Degenerate perturbation theory

The degenerate perturbation theory, an extension of the perturbation theory, is used to find an approximate solution to a quantum-mechanical problem involving non-perturbed states that are degenerate.

In the non-degenerate case, it is unambiguous that $\Psi_n^{\;(0)}=\psi_n^{\;(0)}$ , i.e. $\Psi_n^{\;(0)}$ is described by individual basis states $\psi_n^{\;(0)}$ . However, for a set of degenerate orthonormal states $\left \{\psi_{j,D1}^{\;(0)}\right \}$ ( $\small D1$ denotes the same eigenvalue for all states in the set) of the unperturbed Hamiltonian $\hat{H}^{(0)}$ , where $\hat{H}^{(0)}\psi_{j}^{\;(0)}=E_{D1}^{\;(0)}\psi_{j}^{\;(0)}$ , $\Psi_{n}^{\;(0)}$ can represent any linear combination of $\left \{\psi_{j,D1}^{\;(0)}\right \}$ or the non-degenerate states $\psi_{n}^{\;(0)}$ , where $\psi_{n}^{\;(0)}\notin \left \{\psi_{j,D1}^{\;(0)}\right \}$ . If there are more than one set of degenerate states, each set with a distinct energy level, $\Psi_{n}^{\;(0)}$ represents linear combinations of $\left \{\psi_{j,D1}^{\;(0)}\right \}$ , linear combinations of $\left \{\psi_{j,D2}^{\;(0)}\right \}$ , and so on, and individual states of $\psi_{n}^{\;(0)}$ , where $\psi_{n}^{\;(0)}\notin \left \{\psi_{j,DX}^{\;(0)}\right \}$ .

For each set of degenerate orthonormal states, e.g. $\left \{\psi_{j,D1}^{\;(0)}\right \}$ , of the unperturbed Hamiltonian $\hat{H}^{(0)}$ , the linear combination $\small \Psi_{n}^{\;(0)}=\sum_{j=1}^{g}\alpha_j\psi_{j,D1}^{\;(0)}$ (where $\small g$ is the degeneracy) is an eigenstate of $\hat{H}^{(0)}$ with the same eigenvalue $\small U_{n,D1}^{\;(0)}=E_{n,D1}^{\;(0)}$ :

$\small \hat{H}^{(0)}\Psi_{n}^{\;(0)}=\sum_{j=1}^{g}\alpha_j\hat{H}^{(0)}\psi_{j,D1}^{\;(0)} =E_{n,D1}^{\;(0)}\sum_{j=1}^{g}\alpha_j\psi_{j,D1}^{\;(0)}=E_{n,D1}^{\;(0)}\Psi_{n}^{\;(0)} \; \; \; \; \; \; \; \; 271$

Therefore, the derivation in the previous article branches off from eq267. While $\small \Psi_{n}^{\;(0)}$ is no longer unambiguously described by individual basis states $\small \psi_{n}^{\;(0)}$ , $\small U_{n}^{\;(0)}$ is still $\small E_{n}^{\;(0)}$ . So, eq267 becomes

$\small \hat{H}^{(0)}\Psi_{n}^{\;(1)}+\hat{H}^{(1)}\Psi_{n}^{\;(0)}-E_{n}^{\;(0)}\Psi_{n}^{\;(1)}-U_{n}^{\;(1)}\Psi_{n}^{\;(0)}=0\; \; \; \; \; \; \; \; 272$

Multiplying the above equation from the left by $\small \Psi_{j}^{\;(0)*}$ and integrating,

$( E_{j}^{\;(0)}-E_{n}^{\;(0)} )\langle\Psi_{j}^{\;(0)}\vert\Psi_{n}^{\;(1)}\rangle +\langle\Psi_{j}^{\;(0)}\vert\hat{H}^{(1)}\vert\Psi_{n}^{\;(0)}\rangle =0$

When $E_{j}^{\;(0)}=E_{n}^{\;(0)}$ , we have.

$\langle\Psi_{j}^{\;(0)}\vert\hat{H}^{(1)}\vert\Psi_{n}^{\;(0)}\rangle =0\; \; \; \; \; \; \; \; 273$

Since $\langle\Psi_{j}^{\;(0)}\vert\hat{H}^{(1)}\vert\Psi_{n}^{\;(0)}\rangle$ represents the matrix elements of $\hat{H}^{(1)}$ , eq273 implies that in the presence of degenerate states, the first-order perturbation theory only works if we can select appropriate wavefunctions that diagonalise $\hat{H}^{(1)}$ .

