Download Chapter 3 Approximation Methods in QM

Chapter 3
Approximation Methods in QM
Contents
3.1 Time independent PT (nondegenerate) . . . . . . . . . . . . . . . 52
3.2 Degenerate perturbation theory (PT) . . . . . . . . . . . . . . . . . 59
3.3 Time dependent PT and Fermi’s golden rule . . . . . . . . . . . 62
3.4 Variational method and helium atom . . . . . . . . . . . . . . . . . 67
Basic Questions
A. What causes fine strucutre in hydrogen spectra?
B. What is Fermi’s golden rule?
C. What is the variational principle?
51
52
3.1
3.1.1
CHAPTER 3. APPROXIMATION METHODS IN QM
Time-independent PT (nondegenerate)
Introduction
Two quantum problems can be solved exactly, they are harmonic oscillator and hydrogen (or hydrogen-like) atom. Most other quantum problems can not be solved
exactly. One has to develop approximate method to solve such problem.
Consider a Hamiltonian Ĥ, its Schrödinger eq.
ĤΨp = Ep Ψp ,
p = quantum no.
is difficult to solve. If Ĥ can be written as
Ĥ = Ĥ0 + V̂
where V̂ is time-independent and ”small”, and the Schrödinger eq. of Ĥ0 has been
solved
Ĥ0 Φk = ek Φk → Ĥ0 |Φk i = ek |Φk i
where k is the quantum number labelling the nondegenerate states. We wish to use
the information of Ĥ0 to find the eigenstates and eigenvalues of Ĥ. Ĥ0 is usually
referred to as unperturbed Hamiltonian, V̂ as perturbation potential (operator).
In principle (the completeness principle), any state Ψp can be written as a linear
combination of Φk as
X
|Ψp i =
Cpk |Φk i
k
where {Cpk } are constants to be determined by eq. of Ĥ, together with eigenvalue
Ep . Since the dimensionality (total number of the possible k) is in general infinite,
the exact solution is not easy. However, since the perturbation V̂ is assumed small,
it is not difficult to get approximate solution for |Ψp i and Ep .
3.1.2
Perturbation expansion
Obviously, if V̂ is zero, p = k, |Ψk i = |Φk i, Ek = ek . Hence, for non-zero V̂ , this
solution {p = k, |Ψk i = |Φk i , Ek = ek } is referred to as zero-order perturbation
results. We need to go beyond this, and find nontrivial correction to this zero-order
results. In order to do it in a systematic way, we introduce a parameter λ in the
Hamiltonian
Ĥ (λ) = Ĥ0 + λV̂ , 0 ≤ λ ≤ 1
so that as λ → 0, solutions for Ĥ (λ) is reduced to that of Ĥ0 and what we want is
the results in the limit λ = 1. In perturbation theory, we assume
p=k
3.1. TIME-INDEPENDENT PT (NONDEGENERATE)
53
i.e., the quantum number for the states of Ĥ is same as that of Ĥ0 . Therefore, we
can expand each of the eigenvalues and eigenstates of Ĥ (λ) in terms of powers of λ
as
(0)
(1)
(2)
(3)
(0)
(1)
(2)
(3)
Ek = Ek + λEk + λ2 Ek + λ3 Ek + · · ·
Ψk = Ψk + λΨk + λ2 Ψk + λ3 Ψk + · · ·
(0)
where Ek = ek is the zero-order approximation to the k-th eigenvalue of Ĥ, and
(1)
(2)
(3)
(0)
Ek , Ek , Ek , · · · are the successive correction to this. Similarly, Ψk = Φk is the
(1)
(2)
(3)
zero-order appro. to k-th eigenstate of Ĥ and Ψk , Ψk , Ψk , · · · are the successive
correction to this.
(1)
(2)
(3)
(1)
(2)
(3)
In order to find Ek , Ek , Ek , · · · and Ψk , Ψk , Ψk , · · ·, we substitute the expansions into the original Schrödinger eq.
ĤΨk = Ek Ψk
or
Ĥ0 + λV̂
=
(0)
(0)
(1)
(2)
Ψk + λΨk + λ2 Ψk + · · ·
(1)
(2)
Ek + λEk + λ2 Ek + · · ·
(0)
(1)
(2)
Ψk + λΨk + λ2 Ψk + · · ·
or, each side is arranged according to the powers of λ
(0)
(0)
(1)
Ĥ0 Ψk + λ V̂ Ψk + Ĥ0 Ψk
(0)
(0)
(1)
(0)
(1)
(2)
+ λ2 V̂ Ψk + Ĥ0 Ψk
(0)
(1)
= E k Ψk + λ Ek Ψk + E k Ψk
(2)
(0)
+···
(1)
(1)
(0)
(2)
+ λ 2 Ek Ψk + E k Ψk + E k Ψk
+···
Equating the coefficients of λ on both sides of the eq. (consider λ as an arbitrary
number, this must be true), we have
(0)
(0)
(0)
Ĥ0 Ψk = Ek Ψk
→
(0)
(0)
Ψk = Φk ,
Ek = e k
as zero-order result, and
(0)
(1)
(1)
(0)
(0)
(1)
V̂ Ψk + Ĥ0 Ψk = Ek Ψk + Ek Ψk
(1)
(1)
as first-order eq. for Ek and Ψk and
(1)
(2)
(2)
(0)
(1)
(1)
(0)
(2)
V̂ Ψk + Ĥ0 Ψk = Ek Ψk + Ek Ψk + Ek Ψk
(2)
(2)
as second-order eq. for Ek and Ψk etc.
