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.


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