Perturbation theory
We write
|ni = |n(0) i + λ|n(1) i + λ2 |n(2) i + · · ·
2 (2)
∆n = λ∆(1)
n + λ ∆n + · · · .
Stationary perturbation methods
Let us suppose that
• we have solved completely the problem
H0 |n
(0)
i=
Because
hn(0) |λV − ∆n |ni = 0,
En(0) |n(0) i.
we have, on the other hand
The basis {|n(0) i} is now complete.
∆n = λhn(0) |V |ni.
• the states |n(0) i are non degenerate.
• we want to solve the problem
Thus we get
(H0 + λV )|niλ = En(λ) |niλ .
2 (2)
λ∆(1)
n + λ ∆n + · · ·
= λhn(0) |V |n(0) i + λ2 hn(0) |V |n(1) i + · · · .
Usualy the index λ is dropped off.
When we denote
Equalizing the coefficients of the powers of the parameter
λ we get
∆n ≡ En − En(0) ,
(En(0) − H0 )|ni = (λV − ∆n )|ni.
(0)
Note Because the expression (En −
(0)
undefined the operator (En − H0 )−1
−1
(0)
H0 ) |n i is
is not well defined.
So, in the equation above we cannot invert the operator
(0)
(En − H0 ).
Now
hn(0) |λV − ∆n |ni = hn(0) |En(0) − H0 |ni,
so in the state (λV − ∆n )|ni there is no component along
the state |n(0) i.
We define a projection operator as
X
φn = 1 − |n(0) ihn(0) | =
|k (0) ihk (0) |.
k6=n
Now
1
En(0)
− H0
φn =
1
X
(0)
k6=n En
−
(0)
Ek
|ni = |n(0) i +
= hn
..
.
(0)
|V |n
(N −1)
φn
En(0)
− H0
(λV − ∆n )|ni
×(|n(0) i + λ|n(1) i + · · ·).
Equalizing the coefficients of the linear λ-terms we get in
the first order
O(λ) : |n(1) i
the formal solution is of the form
=
φn
En(0)
− H0
φn
En(0) − H0
V |n(0) i −
∆(1)
n
En(0)
− H0
φn |n(0) i
V |n(0) i,
φn (λV − ∆n )|ni,
because
(0)
φn ∆(1)
i = 0.
n |n
where
lim cn (λ) = 1
We substitute |n(1) i into the expression
and
(0)
∆(2)
|V |n(1) i,
n = hn
cn (λ) = hn(0) |ni.
Diverting from the normal procedure we normalize
hn(0) |ni = cn (λ) = 1.
i
|n(0) i + λ|n(1) i + λ2 |n(2) i + · · ·
φn
2 (2)
= |n(0) i + (0)
(λV − λ∆(1)
n − λ ∆n − · · ·)
En − H0
=
λ→0
.
for the state vector the power series of the state vector
and the energy correction and we get
|ni → |n(0) i,
1
= hn(0) |V |n(0) i
= hn(0) |V |n(1) i
..
.
We substitute into the expression
Since in the limit λ → 0 we must have
En(0) − H0
(N )
∆n
O(λ ) :
..
.
(λV − ∆n )|ni = φn (λV − ∆n )|ni.
|ni = cn (λ)|n(0) i +
∆n
(2)
∆n
N
|k (0) ihk (0) |
and
(1)
O(λ1 ) :
O(λ2 ) :
..
.
the eigenvalue equation to be solved takes the form
so
(0)
∆(2)
|V
n = hn
φn
En(0)
− H0
V |n(0) i.
We substitute this further into the power series of the
state vectors and we get for the coeffients of λ2 the
condition
φn
O(λ2 ) : |n(2) i =
−
φn
En(0)
− H0
En(0)
− H0
φn
V
En(0)
hn(0) |V |n(0) i
− H0
φn
En(0) − H0
V |n(0) i
V |n(0) i.
