Perturbation of a degenerate state

Perturbation theory of degenerate states

Let H=H₀+H'=H₀+λW. Let {|φⁱ_p>} denote an orthonormal eigenbasis of H₀, H₀|φⁱ_p>=E₀^p|φⁱ_p>. Here i denotes the degeneracy. Consider a particular degenerate eigenvalue E₀^p of H₀. Assume that this eigenvalue is g-fold degenerate, i = 1, 2, ...,g. To find E₁^p we use

Multiplying from the left by <φⁱ_p| we obtain

This is an eigenvalue equation for the operator W in the subspace E(0,p) of vectors with the eigenvalue of H₀ equal to E₀^p. We find g eigenvalues E₁^p,i which may or may not be degenerate. We then find the corresponding g eigenvectors. To solve for the eigenvalues we set det(W-E₁^p)=0 in the subspace E(0,p). If we find a non-degenerate eigenvalue E₁^p,i, then the corresponding eigenvector is uniquely defined. If the eigenvalue E₁^p,i is degenerate then the corresponding eigenvector is still not uniquely defined. The degeneracy may or may not be removed in higher order.

Assume that |χ^j_p> is an eigenvector with eigenvalue E₀^p of H₀, but with a non-degenerate eigenvalue E₁^p,j. |χ^j_p> is uniquely defined. First-order perturbation theory has removed the degeneracy. Let |y⁰_p>=|χ^j_p>.

(H₀-E₀^p)|y_p¹> =( E₁^p,j-W)|y_p⁰> .

|y_p⁰> is an eigenstate of the unperturbed Hamiltonian. We may expand

in terms of the basis vectors.

<y_p⁰|y_p¹> = 0.

Multiply from the left by <φⁱ_p''|.

We can solve for all bⁱ_p' but not for the bⁱ_p, they remain undetermined. However, this does not preclude us from finding E₂^p,j , the second-order energy correction.

All the matrix elements multiplying the bⁱ_p are zero, since we have diagonalized the matrix of W. Therefore, if first-order perturbation theory removes the degeneracy of a set of degenerate eigenvalues, we can proceed to find higher-order corrections to the now uniquely defined eigenvalues just like in the non degenerate case.

Problem:

Consider a charged particle on a ring of unit radius with flux φ/f₀=α passing through the ring, where f₀=hc/e is the flux quantum. The Hamiltonian operator can be written as H=H₀+V, where

θ is the angular coordinate. We have chosen units with

(a) Find the complete set of eigenvalues and eigenfunctions of H₀.
(b) Use perturbation theory to find the first and second-order corrections to the ground state energy E⁰ of H₀ due to the perturbation V for 0<α<½ .
(c) For α=½ the ground state energy of H₀ is degenerate. Find the first-order correction to E⁰ for this case.

Solution:
(a)
Try

(b) Let 0 < α < ½. Then all states are non degenerate and the ground state has n=0.

(c) If α=½ then all states are degenerate. The ground state is any linear combination of n=0 and n=1.
We have to diagonalize the matrix of V in the subspace spanned by

	Solution: (a) Try (b) Let 0 < α < ½. Then all states are non degenerate and the ground state has n=0. (c) If α=½ then all states are degenerate. The ground state is any linear combination of n=0 and n=1. We have to diagonalize the matrix of V in the subspace spanned by

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