Time Independent Perturbation Theory | Quantum Mechanics for GATE - GATE Physics PDF Download

 Page 1


Time-Independent Perturbation Theory
1 The central problem in time-independent perturbation theory:
LetH
0
be the unperturbed (a.k.a. `background' or `bare') Hamiltonian, whose eigenvalues and eigenvectors
are known. Let E
(0)
n
be the n
th
unperturbed energy eigenvalue, andjn
(0)
i be the n
th
unperturbed energy
eigenstate. They satisfy
H
0
jn
(0)
i =E
(0)
n
jn
(0)
i (1)
and
hn
(0)
jn
(0)
i = 1: (2)
Let V be a Hermitian operator which `perturbs' the system, such that the full Hamiltonian is
H =H
0
+V: (3)
The standard approach is to instead solve
H =H
0
+V; (4)
and use  as for book-keeping during the calculation, but set  = 1 at the end of the calculation.
The goal is to nd the eigenvalues and eigenvectors of the full Hamiltonian (4). Let E
n
andjni be then
th
eigenvalue of the full Hamiltonian, and its corresponding eigenstate, respectively. They satisfy
Hjni =E
n
jni (5)
and
hnjni = 1: (6)
Because  is a small parameter, it is assumed that accurate results can be obtained by expanding E
n
and
jni in powers of , and keeping only the leading term(s). Formally expanding the perturbed quantities
gives
E
n
=E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::; (7)
and
jni =jn
(0)
i +jn
(1)
i +
2
jn
(2)
i +:::; (8)
where E
(j)
n
andjn
(j)
i are yet-to-be determined expansion coecients Inserting these expansions into the
eigenvalue equation (5) then gives
(H
0
+V )(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::) = (E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::)(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::): (9)
Page 2


Time-Independent Perturbation Theory
1 The central problem in time-independent perturbation theory:
LetH
0
be the unperturbed (a.k.a. `background' or `bare') Hamiltonian, whose eigenvalues and eigenvectors
are known. Let E
(0)
n
be the n
th
unperturbed energy eigenvalue, andjn
(0)
i be the n
th
unperturbed energy
eigenstate. They satisfy
H
0
jn
(0)
i =E
(0)
n
jn
(0)
i (1)
and
hn
(0)
jn
(0)
i = 1: (2)
Let V be a Hermitian operator which `perturbs' the system, such that the full Hamiltonian is
H =H
0
+V: (3)
The standard approach is to instead solve
H =H
0
+V; (4)
and use  as for book-keeping during the calculation, but set  = 1 at the end of the calculation.
The goal is to nd the eigenvalues and eigenvectors of the full Hamiltonian (4). Let E
n
andjni be then
th
eigenvalue of the full Hamiltonian, and its corresponding eigenstate, respectively. They satisfy
Hjni =E
n
jni (5)
and
hnjni = 1: (6)
Because  is a small parameter, it is assumed that accurate results can be obtained by expanding E
n
and
jni in powers of , and keeping only the leading term(s). Formally expanding the perturbed quantities
gives
E
n
=E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::; (7)
and
jni =jn
(0)
i +jn
(1)
i +
2
jn
(2)
i +:::; (8)
where E
(j)
n
andjn
(j)
i are yet-to-be determined expansion coecients Inserting these expansions into the
eigenvalue equation (5) then gives
(H
0
+V )(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::) = (E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::)(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::): (9)
Because of the linear independence of terms in a power series, this equation can only be satised for
arbitrary  if all terms with the same power of  cancel independently. Equating powers of  thus gives

0
: H
0
jn
(0)
i =E
(0)
n
jn
(0)
i; (10)

1
: (H
0
E
(0)
n
)jn
(1)
i = Vjn
(0)
i +E
(1)
n
jn
(0)
i (11)