Likewise we could continue to higher powers of the
parameter λ. This method is known as the
Rayleigh-Schrödinger perturbation theory.
The explicit expression for the second order energy
correction will be
∆(2)
n
= hn
=
(0)
X
|V
φn
V |n
En(0) − H0
hn(0) |V |k (0) ihk (0) |
k,l
=
X
Vnk
k,l6=n
=
X
hk (0) |l(0) i
(0)
En(0) − El
|Vnk |2
(0)
k6=n
En(0) − Ek
(0)
i
− H0
|l(0) ihl(0) |V |n(0) i
so that
hn(0) |niN = Zn1/2 hn(0) |ni = Zn1/2 .
Thus the normalization factor Zn is the probablility for
the perturbed state to be in the unperturbed state.
The normalization condition
N hn|niN
Zn−1
Vln
.
= hn|ni = (hn(0) | + λhn(1) | + λ2 hn(2) | + · · ·)
×(|n(0) i + λ|n(1) i + λ2 |n(2) i + · · ·)
= 1 + λ2 hn(1) |n(1) i + O(λ3 )
X
|Vkn |2
= 1 + λ2
+ O(λ3 ).
(0) 2
(0)
k6=n (En − Ek )
≡ En − En(0)
= λVnn + λ2
Zn ≈ 1 − λ2
|Vnk |2
X
(0)
k6=n
= Zn hn|ni = 1
Up to the order λ2 the probability for the perturbed state
to lie in the unperturbed state is thus
Thus, up to the second order we have
∆n
(0)
gives us
φn
En(0)
(0)
Perturbation expansions converge if |Vij /(Ei − Ej )| is
”small”. In general, no exact convergence criterion is
known.
The state |ni is not normalized. We define the normalized
state
|niN = Zn1/2 |ni,
En(0) − Ek
+ ···.
Correspondingly, up to the second order the state vector
is
X
Vkn
|ni = |n(0) i + λ
|k (0) i
(0)
(0)
En − Ek
k6=n

X
X
Vkl Vln
+λ2
|k (0) i 
(0)
(0)
(0)
(0)
k6=n
l6=n (En − Ek )(En − El )
!
Vnn Vkn
−
(0)
(En(0) − Ek )2
+ ···.
We see that the perturbation mixes in also other states
(than |n(0) i).
We see that
X
|Vkn |2
k6=n
(En(0) − Ek )2
(0)
.
The latter term can be interpreted as the probability for
the ”leakage” of the system from the state |n(0) i to other
states.
Example The quadratic Stark effect.
We consider hydrogen like atoms, i.e. atoms with one
electron outside a closed shell, under external uniform
electric field parallel to the positive z-axis. Now
H0 =
p2
+ V0 (r) and V = −e|E|z.
2m
We suppose that the eigenstates of H0 are non degenerate
(not valid for hydrogen). The energy shift due to the
external field is
∆k = −e|E|zkk + e2 |E|2
|zkj |2
X
(0)
j6=k Ek
(0)
− Ej
+ ···,
• in the 1st order we need only the matrix element Vnn . where
• in the 2nd order the energy levels i and j repel each
(0)
(0)
other. Namely, if Ei < Ej , then the contributions
of one of these states to the energy corrections of the
other are
(2)
∆i
(2)
∆j
=
|Vij |2
(0)
Ei
=
(0)
− Ej
|Vij |2
(0)
Ej
(0)
− Ei
<0
>0
and the energy levels move apart from each other.
zkj = hk (0) |z|j (0) i.
Since we assumed the states |k (0) i to be non degenerate
they are eigenstates of the parity. So, according to the
parity selection rule the matrix element of zkk vanishes.
Indeces k and j are collective indeces standing for the
quantum number triplet (n, l, m). According to the
Wigner-Eckart theorem we have
zkj
= hn0 , l0 m0 |z|n, lmi
= hl1; m0|l1; l0 m0 i
hn0 l0 kT (1) knli
√
,
2l + 1
where we have written the operator z as the spherical
tensor
(1)
T0 = z.
