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 , i.e. is described by individual basis states . However, for a set of degenerate orthonormal states ( denotes the same eigenvalue for all states in the set) of the unperturbed Hamiltonian , where , can represent any linear combination of or the non-degenerate states , where . If there are more than one set of degenerate states, each set with a distinct energy level, represents linear combinations of , linear combinations of , and so on, and individual states of , where .

For each set of degenerate orthonormal states, e.g. , of the unperturbed Hamiltonian , the linear combination (where is the degeneracy) is an eigenstate of with the same eigenvalue :

Therefore, the derivation in the previous article branches off from eq267. While is no longer unambiguously described by individual basis states , is still . So, eq267 becomes

Multiplying the above equation from the left by and integrating,

When , we have.

Since represents the matrix elements of , eq273 implies that in the presence of degenerate states, the first-order perturbation theory only works if we can select appropriate wavefunctions that diagonalise .

Consider the case of just one set of degenerate orthonormal states of the unperturbed Hamiltonian , with the other states of being non-degenerate. If we further analyse the subspace of , we substitute in eq272 to give:

Taking the inner product with and using eq36 on the first term,

where are the matrix elements of .

Rewriting the eigenvalue equation of eq274 in matrix form,

where is the identity matrix.

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

Expanding the determinant (also known as a secular equation), we get the * characteristic equation*, which can be solved for the roots of . Each of these roots is then substituted into eq274 to find the set of coefficients and hence, the exact eigenstates . If these exact eigenstates diagonalises , we refer to them as “good” states, which allow 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 , where is an matrix and is column vector. Let be a matrix representing three position vectors ** **and . We have

So, , and . Now, a non-trivial solution of occurs when . Therefore, we require any non-zero position vector that is orthogonal to . This is fulfilled if the position vectors lie in a plane, which happens when the scalar triplet product **,** which is equal to the volume of the parallelepiped spanned by , is zero. Since the scalar triple product is also equal to the determinant of , i.e. (which can be easily proven by expanding both sides and showing that LHS equals to RHS), has non-trivial solutions if .

Since there are multiple roots of , the perturbation usually lifts the degeneracy of the unperturbed states. An example is the relativistic Hamiltonian’s spin-orbit coupling perturbation , which splits terms into levels . 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).

###### Question

Find the characteristic equation for the degenerate set .

###### Answer

Eq276 becomes,