2
: (H
0
E
(0)
n
)jn
(2)
i =Vjn
(1)
i +E
(2)
n
jn
(0)
i +E
(1)
n
jn
(1)
i (12)
etc::: (13)
This generalizes to

j
: (H
0
E
(0)
n
)jn
(j)
i =Vjn
(j1)
i +
j
X
k=1
E
(k)
n
jn
(jk)
i: (14)
Similiarly, we can expand the normalization equation in powers of , giving
1 =hn
(0)
jn
(0)
i +(hn
(1)
jn
(0)
i +hn
(0)
jn
(1)
i) +
2
(hn
(2)
jn
(0)
i +hn
(1)
jn
(1)
i +hn
(0)
jn
(2)
i) +:::: (15)
Again, we require that all terms of the same power in  cancel independently, resulting in the set of
equations:

0
: hn
(0)
jn
(0)
i = 1; (16)

1
: hn
(1)
jn
(0)
i +hn
(0)
jn
(1)
i = 0; (17)

2
: hn
(2)
jn
(0)
i +hn
(1)
jn
(1)
i +hn
(0)
jn
(2)
i; (18)
which readily generalizes to

j
:
j
X
k=0
hn
(jk)
jn
(k)
i = 0: (19)
Solving these equations can be simplied by choosing a convenient global phase-factor. In this case, we
can choose the phase of the full eigenstates to be such thathn
(0)
jni is real-valued. To enforce our choice
of global phase, we can expandhn
(0)
jni in power series, giving
hn
(0)
jni = 1 +hn
(0)
jn
(1)
i +
2
hn
(0)
jn
(2)
i +:::: (20)
Requiring that the r.h.s. be real-values for any real-valued  requires that each term be independently
real-valued, so that
hn
(0)
jn
(j)
i =hn
(j)
jn
(0)
i: (21)
Page 3


Time-Independent Perturbation Theory
1 The central problem in time-independent perturbation theory:
LetH
0
be the unperturbed (a.k.a. `background' or `bare') Hamiltonian, whose eigenvalues and eigenvectors
are known. Let E
(0)
n
be the n
th
unperturbed energy eigenvalue, andjn
(0)
i be the n
th
unperturbed energy
eigenstate. They satisfy
H
0
jn
(0)
i =E
(0)
n
jn
(0)
i (1)
and
hn
(0)
jn
(0)
i = 1: (2)
Let V be a Hermitian operator which `perturbs' the system, such that the full Hamiltonian is
H =H
0
+V: (3)
The standard approach is to instead solve
H =H
0
+V; (4)
and use  as for book-keeping during the calculation, but set  = 1 at the end of the calculation.
The goal is to nd the eigenvalues and eigenvectors of the full Hamiltonian (4). Let E
n
andjni be then
th
eigenvalue of the full Hamiltonian, and its corresponding eigenstate, respectively. They satisfy
Hjni =E
n
jni (5)
and
hnjni = 1: (6)
Because  is a small parameter, it is assumed that accurate results can be obtained by expanding E
n
and
jni in powers of , and keeping only the leading term(s). Formally expanding the perturbed quantities
gives
E
n
=E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::; (7)
and
jni =jn
(0)
i +jn
(1)
i +
2
jn
(2)
i +:::; (8)
where E
(j)
n
andjn
(j)
i are yet-to-be determined expansion coecients Inserting these expansions into the
eigenvalue equation (5) then gives
(H
0
+V )(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::) = (E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::)(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::): (9)
Because of the linear independence of terms in a power series, this equation can only be satised for
arbitrary  if all terms with the same power of  cancel independently. Equating powers of  thus gives

0
: H
0
jn
(0)
i =E
(0)
n
jn
(0)
i; (10)

1
: (H
0
E
(0)
n
)jn
(1)
i = Vjn
(0)
i +E
(1)
n
jn
(0)
i (11)

