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===test 2===

* problem 1

<math> \frac{- \hbar ^2}{2m} \frac{d^2}{dx^2} \psi(x) = E \psi(x) </math>

<math> \psi(-a/2) = \psi(a/2) = 0 \ </math>

Inside the well we should expect plane wave solutions:

<math>\psi(x) = A e^{i k x} + B e^{-ikx} \ </math>

Where <math> k^2 = \frac{2m}{\hbar ^2} E \ </math>

Applying the BC this gives:

<math> A e^{-ika/2} + B e^{ika/2} = 0\ </math>

<math> A e^{ika/2} + B e^{-ika/2} = 0\  </math>

At this point you have two equatins and two unknowns.  I don't care how you solve them.  Here is one way.
Add the two equations to get:

<math> 2A \cos(ka/2) + 2B \cos(ka/2) = 0 \ </math>

now subtract to get:

<math> 2iA \sin(ka/2) - 2iB \sin(ka/2) = 0\ </math>

These are nice since we can factor out the common sine and cosine terms:

<math>(A+B) \cos(ka/2) = 0 \ </math>

<math>(A-B) \sin(ka/2) = 0 \ </math>


There are only two ways we can satisfy both equations simultaneously.  If A=B then the second equatino is
taken care of, but then A+B cannot = 0, so we have to insist that <math>\cos(ka/2) = 0 \ </math>.  This requires

<math> ka/2 = (n-1/2) \pi  \ </math>

On the other hand, if we choose A = -B, then the first equation is automatically satisfied and then we must
insist that <math>\sin(ka/2) = 0 \ </math>.   This requires

<math> ka/2 = n \pi \ </math>

So we have two families of solutions:

<math> \psi(x) = A \left(e^{i k x} + e^{-ikx} \right) = A \cos \left( \frac{2 \pi (n-1/2)}{a} x \right)   \ </math>


<math> \psi(x) = B \left(e^{i k x} - e^{-ikx} \right) = B \sin \left(\frac{2 \pi n}{a} x \right)   \ </math>


The cosine states are even (have even parity), the sine states are odd (have odd parity)

Now since <math> k^2 = \frac{2m}{\hbar ^2} E \ </math> we have

<math> E_n = \frac{\hbar^2}{2m} \left( \frac{2}{a} \right) ^2 (2 n \pi)^2  = \xi 4(n-1/2)^2  \ </math> 
for the odd parity states and


<math> E_n = \frac{\hbar^2}{2m} \left(\frac{2}{a} \right)^2 (2 (n- 1/2) \pi)^2  = \xi 4 n^2 \ </math> 
for the even parity states.


In numerical order these are:


<math> \xi  \times \{ 1, 4, 9, 16, ... \}  </math>


* problem 5

<math> [x,p_x] \ </math> is an operator equation.  By definition it is:
<math> (x p_x - p_x x) \ </math>.  In order to work this out it is helpful to apply the operator to a test function.

<math> [x,p_x] f(x) = (x p_x - p_x x)f(x) = (x (-i \hbar d/dx)) f(x) - ((-i \hbar d/dx) x) f(x) </math>
<math> = -i \hbar \left( x \prime{f} - (x \prime{f} - f) \right) = i \hbar f  </math>