Consider the case of just one set of degenerate orthonormal states $\left \{ \psi_{l,D1}^{(0)} \right \}$ of the unperturbed Hamiltonian $\hat{H}^{(0)}$ , with the other states of $\hat{H}^{(0)}$ being non-degenerate. If we further analyse the subspace of $\left \{ \psi_{l,D1}^{(0)} \right \}$ , we substitute $\Psi_{n}^{\;(0)}=\sum_{l=1}^{g}\alpha_l\psi_{l,D1}^{\;(0)}$ in eq272 to give:

$\hat{H}^{(0)}\Psi_{n}^{\;(1)}+\hat{H}^{(1)}\sum_{l=1}^{g}\alpha_l\psi_{l,D1}^{\;(0)}-E_{n,D1}^{(0)}\Psi_{n}^{\;(1)}-U_n^{(1)}\sum_{l=1}^{g}\alpha_l\psi_{l,D1}^{\;(0)}=0$

Taking the inner product with $\psi_{j,D1}^{\;(0)}$ and using eq36 on the first term,

$\sum_{l=1}^gW_{jl}\alpha_l=U_n^{(1)}\alpha_l\; \; \; \; \; \; \; \; 274$

where $W_{jl}=\langle\psi_{j,D1}^{(0)}\vert\hat{H}^{(1)}\vert\psi_{l,D1}^{(0)}\rangle$ are the matrix elements of $\hat{H}^{(1)}$ .

Rewriting the eigenvalue equation of eq274 in matrix form,

$\left [ \begin{pmatrix} W_{11} &W_{12} &\cdots &W_{1g} \\ W_{21} &W_{22} &\cdots &W_{2g} \\ \vdots &\vdots & \ddots &\vdots \\ W_{g1} & W_{g2}&\cdots &W_{gg} \end{pmatrix}-U_n^{(1)}I \right ]\begin{pmatrix} \alpha_1\\\alpha_2 \\ \vdots \\ \alpha_g \end{pmatrix}=0\; \; \; \; \; \; \; \; 275$

where $I$ is the $g\times g$ identity matrix.

Eq275 is a linear homogeneous equation, which has non-trivial solutions if

$\left | \begin{pmatrix} W_{11} &W_{12} &\cdots &W_{1g} \\ W_{21} &W_{22} &\cdots &W_{2g} \\ \vdots &\vdots & \ddots &\vdots \\ W_{g1} & W_{g2}&\cdots &W_{gg} \end{pmatrix}-U_n^{(1)}I \right |=0\; \; \; \; \; \; \; \; 276$

Expanding the determinant (also known as a secular equation), we get the characteristic equation, which can be solved for the roots of $U_n^{(1)}$ . Each of these roots is then substituted into eq274 to find the set of coefficients $\alpha_l$ and hence, the exact eigenstates $\Psi_n^{(0)}$ . If these exact eigenstates diagonalises $\hat{H}^{(1)}$ , we refer to them as “good” states, which allow $U_n^{(1)}$ to be calculated (see next article for an example).

Question

What is a linear homogeneous equation? Why does a linear homogeneous equation have non-trivial solutions if the determinant of the operator is zero?

Answer

A linear homogeneous equation is a matrix equation of the form $A\boldsymbol{\mathit{X}}=0$ , where $A$ is an $n\times n$ matrix and $\boldsymbol{\mathit{X}}$ is column vector. Let $A$ be a $3\times3$ matrix representing three position vectors and $\boldsymbol{\mathit{X}}=(x,y,z)$ . We have

$\begin{pmatrix} a_1 &a_2 &a_3 \\ b_1 & b_2 &b_3 \\ c_1 &c_2 &c_3 \end{pmatrix}\begin{pmatrix} x\\y \\ z \end{pmatrix}=\begin{pmatrix} a_1x+a_2y+a_3z\\b_1x+b_2y+b_3z \\ c_1x+c_2y+c_3z \end{pmatrix}=0$

So, $\boldsymbol{\mathit{a}}\cdot\boldsymbol{\mathit{X}}=0$ , $\boldsymbol{\mathit{b}}\cdot\boldsymbol{\mathit{X}}=0$ and $\boldsymbol{\mathit{c}}\cdot\boldsymbol{\mathit{X}}=0$ . Now, a non-trivial solution of $A\boldsymbol{\mathit{X}}=0$ occurs when $\boldsymbol{\mathit{X}}\neq0$ . Therefore, we require any non-zero position vector $\boldsymbol{\mathit{X}}$ that is orthogonal to $\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}}$ . This is fulfilled if the position vectors $\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}}$ lie in a plane, which happens when the scalar triplet product $\boldsymbol{\mathit{a}}\cdot(\boldsymbol{\mathit{b}}\times\boldsymbol{\mathit{c}})$ , which is equal to the volume of the parallelepiped spanned by $\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}}$ , is zero. Since the scalar triple product is also equal to the determinant of $A$ , i.e. $\vert A\vert=\boldsymbol{\mathit{a}}\cdot(\boldsymbol{\mathit{b}}\times\boldsymbol{\mathit{c}})$ (which can be easily proven by expanding both sides and showing that LHS equals to RHS), $A\boldsymbol{\mathit{X}}=0$ has non-trivial solutions if $\vert A\vert=0$ .

Since there are multiple roots of $U_n^{(1)}$ , the perturbation $\hat{H}^{(1)}$ usually lifts the degeneracy of the unperturbed states. An example is the relativistic Hamiltonian’s spin-orbit coupling perturbation $\hat{H}^{(1)}=\frac{1}{2m_e^{\;2}c^2r}\frac{dV}{dr}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}$ , which splits terms $^{2S+1}\textrm{\textit{L}}$ into levels $^{2S+1}\textrm{\textit{L}}_J$ . If there are equal roots, we would need to further analyse the problem with higher order perturbations to completely lift the degeneracy (as is the case when considering the Zeeman effect).