2
: (H
0
E
(0)
n
)jn
(2)
i =Vjn
(1)
i +E
(2)
n
jn
(0)
i +E
(1)
n
jn
(1)
i (12)
etc::: (13)
This generalizes to

j
: (H
0
E
(0)
n
)jn
(j)
i =Vjn
(j1)
i +
j
X
k=1
E
(k)
n
jn
(jk)
i: (14)
Similiarly, we can expand the normalization equation in powers of , giving
1 =hn
(0)
jn
(0)
i +(hn
(1)
jn
(0)
i +hn
(0)
jn
(1)
i) +
2
(hn
(2)
jn
(0)
i +hn
(1)
jn
(1)
i +hn
(0)
jn
(2)
i) +:::: (15)
Again, we require that all terms of the same power in  cancel independently, resulting in the set of
equations:

0
: hn
(0)
jn
(0)
i = 1; (16)

1
: hn
(1)
jn
(0)
i +hn
(0)
jn
(1)
i = 0; (17)

2
: hn
(2)
jn
(0)
i +hn
(1)
jn
(1)
i +hn
(0)
jn
(2)
i; (18)
which readily generalizes to

j
:
j
X
k=0
hn
(jk)
jn
(k)
i = 0: (19)
Solving these equations can be simplied by choosing a convenient global phase-factor. In this case, we
can choose the phase of the full eigenstates to be such thathn
(0)
jni is real-valued. To enforce our choice
of global phase, we can expandhn
(0)
jni in power series, giving
hn
(0)
jni = 1 +hn
(0)
jn
(1)
i +
2
hn
(0)
jn
(2)
i +:::: (20)
Requiring that the r.h.s. be real-values for any real-valued  requires that each term be independently
real-valued, so that
hn
(0)
jn
(j)
i =hn
(j)
jn
(0)
i: (21)
2 The non-degenerate case
We will now describe how to solve these equations in the case where none of the unperturbed energy levels
are degenerate.
Step #1: To obtain the j
th
correction to the n
th
energy eigenvalue, simply hit the 
j
equation from the
left with the brahn
(0)
j and solve for E
(j)
n
, giving
E
(j)
n
=hn
(0)
jVjn
(j1)
i
j1
X
k=1
E
(k)
n
hn
(0)
jn
(jk)
i; (22)
where we have usedhn
(0)
j(H
0
E
(0)
n
) = 0 andhn
(0)
jn
(0)
i = 1. Note that all quantities on the r.h.s are of
order <j, and are thus presumed to be known.
Step #2: To obtain thej
th
correction to then
th
eigenstate, we hit both sides of the 
j
equation with the
brahm
(0)
j where m6=n. This gives
hm
(0)
jn
(j)
i =
hm
(0)
jVjn
(j1)
i
E
(0)
m
E
(0)
n
+
j1
X
k=1
E
(k)
n
hm
(0)
jn
(jk)
i
E
(0)
m
E
(0)
n
; (23)
which is the expansion coecient of the j
th
correction onto the m
th
bare eigenvector.
Step #3: To obtain the remaining unknown quantity, hn
(0)
jn
(j)
i, we simply solve the normalization
equation, giving
hn
(0)
jn
(j)
i =
1
2
j1
X
k=1
hn
(jk)
jn
(k)
i: (24)
Steps 1, 2, and 3 must be solved iteratively, starting with j = 1, and repeating for each order until the
desired accuracy is achieved.
Final Step: Once these quantities are computed, we can reconstruct the n
th
eigenvalue and eigenvector
via
E
n
=
1
X
j=0

j
E
(j)
n
= E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::; (25)
and
jni = jn
(0)
i
1
X
j=0
hn
(0)
jn
(j)
i +
X
m6=n
jm
(0)
i
1
X
j=0
hm
(0)
jn
(j)
i
= jn
(0)
i
h
1 +hn
(0)
jn
(1)
i +
2
hn
(0)
jn
(2)
i +:::
i
+
X
m6=n
jm
(0)
i
h
hm
(0)
jn
(1)
i +
2
hm
(0)
jn
(2)
i +:::
i
:
(26)
Page 4


