$ with that in the region $(C,D)$. The probability to find the particle in $(A,B)$ is
$$\frac{\int_A^B |\psi(x)|^2 dx}{\int_A^D |\psi(x)|^2 dx}.$$
When we compare this to the probability for $(C,D)$, the denominators cancel, so in the approximation we won't need to know the wavefunction for the region $(B,C)$.

Post back if you're still having trouble. Like I mentioned, I think this problem is tricky if you're rusty, since you have to make some assumptions, for example to approximate the ground state energy.

3. Jul 21, 2015

### treynolds147

If then the integrals between AB and CD are equal, would that not mean that the distance between C and D is ΔL as well? I can't imagine that the ground state wavefunction would be altered by CD's non-zero bottom, just the energy levels would be shifted. Since both regions AB and CD are approximately infinite square wells, they'd have the same functional form for the wavefunction, so having equal probabilities must mean that they have the same length as well.

4. Jul 21, 2015

### fzero

Sure, the functional form is the same, but the values of the integration over the same distance are not the same, because of the dependence of the wavefunction on the different energy. Try to work out the expressions for the two integrals that you have, using an unknown, say $d$ for the distance CD.

5. Jul 21, 2015

### treynolds147

Alright, so we'd have a wavefunction that looks like ψtotal=aψAB+aψCD, where c is the expansion coefficient, right? They'd have the same coefficient since the probability of obtaining the particle in either well is supposed to be equal. And since the two wells are approximately independent we can ignore any interaction terms that pop up in the integral, I'm guessing. Then we do the integral of the modulus squared of ψtotal over AB and over CD and equate them, yes? Doing this I end up getting an implicit formula for d though, so I must have gone wrong somewhere. Unless this is just an unusually cruel problem!

6. Jul 22, 2015

### fzero

The wavefunction isn't the sum of terms, rather it's defined piecewise:
$$\psi(x) = \begin{cases} \psi_{AB}(x), & A \leq x < B, \\ \psi_{BC}(x), & B \leq x < C, \\ \psi_{CD}(x), & C \leq x \leq D, \end{cases}$$
so it's for this reason that you don't have cross terms. $\psi_{AB} = \sin k_{AB} x$ and $\psi_{CD = \sin k_{CD} x}$ are infinite square well sin functions, under the assumptions. In general, the coefficients could be different numbers, but I think you have to assume they are approximately the same in order to complete the problem.

Now, the boundary conditions would be very important if we were solving the full problem. For example, the wavefunctions have to be zero at points $A$ and $D$, where the potential is infinite. For this approximate solution, we can choose point $A$ to be $x=0$, so $\psi_{AB}(A) = 0$ automatically. $\psi_{CD}(D)=0$ is more problematic. Point $D$ is $x=11\Delta L +d$ and so we'd ordinarily require
$$k_{CD} D = \frac{\sqrt{ 2m(E-V_{CD}}}{\hbar } (11\Delta L +d)$$
to be a multiple of $\pi$. However, this gives a mess that relates the ground state energy $E$ to $d$. If we use this, we'll get an implicit equation for $d$ like you suggest.

So I think we need to be simpler in our assumptions. When we integrate, we'll find a term
$$\frac{1}{4 k_CD} \sin ( 2 k_{CD} D).$$
This vanishes if $k_{CD} D$ is a multiple of $\pi$, so we can drop it. We still don't know the ground state energy $E$, but since the potential in the $CD$ well $V_{CD}$ is 20% of the ground state energy for a well of width $\Delta L$, we can just approximate the ground state energy for the two well problem by the ground state energy of the $AB$ well. If we do this, setting our integrals equal will give us 3 terms :
$$\frac{d_{AB}}{2} = \frac{d_{CD}}{2} + \frac{1}{k_{CD}} \sin (2k_{CD} C) .$$
I've dropped a term $\sin (2 k_{AB} B)$ since this vanishes for our choice of ground state energy.

With these assumptions, $d_{CD}=d$ only appears linearly. The argument of the sin that appears has $C = 11\Delta L$, and the other terms just depend on $m$ and $\Delta L$.

Having never solved the full problem, I don't know how good this approximation is. I think, given the way the problem is stated, that the assumptions I suggest are the ones that make the problem tractable.

7. Jul 22, 2015

### treynolds147

Ah, that was my hold up, I was treating the the sine of C as being a vanishing quantity. Yes, this makes much more sense, thank you so much. I didn't think this problem would trip me up as much as it did, I guess I'm really out of practice!