PHY 6646 Spring 2003
K. Ingersent
Degenerate Rayleigh-Schrödinger Perturbation Theory
• Assume that we know the stationary states of the unperturbed Hamiltonian H0 , namely
the kets |n, ri satisfying H0 |n, ri = εn |n, ri. The integer index r (1 ≤ r ≤ gn ) is used to
distinguish among the gn eigenstates of energy εn . (For simplicity, we assume that the
vector space is has a finite or countably infinite dimension. The extension to continuous
vector spaces is straightforward.)
• We seek stationary solutions |ψn,r i of the perturbed problem
(H0 + λH1 )|ψn,r i = En,r |ψn,r i
(1)
in the form of power-series expansions
|ψn,r i =
∞
X
j=0
(j)
λj |ψn,r
i,
En,r =
∞
X
j=0
(j)
λj En,r
.
(2)
Let us insert Eqs. (2) into Eq. (1), and collect terms having the same power of λ.
(0)
(0)
• At order λ0 we have (H0 − En,r
)|ψn,r
i = 0, which is satisfied by any linear combination
of the unperturbed eigenkets of energy εn , i.e.,
(0)
|ψn,r
i
=
gn
X
(cn,r )t |n, ti,
(0)
En,r
= εn .
with
t=1
(0
(0)
Orthonormality requires that hψn,r
|ψn,s
i=
Pgn
∗
t=1 (cn,r )t (cn,s )t
= δr,s .
(0)
(1)
(1)
(0)
)|ψn,r
i = (En,r
− H1 )|ψn,r
i. Acting from the left with
• At order λ1 we find (H0 − En,r
hm, s|, we obtain
(1)
(1)
i = δm,n En,r
(cn,r )s −
(εm − εn )hm, s|ψn,r
gn
X
t=1
hm, s|H1 |n, ti(cn,r )t .
(3)
• For m = n, Eq. (3) yields
gn
X
t=1
(1)
hn, s|H1 |n, ti(cn,r )t = En,r
(cn,r )s ,
which is the matrix eigenequation for H1 in the gn -dimensional subspace spanned by
the unperturbed states of energy εn . It is perfectly consistent with the λ0 result to
(0)
i’s to be the eigenkets of this problem. We will assume henceforth that
choose the |ψn,r
this is the case, so that
(0)
(0)
(1)
hψn,s
|H0 + λH1 |ψn,r
i = (εn + λEn,r
)δs,r ,
(4)
where the first-order correction to the unperturbed energy is
(1)
(0)
(0)
En,r
= hψn,r
|H1 |ψn,r
i.
1
(5)
(0)
Note that we cannot assume that |ψn,r
i is an eigenket of H1 in the full vector space.
Equation (4) implies only that
(0)
H1 |ψn,r
i
=
(1) (0)
En,r
|ψn,r i
gm
X X
+
m6=n t=1
(0)
(0)
(0)
|ψm,t ihψm,t |H1 |ψn,r
i.
(6)
• For m 6= n, Eq. (3) yields
(1)
hm, s|ψn,r
i=
gn
X
hm, s|H1 |n, ti
t=1
(cn,r )t =
εn − εm
or
(0)
(1)
|ψn,r
i=
hψm,s
(0)
(0)
hψm,s
|H1 |ψn,r
i
,
εn − εm
(0)
hm, s|H1 |ψn,r
i
.
εn − εm
m 6= n.
(7)
• If gn = 1, we can drop the second label for each eigenket. Then Eqs. (5) and (7) reduce
to the standard results of nondegenerate perturbation theory.
• Conversely, it appears from Eqs. (5) and (7) that the perturbative solution of the degenerate problem to order λ1 can be obtained from that of a nondegenerate problem by
P
P P m
(0)
i and m → m gt=1
. However, this conclusion is premature
substituting |ni → |ψn,r
(1)
(0)
(1)
because Eq. (3) does not determine hn, s|ψn,r i, or alternatively, hψn,s
|ψn,r
i. We will
now correct this omission.
(0)
• Following a convention from the nondegenerate theory, we enforce hψn,r
|ψn,r i = 1.
(0)
(j)
Thus, hψn,r |ψn,r i = 0 for all j > 0, which includes as a special case
(0)
(1)
|ψn,r
i = 0.
hψn,r
(0)
(1)
• To determine hψn,s
|ψn,r
i for s 6= r, it is necessary to proceed to order λ2 in the expansion
of Eq. (1):
(0)
(2)
(1)
(1)
(2) (0)
)|ψn,r
i = (En,r
− H1 )|ψn,r
i + En,r
|ψn,r i.
(H0 − En,r
(0)
(1)
Acting from the left with hψn,s
|, we eliminate all but the term involving |ψn,r
i. Then,
using the adjoint of Eq. (6) with s replacing r, we obtain
0=
(1)
(En,r
−
(1)
(0)
(1)
En,s
)hψn,s
|ψn,r
i
−
gm
X X
m6=n t=1
(0)
(0)
(0)
(1)
hψn,s
|H1 |ψm,t ihψm,t |ψn,r
i,
where the last inner product on the right-hand side can be evaluated using Eq. (7).
(1)
(1)
Provided that En,r
6= En,s
, we can conclude that
(0)
(1)
hψn,s
|ψn,r
i
(0)
=
(0)
gm
(0)
(0)
X X
hψn,s
|H1 |ψm,t ihψm,t |H1 |ψn,r
i
m6=n t=1
(1)
(1)
(En,r − En,s )(εn − εm )
,
s 6= r.
• Summary: In cases of degeneracy, it is necessary to work at least to second order in
λ to obtain |ψn,r i correct to first order. [If H1 does not lift the degeneracy between
(0)
(0)
(1)
(1)
i and |ψn,s
i (i.e., En,r
= En,s
), then one must work to third order or higher.]
|ψn,r
2