Time-Independent Perturbation Theory
1 The central problem in time-independent perturbation theory:
LetH
0
be the unperturbed (a.k.a. `background' or `bare') Hamiltonian, whose eigenvalues and eigenvectors
are known. Let E
(0)
n
be the n
th
unperturbed energy eigenvalue, andjn
(0)
i be the n
th
unperturbed energy
eigenstate. They satisfy
H
0
jn
(0)
i =E
(0)
n
jn
(0)
i (1)
and
hn
(0)
jn
(0)
i = 1: (2)
Let V be a Hermitian operator which `perturbs' the system, such that the full Hamiltonian is
H =H
0
+V: (3)
The standard approach is to instead solve
H =H
0
+V; (4)
and use  as for book-keeping during the calculation, but set  = 1 at the end of the calculation.
The goal is to nd the eigenvalues and eigenvectors of the full Hamiltonian (4). Let E
n
andjni be then
th
eigenvalue of the full Hamiltonian, and its corresponding eigenstate, respectively. They satisfy
Hjni =E
n
jni (5)
and
hnjni = 1: (6)
Because  is a small parameter, it is assumed that accurate results can be obtained by expanding E
n
and
jni in powers of , and keeping only the leading term(s). Formally expanding the perturbed quantities
gives
E
n
=E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::; (7)
and
jni =jn
(0)
i +jn
(1)
i +
2
jn
(2)
i +:::; (8)
where E
(j)
n
andjn
(j)
i are yet-to-be determined expansion coecients Inserting these expansions into the
eigenvalue equation (5) then gives
(H
0
+V )(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::) = (E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::)(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::): (9)
Because of the linear independence of terms in a power series, this equation can only be satised for
arbitrary  if all terms with the same power of  cancel independently. Equating powers of  thus gives

0
: H
0
jn
(0)
i =E
(0)
n
jn
(0)
i; (10)

1
: (H
0
E
(0)
n
)jn
(1)
i = Vjn
(0)
i +E
(1)
n
jn
(0)
i (11)

2
: (H
0
E
(0)
n
)jn
(2)
i =Vjn
(1)
i +E
(2)
n
jn
(0)
i +E
(1)
n
jn
(1)
i (12)
etc::: (13)
This generalizes to

j
: (H
0
E
(0)
n
)jn
(j)
i =Vjn
(j1)
i +
j
X
k=1
E
(k)
n
jn
(jk)
i: (14)
Similiarly, we can expand the normalization equation in powers of , giving
1 =hn
(0)
jn
(0)
i +(hn
(1)
jn
(0)
i +hn
(0)
jn
(1)
i) +
2
(hn
(2)
jn
(0)
i +hn
(1)
jn
(1)
i +hn
(0)
jn
(2)
i) +:::: (15)
Again, we require that all terms of the same power in  cancel independently, resulting in the set of
equations:

0
: hn
(0)
jn
(0)
i = 1; (16)

1
: hn
(1)
jn
(0)
i +hn
(0)
jn
(1)
i = 0; (17)

2
: hn
(2)
jn
(0)
i +hn
(1)
jn
(1)
i +hn
(0)
jn
(2)
i; (18)
which readily generalizes to

