Degenerate Perturbation Theory (Particle in 3D box)

[d3nat]

Homework Statement

Consider a particle confined in a cubical box with the sides of length L each.
Obtain the general solution to the eigenvalues and the corresponding eigenfunctions.

Compute the degeneracy of the first excited state.

A perturbation is applied having the form

H' = V from 0 <x < $\frac{L}{2}$ and 0<y<$\frac{L}{2}$
= 0 elsewhere

Compute the first order shifts in the energy levels of the first excited state due to the perturbation.
Construct the corresponding eigenstates.

Homework Equations

The Attempt at a Solution

I'm struggling with perturbation theory.

I solved the first part of the problem.
the first excited state is triply degenerate
and has Energy equal to:
$E = \frac{\pi^2\hbar^2}{2mL^2}(n_1^2+n_2^2+n3^2) = \frac{6\pi^2\hbar^2}{3mL^2}$$\psi = (\frac{2}{L})^\frac{3}{2}sin(k_1x)sin(k_2y)sin(k_3z)$

the quantum numbers have to start at n = 1

so for the excited state the degeneracy is

$n_1 = 1, n_2 = 1, n_3 = 2$ corresponds to $\psi_1$ and $E_1$
$n_1 = 1, n_2 = 2, n_3 = 1$ corresponds to $\psi_2$ and $E_2$
$n_1 = 2, n_2 = 1, n_3 = 1$ corresponds to $\psi_3$ and $E_3$
Now, I've used projection operators on the Hamiltonian and have gotten through a series of steps the following matrix:

\begin{array}{cc}
E_1 + <1|V|1> & <1|V|2> & <1|V|3> \\
<2|V|1> & E_2 + <2|V|2> & <2|V|3>\\
<3|V|1> & <3|V|2> & E_3+<3|V|3>
\end{array}

which is equal to the matrix:

\begin{array}{cc}
<1|i>\\
<2|i>\\
<3|i>
\end{array}

with $E_i$ multiplied in front (sorry, I don't know how to write that here)

Now, I think I should bring that over and solve like a normal

$(A-\lambda I)x = 0$ problem

but what do I do with the subscript i?
Do I drop it?

I also don't know what to do from here.
I tried asking my professor, but he just told me to "solve the integrals."

What integrals??
I'm so lost.

Any help would really be appreciated.
I know that matrix with the bras and kets should have lots of zeros, but I don't know how to get it like that...

 

[vela]

Homework Statement

Consider a particle confined in a cubical box with the sides of length L each.
Obtain the general solution to the eigenvalues and the corresponding eigenfunctions.

Compute the degeneracy of the first excited state.

A perturbation is applied having the form

H' = V from 0 <x < $\frac{L}{2}$ and 0<y<$\frac{L}{2}$
= 0 elsewhere

Compute the first order shifts in the energy levels of the first excited state due to the perturbation.
Construct the corresponding eigenstates.

Homework Equations

The Attempt at a Solution

I'm struggling with perturbation theory.

I solved the first part of the problem.
the first excited state is triply degenerate
and has Energy equal to:
$E = \frac{\pi^2\hbar^2}{2mL^2}(n_1^2+n_2^2+n3^2) = \frac{6\pi^2\hbar^2}{3mL^2}$


$\psi = (\frac{2}{L})^\frac{3}{2}sin(k_1x)sin(k_2y)sin(k_3z)$

the quantum numbers have to start at n = 1

so for the excited state the degeneracy is

$n_1 = 1, n_2 = 1, n_3 = 2$ corresponds to $\psi_1$ and $E_1$
$n_1 = 1, n_2 = 2, n_3 = 1$ corresponds to $\psi_2$ and $E_2$
$n_1 = 2, n_2 = 1, n_3 = 1$ corresponds to $\psi_3$ and $E_3$



Now, I've used projection operators on the Hamiltonian and have gotten through a series of steps the following matrix:

\begin{array}{cc}
E_1 + <1|V|1> & <1|V|2> & <1|V|3> \\
<2|V|1> & E_2 + <2|V|2> & <2|V|3>\\
<3|V|1> & <3|V|2> & E_3+<3|V|3>
\end{array}

which is equal to the matrix:

\begin{array}{cc}
<1|i>\\
<2|i>\\
<3|i>
\end{array}

with $E_i$ multiplied in front (sorry, I don't know how to write that here)

How can a 3x3 matrix be equal to a 3x1 matrix? What did you get for the matrix elements?

Now, I think I should bring that over and solve like a normal

$(A-\lambda I)x = 0$ problem

but what do I do with the subscript i?
Do I drop it?

I also don't know what to do from here.
I tried asking my professor, but he just told me to "solve the integrals."

What integrals??
I'm so lost.

Any help would really be appreciated.
I know that matrix with the bras and kets should have lots of zeros, but I don't know how to get it like that...