Qal QuantP1S

Solution:

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

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)

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

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

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

$w(x) = A\sin(mx)$
$w(y) = C\sin(ky)$
$w(z) = E\sin(qz)$

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

A, C and E are normalization constants

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

The eigenfunction for each component will be

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

The eigenenergies

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

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

2.)Solution:

a.) $H=\begin{pmatrix}w_1&v\\v&w_2\end{pmatrix}$

$H=\begin{pmatrix}(w_1-\lambda)&v\\v&(w_2-\lambda)\end{pmatrix}$

$(w_1 - \lambda)(w_2 - \lambda) - v^2=0$

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