j
:
j
X
k=0
hn
(jk)
jn
(k)
i = 0: (19)
Solving these equations can be simplied by choosing a convenient global phase-factor. In this case, we
can choose the phase of the full eigenstates to be such thathn
(0)
jni is real-valued. To enforce our choice
of global phase, we can expandhn
(0)
jni in power series, giving
hn
(0)
jni = 1 +hn
(0)
jn
(1)
i +
2
hn
(0)
jn
(2)
i +:::: (20)
Requiring that the r.h.s. be real-values for any real-valued  requires that each term be independently
real-valued, so that
hn
(0)
jn
(j)
i =hn
(j)
jn
(0)
i: (21)
2 The non-degenerate case
We will now describe how to solve these equations in the case where none of the unperturbed energy levels
are degenerate.
Step #1: To obtain the j
th
correction to the n
th
energy eigenvalue, simply hit the 
j
equation from the
left with the brahn
(0)
j and solve for E
(j)
n
, giving
E
(j)
n
=hn
(0)
jVjn
(j1)
i
j1
X
k=1
E
(k)
n
hn
(0)
jn
(jk)
i; (22)
where we have usedhn
(0)
j(H
0
E
(0)
n
) = 0 andhn
(0)
jn
(0)
i = 1. Note that all quantities on the r.h.s are of
order <j, and are thus presumed to be known.
Step #2: To obtain thej
th
correction to then
th
eigenstate, we hit both sides of the 
j
equation with the
brahm
(0)
j where m6=n. This gives
hm
(0)
jn
(j)
i =
hm
(0)
jVjn
(j1)
i
E
(0)
m
E
(0)
n
+
j1
X
k=1
E
(k)
n
hm
(0)
jn
(jk)
i
E
(0)
m
E
(0)
n
; (23)
which is the expansion coecient of the j
th
correction onto the m
th
bare eigenvector.
Step #3: To obtain the remaining unknown quantity, hn
(0)
jn
(j)
i, we simply solve the normalization
equation, giving
hn
(0)
jn
(j)
i =
1
2
j1
X
k=1
hn
(jk)
jn
(k)
i: (24)
Steps 1, 2, and 3 must be solved iteratively, starting with j = 1, and repeating for each order until the
desired accuracy is achieved.
Final Step: Once these quantities are computed, we can reconstruct the n
th
eigenvalue and eigenvector
via
E
n
=
1
X
j=0

j
E
(j)
n
= E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::; (25)
and
jni = jn
(0)
i
1
X
j=0
hn
(0)
jn
(j)
i +
X
m6=n
jm
(0)
i
1
X
j=0
hm
(0)
jn
(j)
i
= jn
(0)
i
h
1 +hn
(0)
jn
(1)
i +
2
hn
(0)
jn
(2)
i +:::
i
+
X
m6=n
jm
(0)
i
h
hm
(0)
jn
(1)
i +
2
hm
(0)
jn
(2)
i +:::
i
:
(26)
2.1 First-order terms:
Introducing a more compact notation via
V
mn
=hm
(0)
jVjn
(0)
i: (27)
and
E
mn
:=E
(0)
m
E
(0)
n
; (28)
we can compute the rst-order terms from Eqs. (22), (23), and (24) with j = 1, yielding
E
(1)
n
=V
nn
; (29)
hm
(0)
jn
(1)
i =
V
mn
E
mn
; (30)
and
hn
(0)
jn
(1)
i = 0: (31)
2.2 Second-order terms:
By setting j = 2, and inserting the rst-order solutions, eqs (22), (23), and (24) give us
E
(2)
n
= hn
(0)
jVjn
(1)
i
= 
X
m6=0
jV
mn
j
2
E
mn
; (32)
hm
(0)
jn
(2)
i = 
hm
(0)
jVjn
(1)
i
E
mn

V
nn
V
mn
E
2
mn
=
X
m
0
6=n
V
mm
0V
m
0
n
E
mn
E
m
0
n

V
mn
V
nn
E
2
mn
(33)
and
hn
(0)
jn
(2)
i = 
1
2
hn
(1)
jn
(1)
i
= 
1
2
X
m6=n
jV
mn
j
2
E
2
mn
(34)
Page 5