In order to satisfy zkj 6= 0 we must have
(0)
We want to solve the problem
• l0 = l − 1, l, l + 1. From these l0 = l is not suitable
due to the parity selection rule.
So we get
0
0
hn , l m |z|n, lmi = 0 unless
l0 = l ± 1
.
m0 = m
We define the polarizability α so that
1
∆ = − α|E|2 .
2
where the summing must be extended also over the
continuum states.
Let us suppose that
(0)
(H0 + λV )|li = El |li
with the boundary condition
X
lim |li →
hm(0) |l(0) i|m(0) i,
λ→0
m∈D
i.e. we are looking for corrections to the degenerated
states. With the help of the energy correction we have to
solve
(0)
(ED − H0 )|li = (λV − ∆l )|li.
We write again
As a special case we consider the ground state
|0(0) i = |1, 00i of hydrogen atom which is non degenerate
when we ignore the spin. The perturbation expansion
gives
∞
X
|hk (0) |z|1, 00i|2
,
α = −2e2
(0)
(0)
k6=0 E0 − Ek
(0)
E0 − Ek ≈ constant,
so that
X
(0)
Let’s suppose that the energy state ED is g-foldly
degenerated:
H0 |m(0) i = ED |m(0) i, ∀|m(0) i ∈ D, dimD = g.
• m0 = m and
0
Degeneracy
|li = |l(0) i + λ|l(1) i + λ2 |l(2) i + · · ·
(1)
∆l
= λ∆l
(2)
+ λ2 ∆l
+ ···,
so we get
(0)
(ED − H0 )(|l(0) i + λ|l(1) i + λ2 |l(2) i + · · ·)
(1)
(2)
− λ2 ∆l
= (λV − λ∆l
×(|l
(0)
i + λ|l
(1)
− · · ·)
2 (2)
i + λ |l
i + · · ·).
Equalizing the coefficients of the powers of λ we get in
the first order
(0)
(ED − H0 )|l(1) i
(1)
|hk
(0)
2
|z|1, 00i|
=
k6=0
X
|hk
(0)
2
|z|1, 00i|
all k’s
= (V − ∆l )|l(0) i
"
#
X
(1)
(0)
(0) (0)
= (V − ∆l )
|m ihm |l i .
m∈D
= h1, 00|z 2 |1, 00i.
In the spherically symmetric ground state we obviously
have
1
hz 2 i = hx2 i = hy 2 i = hr2 i.
3
Using the explicit wave functions we get
hz 2 i = a20 .
Taking the scalar product with the vector hm0
recalling that
hm0
(0)
(0)
| and
(0)
|(ED − H0 ) = 0,
we end up with the simultaneous eigenvalue equations
X
(0)
(1)
Vm0 m hm(0) |l(0) i = ∆l hm0 |l(0) i.
m
Now
(1)
(0)
(0)
(0)
(0)
−E0 + Ek ≥ −E0 + E1
2
e
1
=
1−
,
2a0
4
so
8a0
16a30
≈ 5.3a30 .
2 =
3
3e
The exact summation gives
α < 2e2 a20
α=
9a30
= 4.5a30 .
2
The energy corrections ∆l
From the equation
are obtained as eigenvalues.
(0)
(ED − H0 )|l(1) i = (λV − ∆l )|l(0) i
we also see that
(1)
hm(0) |V − ∆l |l(0) i = 0 ∀|m(0) i ∈ D.
(1)
Thus the vector (V − ∆l )|l(0) i has no components in the
subspace D. Defining a projection operator as
φD = 1 −
g
X
m∈D
|m(0) ihm(0) | =
X
k6∈D
|k (0) ihk (0) |
we can write
(1)
(1)
(V − ∆l )|l(0) i = φD (V − ∆l )|l(0) i = φD V |l(0) i.
We get the equation
We put the atom in external electric field parallel to the
z-axis:
V = −ez|E|.
