Degenerate perturbation theory (an example)

Consider the eigenvalue equation of eq262 for a three-dimensional infinite cubical well, where

$\hat{H}^{(0)}(x,y,z)=\left\{\begin{matrix} 0 & if\: 0<x<a,0<y<a,0<z<a\\ \infty & otherwise \end{matrix}\right.\; \; \; \; \; \; \; \; 277$

$\psi_{n_x,n_y,n_z}^{(0)}=\left ( \frac{2}{a} \right )^{\frac{3}{2}}sin\left ( \frac{n_x\pi}{a}x \right )sin\left ( \frac{n_y\pi}{a}y \right )sin\left ( \frac{n_z\pi}{a}z \right )\; \; \; \; \; \; \; \; 278$

$E_{n_x,n_y,n_z}^{(0)}=\frac{\pi^{2}\hbar^2}{2ma^2}(n_x^{\;2}+n_y^{\;2}+n_z^{\;2})\; \; \; \; \; \; \; \; 279$

where $n_x,n_y,n_z\in \mathbb{Z}^+$ .

From eq278, the ground state is $\psi_{111}^{(0)}$ , while the first excited state is characterised by $\psi_1=\psi_{112}^{(0)}$ , $\psi_2=\psi_{121}^{(0)}$ and $\psi_3=\psi_{211}^{(0)}$ . It is obvious, from eq279, that the first excited state is degenerate. Let’s assume that the potential in the well is perturbed such that first-order correction to $\hat{H}^{(0)}$ is

$\hat{H}^{(1)}=\left\{\begin{matrix} V_0,&if\: 0<x<\frac{a}{2},0<y<\frac{a}{2},0<z<a\\0& otherwise \end{matrix}\right.\; \; \; \; \; \; \; \; 280$

We shall analyse the degenerate subspace of the first excited state. The first step is to compute the matrix elements $W_{jl}=\langle\psi_{j,D1}^{(0)}\vert\hat{H}^{(1)}\vert\psi_{l,D1}^{(0)}\rangle$ and substitute them in eq274, with the assumption that $\Psi_n^{(0)}=\alpha_1\psi_1+\alpha_2\psi_2+\alpha_3\psi_3$ :

$\begin{pmatrix} \frac{V_0}{4} &0 &0 \\ 0 & \frac{V_0}{4} &\frac{16V_0}{9\pi^2} \\ 0 &\frac{16V_0}{9\pi^2} &\frac{V_0}{4} \end{pmatrix}\begin{pmatrix} \alpha_1\\\alpha_2 \\ \alpha_3 \end{pmatrix}=U_n^{(1)}\begin{pmatrix} \alpha_1\\\alpha_2 \\ \alpha_3 \end{pmatrix} \; \; \; \; \; \; \; \; 281$

Finding the non-trivial solutions of the above linear homogeneous equation,

$\begin{vmatrix} \frac{V_0}{4}-U_n^{(1)}&0 &0 \\ 0 & \frac{V_0}{4}-U_n^{(1)}&\frac{16V_0}{9\pi^2} \\ 0 &\frac{16V_0}{9\pi^2} &\frac{V_0}{4}-U_n^{(1)} \end{vmatrix}=0$

$\left ( \frac{V_0}{4}-U_n^{(1)}\right ) \left [\left ( \frac{V_0}{4}-U_n^{(1)}\right )^2-\frac{256V_0^2}{81\pi^4} \right ] =0$

The roots are $U_1^{(1)}=\frac{V_0}{4}$ , $U_2^{(1)}= \frac{V_0}{4}+\frac{16V_0}{9\pi^2}$ and $U_3^{(1)}= \frac{V_0}{4}-\frac{16V_0}{9\pi^2}$ , which means that the perturbation $\hat{H}^{(1)}$ lifts the degeneracy of the unperturbed states. To find the exact eigenstates $\Psi_n^{(0)}$ , we substitute the roots back in eq281. For the first root, we have

$\begin{pmatrix} \frac{V_0}{4}\alpha_1\\\frac{V_0}{4}\alpha_2+ \frac{16V_0}{9\pi^2}\alpha_3\\ \frac{16V_0}{9\pi^2}\alpha_2+ \frac{V_0}{4}\alpha_3 \end{pmatrix}=\frac{V_0}{4}\begin{pmatrix} \alpha_1\\\alpha_2 \\ \alpha_3 \end{pmatrix}$

Comparing both sides of the above equation, $\alpha_1$ can be any number but $\alpha_2=0$ and $\alpha_3=0$ . So, the normalised wavefunction associated with $U_1^{(1)}$ is $\Psi_1^{(0)}=\psi_1$ . For the second root, we have

$\begin{pmatrix} \frac{V_0}{4}\alpha_1\\\frac{V_0}{4}\alpha_2+ \frac{16V_0}{9\pi^2}\alpha_3\\ \frac{16V_0}{9\pi^2}\alpha_2+ \frac{V_0}{4}\alpha_3 \end{pmatrix}=\left ( \frac{V_0}{4}+\frac{16V_0}{9\pi^2}\right )\begin{pmatrix} \alpha_1\\\alpha_2 \\ \alpha_3 \end{pmatrix}$

Comparing both sides, $\alpha_1=0$ since $V_0\neq0$ , and $\alpha_2=\alpha_3=1$ ( $\alpha_2$ and $\alpha_3$ are not equal to zero because that will result in $\Psi_2^{(0)}$ being a zero eigenfunction). The normalised wavefunction associated with $U_2^{(1)}$ is $\Psi_2^{(0)}=\frac{1}{\sqrt{2}}(\psi_2+\psi_3)$ . Similarly, for $U_3^{(1)}$ , we have $\Psi_3^{(0)}=\frac{1}{\sqrt{2}}(\psi_2-\psi_3)$ .

Finally, by substituting $\Psi_1^{(0)}$ , $\Psi_2^{(0)}$ and $\Psi_3^{(0)}$ in eq273, we find that they diagonalise $\hat{H}^{(1)}$ . We call these linearly combined states that diagonalise $\hat{H}^{(1)}$ “good” states.

Degenerate perturbation theory (an example)

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