Time-Independent Perturbation Theory
1 The central problem in time-independent perturbation theory:
LetH
0
be the unperturbed (a.k.a. `background' or `bare') Hamiltonian, whose eigenvalues and eigenvectors
are known. Let E
(0)
n
be the n
th
unperturbed energy eigenvalue, andjn
(0)
i be the n
th
unperturbed energy
eigenstate. They satisfy
H
0
jn
(0)
i =E
(0)
n
jn
(0)
i (1)
and
hn
(0)
jn
(0)
i = 1: (2)
Let V be a Hermitian operator which `perturbs' the system, such that the full Hamiltonian is
H =H
0
+V: (3)
The standard approach is to instead solve
H =H
0
+V; (4)
and use  as for book-keeping during the calculation, but set  = 1 at the end of the calculation.
The goal is to nd the eigenvalues and eigenvectors of the full Hamiltonian (4). Let E
n
andjni be then
th
eigenvalue of the full Hamiltonian, and its corresponding eigenstate, respectively. They satisfy
Hjni =E
n
jni (5)
and
hnjni = 1: (6)
Because  is a small parameter, it is assumed that accurate results can be obtained by expanding E
n
and
jni in powers of , and keeping only the leading term(s). Formally expanding the perturbed quantities
gives
E
n
=E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::; (7)
and
jni =jn
(0)
i +jn
(1)
i +
2
jn
(2)
i +:::; (8)
where E
(j)
n
andjn
(j)
i are yet-to-be determined expansion coecients Inserting these expansions into the
eigenvalue equation (5) then gives
(H
0
+V )(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::) = (E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::)(jn
(0)
i+jn
(1)
i+
2
jn
(2)
i+::::): (9)
Because of the linear independence of terms in a power series, this equation can only be satised for
arbitrary  if all terms with the same power of  cancel independently. Equating powers of  thus gives

0
: H
0
jn
(0)
i =E
(0)
n
jn
(0)
i; (10)

1
: (H
0
E
(0)
n
)jn
(1)
i = Vjn
(0)
i +E
(1)
n
jn
(0)
i (11)

2
: (H
0
E
(0)
n
)jn
(2)
i =Vjn
(1)
i +E
(2)
n
jn
(0)
i +E
(1)
n
jn
(1)
i (12)
etc::: (13)
This generalizes to

j
: (H
0
E
(0)
n
)jn
(j)
i =Vjn
(j1)
i +
j
X
k=1
E
(k)
n
jn
(jk)
i: (14)
Similiarly, we can expand the normalization equation in powers of , giving
1 =hn
(0)
jn
(0)
i +(hn
(1)
jn
(0)
i +hn
(0)
jn
(1)
i) +
2
(hn
(2)
jn
(0)
i +hn
(1)
jn
(1)
i +hn
(0)
jn
(2)
i) +:::: (15)
Again, we require that all terms of the same power in  cancel independently, resulting in the set of
equations:

0
: hn
(0)
jn
(0)
i = 1; (16)

1
: hn
(1)
jn
(0)
i +hn
(0)
jn
(1)
i = 0; (17)

2
: hn
(2)
jn
(0)
i +hn
(1)
jn
(1)
i +hn
(0)
jn
(2)
i; (18)
which readily generalizes to

j
:
j
X
k=0
hn
(jk)
jn
(k)
i = 0: (19)
Solving these equations can be simplied by choosing a convenient global phase-factor. In this case, we
can choose the phase of the full eigenstates to be such thathn
(0)
jni is real-valued. To enforce our choice
of global phase, we can expandhn
(0)
jni in power series, giving
hn
(0)
jni = 1 +hn
(0)
jn
(1)
i +
2
hn
(0)
jn
(2)
i +:::: (20)
Requiring that the r.h.s. be real-values for any real-valued  requires that each term be independently
real-valued, so that
hn
(0)
jn
(j)
i =hn
(j)
jn
(0)
i: (21)
2 The non-degenerate case
We will now describe how to solve these equations in the case where none of the unperturbed energy levels
are degenerate.
Step #1: To obtain the j
th
correction to the n
th
energy eigenvalue, simply hit the 
j
equation from the
left with the brahn
(0)
j and solve for E
(j)
n
, giving
E
(j)
n
=hn
(0)
jVjn
(j1)
i
j1
X
k=1
E
(k)
n
hn
(0)
jn
(jk)
i; (22)
where we have usedhn
(0)
j(H
0
E
(0)
n
) = 0 andhn
(0)
jn
(0)
i = 1. Note that all quantities on the r.h.s are of
order <j, and are thus presumed to be known.
Step #2: To obtain thej
th
correction to then
th
eigenstate, we hit both sides of the 
j
equation with the
brahm
(0)
j where m6=n. This gives
hm
(0)
jn
(j)
i =
hm
(0)
jVjn
(j1)
i
E
(0)
m
E
(0)
n
+
j1
X
k=1
E
(k)
n
hm
(0)
jn
(jk)
i
E
(0)
m
E
(0)
n
; (23)
which is the expansion coecient of the j
th
correction onto the m
th
bare eigenvector.
Step #3: To obtain the remaining unknown quantity, hn
(0)
jn
(j)
i, we simply solve the normalization
equation, giving
hn
(0)
jn
(j)
i =
1
2
j1
X
k=1
hn
(jk)
jn
(k)
i: (24)
Steps 1, 2, and 3 must be solved iteratively, starting with j = 1, and repeating for each order until the
desired accuracy is achieved.
Final Step: Once these quantities are computed, we can reconstruct the n
th
eigenvalue and eigenvector
via
E
n
=
1
X
j=0