Now z is the q = 0 component of a spherical tensor:
(0)
(1)
(ED − H0 )|l(1) i = φD (λV − ∆l )|l(0) i,
(0)
where now the operator (ED − H0 ) can be inverted:
φD
|l(1) i =
(0)
ED
X
=
− H0
V |l(0) i
|k (0) iVkl
(0)
k6∈D ED
−
(0)
Ek
z = T0 .
According to the parity selection rule the operator V now
has nonzero matrix elements only between states with
l = 0 and l = 1, and due to the m-selection rule all states
must have the same m:
2s 2p, 0

2s
0
×
2p, 0 
×
0

V =
2p, 1  0
0
2p, −1 0
0
.
When we again normalize
hl(0) |li = 1,
2p, 1
0
0
0
0
2p, −1

0
0 
.
0 
0
The nonzero matrix elements are
we get from the equation
h2s|V |2p, m = 0i = h2p, m = 0|V |2si = 3ea0 |E|.
(0)
(ED − H0 )|li = (λV − ∆l )|li
The eigenvalues of the perturbation matrix are
for the energy shift
(1)
∆± = ±3ea0 |E|
∆l = λhl(0) |V |li.
and the eigenvectors
We substitute the power series and get
1
|±i = √ (|2s, m = 0i ± |2p, m = 0i).
2
λhl(0) |V (|l(0) i + λ|l(1) i + λ2 |l(2) i + · · ·)
(1)
= λ∆l
(2)
+ λ2 ∆l
+ ···.
Note The energy shift is a linear function of the electric
The second order energy correction is now
(2)
∆l
= hl(0) |V |l(1) i = hl(0) |V
=
X
|Vkl |2
(0)
k6∈D
φD
(0)
ED
− H0
V |l(0) i
.
(0)
Diagonalize the perturbation matrix.
◦
The resulting eigenvalues are first order corrections
for the energy shifts. The corresponding eigenvectors
are those zeroth order eigenvectors to which the
corrected eigenvectors approach when λ → 0.
3
Let the states m ∈ D to be almost degenerate. We write
V = V1 + V 2 ,
where
Identify the degenerated eigenstates. We suppose
that their count is g. Construct the
g × g-perturbation matrix V .
◦
2
Nearly degenerated states
ED − Ek
Thus the perturbation calculation in a degenerate system
proceeds as follows:
1◦
field. The states |±i are not parity eigenstates so it is
perfectly possible that they have permanent electric
dipole moment hzi =
6 0.
4◦ Evaluate higher order corrections using non
degenerate perturbation methods but omit in the
summations all contributions coming from the
degenerated state vectors of the space D.
Example The Stark efect in the hydrogen atom.
The hydrogen 2s (n = 2, l = 0, m = 0) and 2p
(n = 2, l = 1, m = −1, 0, 1) states are degenerate. Their
energy is
(0)
ED = −e2 /8a0 .
V1
=
X X
|m(0) ihm(0) |V |m0(0) ihm0(0) |
m∈D m0 ∈D
V2
= V − V1 .
We proceed so that
1. we diagonalize the Hamiltonian H0 + V1 exactly in
the basis {|m(0) i} and
2. handle the term V2 like in an ordinary non
degenerate perturbation theory. This is possible since
hm0(0) |V2 |m(0) i = 0
∀m, m0 ∈ D.
Example Weak periodic potential.
Now
H0 =
p2
2m
and for the perturbation
V (x) = V (x + a).
We denote the unperturbed eigenstates by the wave
vector:
h̄2 k 2
H0 |ki =
|ki,
2m
so that
h̄2 k 2
(0)
.
Ek =
2m
We impose the periodic boundary conditions
hx0 |ki = ψk (x0 ) = hx0 + L|ki = ψk (x0 + L),
Correspondingly the energy up to the second order is
(2)
|Vn |2
(0)
n6=0 Ek
− Ek+nK
(0)
.
Let us suppose now that
k≈−
nK
.
