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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.

The main purpose of today's lecture was to solve for the odd solutions to a finite square well. A first point to consider is that an odd solution will be antisymmetric across the origin. This is what we focused on today.

For the finite square well we choose the axis in the middle of the well instead of the edge of the well like in the infinite square well this is standard convention. As will be seen the algebra is easier because of this convention. So, the well starts at -a and goes to +a and has a depth of V_0 and E<0. For this well we seperate it into three regions so we can evaluate this problem. Then for each region we have:

Since E < 0 and we are only looking for odd solutions we can simplify our representation of the the waves listed above.

We know that, in region 1 as x→ -∞ <math>\psi_1</math> must go to 0. A' must be zero or else <math>\psi_1</math> would → ∞. Cant have it. Also, for the same logic, in region 3 as x→ ∞, <math>\psi_{2}</math> must go to zero D'=0

Where,
k= <math> \frac{sqrt{-2mE}}{\hbar} </math>
And,
k'= <math> \frac{sqrt{(V_0+E)2m}}{\hbar} </math>
b/c E<0
Since we know that k and k' are related to E we can say that with the wavefunctions we have 4 unknowns. At this point we must be careful due to the fact that we are only looking for odd solutions. These are:
<math>\psi_1 (-a)=\psi_2 (-a) </math>
⇒<math>Ae^{k(-a)}=Be^{ik'(-a)}+Ce^{-ik'(-a)}</math>
<math>\partial_x\psi_1 (-a)=\partial_x\psi_2 (-a) </math>
⇒<math>kAe^{k(-a)}=ik(Be^{ik'(-a)}-Ce^{-ik'(-a)})</math>
At this point we impose only odd solutions boundary conditions:
<math>\psi_2 (0)=0</math>
Due to the symmetry of the potential well and the fact that we are looking for antisymmetric solutions.
⇒B=C
And
<math>\psi_{1}=-\psi_{3}</math>
⇒<math>Ae^{kx}=-Ae^{-kx}</math> incomplete at this time

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