Qal QuantP1S

Solution:

• [math] [- \frac{\hbar^2}{2M}\Delta^2 + V]W(x,y,z)=E W(x,y,z) [/math]
  

In our case, using separation of variables, we will get 3 differential equations for x, y and z. W(x,y,z)=w(x)w(y)w(z)

[math]\frac{d^2 w(x)}{dx^2} - m^2 w(x) = 0[/math](1)
The same will be for y and z.

Solution of equation (1) is following
[math]w(x) = A\sin(mx)+B\cos(mx)[/math]
[math]w(y) = C\sin(ky)+D\cos(ky)[/math]
[math]w(z) = E\sin(qz)+F\cos(qz)[/math]

• Applying B. C. at x=y=z=0 wave function should be zero, that means B=D=F=0. We have
  

[math]w(x) = A\sin(mx) [/math]
[math]w(y) = C\sin(ky) [/math]
[math]w(z) = E\sin(qz) [/math]

Also, w(a)=0 which gives [math]A\sin(ma)=0, m=\frac{\pi n_x}{a}[/math]. For y component [math]C\sin(kb)=0, k=\frac{\pi n_y}{b}[/math] and for z [math]E\sin(qc)=0, q=\frac{\pi n_z}{c}[/math]

A, C and E are normalization constants

[math]\frac{1}{A^2}=\int\sin^2 (\pi nx/a) dx = \frac{a}{2} [/math], limits are from 0 to a.

The eigenfunction for each component will be

[math]w(x) = \sqrt{\frac{2}{a}} \sin(\pi n_x x/a)[/math]
[math]w(y) = \sqrt{\frac{2}{b}} \sin(\pi n_y y/b)[/math]
[math]w(z) = \sqrt{\frac{2}{c}} \sin(\pi n_z z/c)[/math]

The eigenenergies

[math]E_{n_x} = \frac{\pi^2 \hbar^2 {n_x}^2}{2Ma^2}[/math], [math]E_{n_y} = \frac{\pi^2 \hbar^2 {n_y}^2}{2Mb^2}[/math], [math]E_{n_z} = \frac{\pi^2 \hbar^2 {n_z}^2}{2Mc^2}[/math]
Total energy is sum of these energies.

• [math]E = \frac{\pi^2 \hbar^2 n^2}{2M^2 a^2}[/math], where [math]n^2=n_x ^2 + n_y ^2 + n_z ^2[/math], n=1,2,3...

2.)Solution:

a.) [math]H=\begin{pmatrix}w_1&v\\v&w_2\end{pmatrix}[/math]

[math]H=\begin{pmatrix}(w_1-\lambda)&v\\v&(w_2-\lambda)\end{pmatrix}[/math]

[math](w_1 - \lambda)(w_2 - \lambda) - v^2=0[/math]

[math]\lambda_{1,2} = \frac{(w_1 + w_2)+/- \sqrt{(w_1 + w_2)^2 - 4(w_1 w_2 - v^2)}}{2}[/math]

Dr. Forest: We have not had Perturbation Theory.