Consider a half infinite well with U(x) = U0 for x ≥ L, U(x)...

Consider a half infinite well with $U(x) = U_0$ for $x \ge L$, $U(x) = 0$ for $0 < x < L$, and $U(x) = \infty$ for $x \le 0$. • (a) Sketch the potential. (0.5 points) • (b) Write generic solutions to the Schroedinger equation appropriate in the various regions. (1.5 points) • (c) Impose the continuity conditions, and obtain the energy quantization condition (2 points): $\sqrt{E} \cot \left(\frac{\sqrt{2mE}}{\hbar}L\right) = -\sqrt{U_0 - E}$. (3)

Transcript

00:01 All right, so we're looking at the time -independent shortage of equation in three dimensions, and we're asked to separate this into three equations for each dimension.

00:12 So i have x, y, and z.

00:13 So what we're going to do is look at our initial wave function, just dependent on x, y, and z.

00:23 We're going to split it up into three wave functions, which we'll just call call x like capital x so this is our x part capital y is our y part and capital z is our z part so i couldn't think of any better lettering to do so we're going to break the wave function up into a product of three smaller ones that only depend on their respective coordinates and so if we look at the time independent schrodinger equation it looks something like this um right here and i've dropped the x y and z dependence for notation's sake for now and so if we plug this in here what are we going to get we're going to get well this nabla squared of this laploccian operator looks like this i'm sure you're familiar with right it's the second partial derivative with respect to x plus the second partial derivative with respect to y plus second partial derivative with respect to z so if we plug that in here what we're going to get is you know negative h bar squared to m and then we're going to get this x part plus this y part plus the z part the laploccian operator acting on x y and z so i'm using capital letters to denote the wave functions in each coordinate plus v of x y and z i probably should have made the notation explicit there so v is presumably dependent on all three coordinates, and this is still multiplied by x, y, and z, and then this is equal to e times si, which is, of course, is x, y, z.

02:15 And so we notice, since this is a linear operator, it's not going to combine x, y, and z in any strange way.

02:22 So it's, when we expand out what this operator does, the ploshan operator, this part, actually might be better, if i write this a little bit differently.

02:34 The first part only acts on the x function.

02:37 So the y and z functions are left unchanged.

02:40 So what we've got is our second derivative of the x function like this.

02:45 And then the next part only acts on the y part.

02:48 So it leaves the x and z parts unchanged.

02:54 And then similarly, the last part only acts on the z part...

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