2
We diagonalize the Hamiltonian in the basis
for the wave function and get
{|ki, |k + nKi},
0
1
2π
ψk (x ) = √ eikx , k =
n, n ∈ I.
L
L
0
i.e. we diagonalize the matrix
Because the potential V is periodic it can be represented
as the Fourier series
∞
X
V (x) =
X
(0)
Ek = Ek + V 0 +
inKx
e
Vn ,
n=−∞
|ki
(0)
Ek
|ki
|k + nKi
|k + nKi
+ V0
Vn
Vn∗
(0)
Ek + V 0
!
.
Its eigenvalues are
where
(0)
K = 2π/a
is the reciprocal lattice vector. The only nonzero matrix
elements are now
hk + nK|V |ki = Vn ,
(0)
because
(0)
E + Ek+nK
Ek± = V0 + k
2
v
!2
u
u E (0) − E (0)
t
k
k+nK
+ |Vn |2 .
±
2
(0)
When |Ek − Ek+nK | |Vn |, it reduces to two solutions
Z
0 0
0
0
1X
Vn dx0 e−ik x einKx einkx
L n
X
=
Vn δk+nK,k0 .
hk 0 |V |ki =
n
(0)
Ek+
= Ek + V 0
Ek−
= Ek+nK + V0 ,
(0)
which are first order corrected energies. In the limiting
case we get
So the potential couples states
|ki, |k + Ki, . . . , |k + nKi, . . . .
(0)
lim
k→−nK/2
Ek± = EnK/2 + V0 ± |Vn |.
The corresponding energy denominators are
(0)
(0)
Brillouin-Wigner perturbation theory
Ek − Ek+nK .
We start with the Schrödinger equation
In general
(0)
(En − H0 )|ni = λV |ni,
(0)
Ek 6= Ek+nK
except when
and take on both sides the scalar product with the state
|m(0) i, and get
nK
.
2
We suppose that the condition
k≈−
k 6= −
(0)
(En − Em
)hm(0) |ni = λhm(0) |V |ni.
nK
2
holds safely. The first order state vectors are then
|k (1) i = |ki +
X
n6=0
|k + nKi
Vn
(0)
Ek
(0)
− Ek+nK
,
We correct the state |n(0) i. We write the corrected state
|ni in the form
X
|ni =
|m(0) ihm(0) |ni = |n(0) ihn(0) |ni + φn |ni
m
= |n(0) i +
X
|m(0) ihm(0) |ni,
m6=n
and the wave functions
(1)
ψk (x0 )
X 1
0
0
1
V
√ ei(k+nK)x (0) n (0) .
= √ eikx +
L
L
Ek − Ek+nK
n6=0
which has been normalized, like before,
hn(0) |ni = 1.
We substitute into this the scalar products
hm(0) |ni =
λhm(0) |V |ni
(0)
En − E m
,
and end up with the fundamental equation of the
Brillouin-Wigner method
X
|ni = |n(0) i +
|m(0) i
m6=n
λ
(0)
En − E m
hm(0) |V |ni.
Iteration gives us the series
|ni = |n(0) i + λ
X
|m(0) i
m6=n
+ λ2
XX
|l(0) i
m6=n l6=n
1
(0)
E n − Em
1
(0)
En − El
hm(0) |V |n(0) i
hl(0) |V |m(0) i
1
hm(0) |V |n(0) i
(0)
En − Em
X XX
1
hk (0) |V |l(0) i
+ λ3
|k (0) i
(0)
E
−
E
n
m6=n l6=n k6=n
k
×
×
1
En −
+ ···.
(0)
El
hl(0) |V |m(0) i
1
(0)
E n − Em
hm(0) |V |n(0) i
Note This is not a power series of the parameter λ
because the energy denominators
(0)
(0)
En − Em
= En(0) − Em
+ ∆n
depend also on the parameter λ according to the equation
2 (2)
∆n = λ∆(1)
n + λ ∆n + · · ·