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Please try to include the following

• main points understood, and expand them - what is your understanding of what the points were.
  • expand these points by including many of the details the class discussed.
• main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
  • Other classmates can step in and clarify the points, and expand them.
• How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture.
• wonderful tricks which were used in the lecture.

Today we continued with the finite square well potential. From lecture we have:

<math>\psi_I(x) = Ae^{\kappa x}</math> for <math>x < -a</math>

<math>\psi_{II}(x) = B(e^{ikx} - e^{-ikx}) = \beta sin(kx)</math>, where <math>\beta = 2iB</math> for <math> -a < x < a </math>

<math>\psi_{III}(x) = -Ae^{-\kappa x}</math> for <math>x > a</math>

Then we can apply the boundary conditions:

1. Continuity of the wavefunction across the boundary:

<math>\psi_I(-a) = \psi_{II}(-a)</math>

2. Continuity of the wavefunction's derivative across the boundary:

<math>\left. \frac{\partial\psi_{I}(x)}{\partial x}\right|_{x = -a} = \left. \frac{\partial\psi_{II}(x)}{\partial x}\right|_{x = -a}</math>

Our unknowns are A, B, k, and <math>\kappa</math>. It is most important to find k and <math>\kappa</math> since they are related to the energy. Thus, we will not worry about A and B/<math>\beta</math> for the moment.

If you forget how k and <math>\kappa</math> relate to the energy, you can think of it in terms of the momentum first:

Outside of the well, the momentum p is equal to <math>\hbar\kappa</math>. Remember that <math>KE = E - PE \Rightarrow KE = 0 - E = -E</math>. Since <math>KE = \frac{p^2}{2m} \Rightarrow -E = \frac{\hbar^2\kappa^2}{2m}</math>. Thus, <math>\kappa = \frac{\sqrt{-2mE}}{\hbar}</math>.

We can apply the same idea for inside the well. <math>KE = E - PE = E - (-V_0) = E + V_0</math>. The momentum in the well is <math>\hbar k</math>. By applying the above relation, we get <math>k = \frac{\sqrt{2m(E + V_0)}}{\hbar}</math>.

By applying the boundary conditions, we get:

1. <math>\beta\sin{ka} = Ae^{\kappa a}</math>

2. <math>k\beta\sin{ka} = -\kappa A e^{-\kappa a}</math>

If we divide Equation 2 by Equation 1, we get:

<math>k\cot{ka} = -\kappa \Rightarrow \cot{ka} = -\frac{\kappa}{k}</math>

To avoid the confusion of k and <math>\kappa</math>, we will introduce new different variables Z and <math>Z_0</math>.

Since <math>k^2=\frac{-2mE}{h^2}</math> and <math>\kappa^2=\frac{-2m(E+V_0)}{h^2}</math>,

we define<math>Z_0^2=\frac{-2ma^2V_0}{h^2}</math> and <math>Z^2=k^2a^2=\frac{-2ma^2(E+V_0)}{h^2}</math>

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