Exam 2, more harmonic oscillators

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Latest revision as of 16:20, 29 March 2006

test 2

problem 1

$\frac{- \hbar ^2}{2m} \frac{d^2}{dx^2} \psi(x) = E \psi(x)$

$\psi(-a/2) = \psi(a/2) = 0 \$

Inside the well we should expect plane wave solutions:

$\psi(x) = A e^{i k x} + B e^{-ikx} \$

Where $k^2 = \frac{2m}{\hbar ^2} E \$

Applying the BC this gives:

$A e^{-ika/2} + B e^{ika/2} = 0\$

$A e^{ika/2} + B e^{-ika/2} = 0\$

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:

$2A \cos(ka/2) + 2B \cos(ka/2) = 0 \$

now subtract to get:

$2iA \sin(ka/2) - 2iB \sin(ka/2) = 0\$

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

$(A+B) \cos(ka/2) = 0 \$

$(A-B) \sin(ka/2) = 0 \$

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 $\cos(ka/2) = 0 \$ . This requires

$ka/2 = (n-1/2) \pi \$

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

$ka/2 = n \pi \$

So we have two families of solutions:

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

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

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

Now since $k^2 = \frac{2m}{\hbar ^2} E \$ we have

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

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

In numerical order these are:

$\xi \times \{ 1, 4, 9, 16, ... \}$

problem 5

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

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

@@ Line 1: / Line 1: @@
+===test 2===
+* problem 1
 <math> \frac{- \hbar ^2}{2m} \frac{d^2}{dx^2} \psi(x) = E \psi(x) </math>
@@ Line 4: / Line 8: @@
 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:
@@ Line 40: / Line 47: @@
 So we have two families of solutions:
-<math> \psi(x) = A \left(e^{i k x} + e^{-ikx} \right) = A \cos( \frac{2 \pi (n-1/2)}{a} x)   \ </math>
+<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>
-<math> \psi(x) = B \left(e^{i k x} - e^{-ikx} \right) = B \sin( \frac{2 \pi n}{a} x)   \ </math>
+[[more on harmonic oscillators]]

Exam 2, more harmonic oscillators

Latest revision as of 16:20, 29 March 2006

test 2

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