j
E
(j)
n
= E
(0)
n
+E
(1)
n
+
2
E
(2)
n
+:::; (25)
and
jni = jn
(0)
i
1
X
j=0
hn
(0)
jn
(j)
i +
X
m6=n
jm
(0)
i
1
X
j=0
hm
(0)
jn
(j)
i
= jn
(0)
i
h
1 +hn
(0)
jn
(1)
i +
2
hn
(0)
jn
(2)
i +:::
i
+
X
m6=n
jm
(0)
i
h
hm
(0)
jn
(1)
i +
2
hm
(0)
jn
(2)
i +:::
i
:
(26)
2.1 First-order terms:
Introducing a more compact notation via
V
mn
=hm
(0)
jVjn
(0)
i: (27)
and
E
mn
:=E
(0)
m
E
(0)
n
; (28)
we can compute the rst-order terms from Eqs. (22), (23), and (24) with j = 1, yielding
E
(1)
n
=V
nn
; (29)
hm
(0)
jn
(1)
i =
V
mn
E
mn
; (30)
and
hn
(0)
jn
(1)
i = 0: (31)
2.2 Second-order terms:
By setting j = 2, and inserting the rst-order solutions, eqs (22), (23), and (24) give us
E
(2)
n
= hn
(0)
jVjn
(1)
i
= 
X
m6=0
jV
mn
j
2
E
mn
; (32)
hm
(0)
jn
(2)
i = 
hm
(0)
jVjn
(1)
i
E
mn

V
nn
V
mn
E
2
mn
=
X
m
0
6=n
V
mm
0V
m
0
n
E
mn
E
m
0
n

V
mn
V
nn
E
2
mn
(33)
and
hn
(0)
jn
(2)
i = 
1
2
hn
(1)
jn
(1)
i
= 
1
2
X
m6=n
jV
mn
j
2
E
2
mn
(34)
2.3 Eigenvectors and eigenvalues to second-order:
Putting the rst- and second-order terms together then gives the results
E
n
=E
(0)
n
+V
nn

2
X
m6=n
jV
mn
j
2
E
mn
+O(
3
); (35)
and
jni = jn
(0)
i
2
4
1

2
2
X
m6=n
jV
mn
j
2
E
2
mn
+O(
3
)
3
5
+
X
m6=n
jm
(0)
i
2
4

V
mn
E
mn
+
2
0
@
X
m
0
6=n
V
mm
0V
m
0
n
E
mn
E
m
0
n

V
mn
V
nn
E
2
mn
1
A
+O(
3
)
3
5
: (36)
Truncating the series, and setting  = 1 then gives the approximations
E
n
E
(0)
n
+V
nn

X
m6=n
jV
mn
j
2
E
mn
; (37)
and
jnijn
(0)
i
2
4
1
1
2
X
m6=n
jV
mn
j
2
E
2
mn
3
5
+
X
m6=n
jm
(0)
i
2
4

V
mn
E
mn
+
X
m
0
6=n
V
mm
0V
m
0
n
E
mn
E
m
0
n

V
mn
V
nn
E
2
mn
3
5
: (38)