===== Oct 14 (Wed) Finite Square Well=====
** Responsible party: Can, prest121 ** 

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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.

====Main points ====
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 2 new variables Z and <math>Z_0</math>. 


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


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


thus <math>\kappa^2a^2=Z^2-Z_0^2</math>


thus <math>\frac{\kappa}{k}=\sqrt{\frac{\kappa^2a^2}{k^2a^2}}=\sqrt{\frac{Z^2-Z_0^2}{Z^2}}=\sqrt{(\frac{Z_0}{Z})^2-1}\</math>  


<math>\Rightarrow -\cot{Z}=\frac{\kappa}{k}= \sqrt{(\frac{Z_0}{Z})^2-1}</math>


thus we are looking for the solution of Z for the 2 transcendental equation <math>\Rightarrow -\cot{Z}=\frac{\kappa}{k}= \sqrt{(\frac{Z_0}{Z})^2-1}</math>

Since it is really hard to solve it analytically, we can solve it graphically. 

{{:classes:2009:fall:phys4101.001:qm_lec_pic.jpg|}}\\  **the cotangent part of the graph does not look quite right! //Yuichi//**

as we can see from the graph, the periodic functions are CotZ, and the other one is <math>\sqrt{(\frac{Z_0}{Z})^2-1}</math>


The physical meaning of the solution, which is the intersection point, represents one bound state. In other words, the total number of intersection points tell us the number of bound state of the system.
Say when <math>Z_0=1</math>, it yields one intersection, which means we only have one bound state.

The other line is when <math>Z_0=10</math>, which has more intersections denoting more bound states. 

Remember that <math>Z_0^2=\frac{2ma^2V_0}{\hbar^2}</math>, the magnitude of Z0 is determined by the production of a^2 and V0, a is the width of potential well, and V0 is the depth of the potential well. If we keep a constant , raise the potential to infinity, Z0 goes to infinity, then we have infinite square well, which corresponds with infinite intersection on graph, and we would have infinite bound states.

If we keep potential constant, and increase the width of the well, it would also increase the number of bound states. 


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