Perturbation theory of degenerate states

Let H=H0+H'=H0+λW. Let {|φip>} denote an orthonormal eigenbasis of H0, H0|φip>=E0p|φip>.  Here i denotes the degeneracy.  Consider a particular degenerate eigenvalue E0p of H0.  Assume that this eigenvalue is g-fold degenerate, i = 1, 2, ...,g.  To find E1p we use

Multiplying from the left by <φip| 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 E0p.  We find g eigenvalues E1p,i which may or may not be degenerate.  We then find the corresponding g eigenvectors.  To solve for the eigenvalues we set det(W-E1p)=0 in the subspace E(0,p).  If we find a non-degenerate eigenvalue E1p,i, then the corresponding eigenvector is uniquely defined.  If the eigenvalue E1p,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 |χjp> is an eigenvector with eigenvalue E0p of H0, but with a non-degenerate eigenvalue E1p,j. |χjp> is uniquely defined.  First-order perturbation theory has removed the degeneracy.  Let |y0p>=|χjp>.

(H0-E0p)|yp1> =( E1p,j-W)|yp0> .

|yp0> is an eigenstate of the unperturbed Hamiltonian.  We may expand

in terms of the basis vectors.

<yp0|yp1> = 0.

Multiply from the left by <φip''|.

We can solve for all bip' but not for the bip, they remain undetermined.  However, this does not preclude us from finding E2p,j , the second-order energy correction.

.

All the matrix elements multiplying the bip 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

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