5. [6 Marks] Consider the finite square well potential V(x)...

5. [6 Marks] Consider the finite square well potential \(V(x) = \begin{cases} -V_0 & \text{for } -a \le x \le a \\ 0 & \text{for } |x| > a \end{cases} \), for bound states (E<0). a) By considering solutions to the Schrödinger equation, apply the continuity of \(\Psi(x)\) and d\(\Psi\)/dx at x = a and x = -a, to obtain \(\Psi(x) = \begin{cases} Fe^{-\kappa x} & \text{for } x > a \\ D\cos(\ell x) & \text{for } 0 < x < a, \\ \Psi(-x) & \text{for } x < 0 \end{cases} \) where \(\ell = \frac{\sqrt{2m(E+V_0)}}{\hbar}\) and \(\kappa = \frac{\sqrt{-2mE}}{\hbar}\). b) By applying the continuity of \(\Psi(x)\) and d\(\Psi\)/dx at x = a obtain a formulation for the allowed energies letting Z = \ell a and Z_0 = (a/\hbar)\sqrt{2mV_0}. How might this formulation be solved? c) Consider the two limiting cases of a 1) wide, deep well and 2) a shallow narrow well. In Case 1 what is the limit as V_0 \to \infty? As Z_0 \to 0 in Case 2 what is the minimum number of bound states?

Transcript

00:01 At what's called the semi -infinite square well.

00:06 Namely, we're going to have a potential that's infinite on one side and finite on the other side and zero in between.

00:14 So let's draw a little picture of what we're talking about.

00:18 We have something that looks like this, where it's zero in this region from x equals zero to x equals l.

00:25 It's infinite at x equals zero and some finite value beyond l.

00:31 So we have to do a number of things here.

00:34 First, let's write this up a bit.

00:36 So, si, well, rather, u for x less than zero is going to be infinity.

00:46 U for x between l and zero is zero.

00:51 And u is some finite value, call it u not.

01:00 So we're asked for a number of things.

01:02 We're first asked for the form for si in the region where a u is zero.

01:12 We might already be somewhat familiar with this from looking at the infinite square well.

01:16 And then we're asked for what side needs to look like for x greater than l.

01:23 Once we establish a form for side one and side two, we need to impose boundary conditions, namely that si one needs to be continuous with side two at l.

01:33 And in addition, its derivative needs to be continued.

01:37 That will impose some conditions that will allow us to patch together these two functions at the boundary.

01:45 And then finally, we need to prove a kind of complex relationship, namely that given little k equal to, equal to root 2me over each bar squared, and big k, we'll write a little script k, equal to root negative 2m over each bar squared, e minus mu not.

02:15 Given these two, we need to prove that this relation, little k, cotangent, little k l, equals big k we need to prove that this relationship is true and it's going to come as a consequence of these two boundary conditions so let's jump into it let's take a look at the region where u equals zero so when we're talking about a zero potential this is the most like a infinite square well so a natural choice for for our wave function is some combination of signs and cosines so that it's not blowing up anywhere it's oscillatory it's wave like it's probably going to be a solution to the time independent schrodinger equation.

03:00 In the same way that we did the infinite square well, we can insert k as an argument to the sign function.

03:09 That's going to serve us well.

03:11 The reason why i am doing this is because it's been done in the past with the infinite square well.

03:15 So it's kind of that's our intuition.

03:16 That's where our little k is going to appear.

03:19 We can set equal to 2m .e over h bar squared.

03:26 So if we were to plug this into the schrodinger equation for u equal zero, this will be a solution.

03:31 With this value of k, it will work out.

03:35 You can verify that for yourself if you want.

03:37 So we can impose one of the boundary conditions, the left infinite boundary condition on this solution, namely that size 0 needs to be 0 at that edge.

03:48 If we do this, we see that a sign of 0 is going to just zero out a, and b, cosine of 0 will not be zeroed out.

04:00 Thus, b must be equal to 0.

04:06 So we can eliminate the cosine portion, and we're left with the sign portion.

04:11 So we have that psi 1 x equals a sine kx, a solution of the shardering equation in its own domain.

04:21 So now let's talk about the region where the potential is some finite value.

04:29 So this is when x is greater than now.

04:32 Man, so a little bit of lag in the pen.

04:38 So in this case, we cannot have oscillatory solutions.

04:44 We also cannot allow the function to blow up to infinity as x grows to infinity.

04:54 So instead of trying oscillatory solutions, another good thing to try are simply exponential solutions.

05:08 You can say c, e to the, here's where i'm going to insert kappa, just as a hunch.

05:17 So you have some exponential solutions where we can call kappa, whatever we want.

05:23 But we don't even have to define it yet.

05:26 I'm actually show this one.

05:29 So in order for these to be a solution of the schrodinger equation in this region, we're going to have to determine what kappa is in terms of mass and energy, and we'll actually plug it into the schroaringer equation.

05:38 Before we do that, we can actually eliminate this portion of the exponentials using the, i guess we go on a boundary condition.

05:45 As x goes to infinity, we cannot allow the wave function to explode.

05:54 Thus, it must be the case...

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