54
3.1.3
CHAPTER 3. APPROXIMATION METHODS IN QM
Solution for perturbation eq.: 1st-order
Now consider the solution to the above sequence of the perturbation eqs. The 1storder eq. is
(0)
(1)
(1) (0)
(0) (1)
V̂ Ψk + Ĥ0 Ψk = Ek Ψk + Ek Ψk
or, using the zero-order results
(1)
V̂ − Ek
(1)
Φk = ek − Ĥ0 Ψk
(1)
By the completeness principle, we can always express Ψk in terms of a linear combination of Φk as
X (1)
(1)
Ckk0 Φk0
Ψk =
k0
(1)
where Ckk0 are coefficients to be determined and the equation becomes, using Ĥ0 Φk0 =
ek 0 Φk 0
(1)
V̂ − Ek
V̂ −
(1)
Ek
Φk =
X
k0
|Φk i =
X
k0
(1)
(ek − ek0 ) Ckk0 Φk0
(1)
(ek − ek0 ) Ckk0 |Φk0 i
(1)
this is the eq. for Ek and Ckk0 . The standard technique to solve this kind eq. is to
take inner product with state Φk00 on both sides of the eq., i.e., multiplying Φ∗k00 and
integrate
Z
Z
X
(1)
(1)
d3 rΦ∗k00 V̂ − Ek Φk =
(ek − ek0 ) Ckk0 d3 rΦ∗k00 Φk0
k0
using the orthonormal relation
Z
d3 rΦ∗k Φk0 = δkk0
we have
(1)
Vk00 k − Ek δkk00 =
Vk00 k ≡
X
Zk
0
(1)
d3 rΦ∗k00 V̂ Φk
Choosing k 00 = k we have
(1)
(1)
(ek − ek0 ) Ckk0 δk00 k0 = (ek − ek00 ) Ckk00
Ek = Vkk =
Z
d3 rΦ∗k V̂ Φk
3.1. TIME-INDEPENDENT PT (NONDEGENERATE)
Choosing k 00 6= k we have
55
Vk00 k
.
ek − ek00
(1)
Ckk00 =
Note that Ckk is still undetermined and can be chosen as zero.
The 1st-order results are, letting λ = 1,
Ek ≈ e k +
(1)
Ek
= Vkk =
Z
d3 rΦ∗k V̂ Φk
for the energy and
Ψk ≈ Φ k +
Vk 0 k
Φk 0
0
e
−
e
0
k
k
k 6=k
X
for the state. Note: if we choose Ckk 6= 0, the coefficient of Φk in the above formula
will not be unity and Ψk is then not properly normalized in the limit Vk0 k → 0.
(1)
Note that the determine of the 1st-order energy Ek only depends on the zero(0)
(1)
order state function Ψk = Φk , and the 1st-order statefunction Ψk is not necessary.
(2)
However, in order to determine the 2nd -order energy correction Ek , the 1st-order
(1)
statefunction correction Ψk is required. This is generally true: nth-order energy
correction requires the (n − 1)th-order statefunction correction.
3.1.4
Solution for perturbation eq.: 2st-order
The 2nd-order equation is
(1)
(2)
(2)
(0)
(1)
(1)
(0)
(2)
V̂ Ψk + Ĥ0 Ψk = Ek Ψk + Ek Ψk + Ek Ψk
or
(2)
(0)
(1)
Ek Ψk = V̂ − Ek
(0)
(1)
(0)
(1)
(1)
(0)
Ψk + Ĥ0 − Ek
(2)
Ψk
(2)
(2)
where Ek , Ek , Ψk and Ψk are known. We only need to determine Ek and Ψk .
(2)
Again, Ψk can be written as a linear combination of Φk
(2)
Ψk =
X
(2)
Ckk0 Φk0
k0
substituting this into the eq., using the zero- and 1st-order results, we have
(2)
Ek Φk =
X
k0
=
X
k0
(1)
(1)
Ckk0 V̂ − Ek
(1)
(1)
Ckk0 V̂ − Ek
Φk 0 +
X
k0
Φk 0 +
X
k0
(2)
Ckk0 Ĥ0 − ek Φk0
(2)
Ckk0 (ek0 − ek ) Φk0
56
CHAPTER 3. APPROXIMATION METHODS IN QM
as before, taking inner product with state Φk00 on both sides of the eq. and using the
orthonormality of Φk we have
(2)
Ek δk00 k =
X
k0
(1)
(1)
(2)
(1)
Ckk0 V̂k00 k0 − Ek δk00 k0 + Ckk00 (ek00 − ek ) .
(2)
Choosing k 00 = k, we obtain Ek
(2)
Ek
=
X
k0
=
X
(1)
Ckk0 Vkk0 − Ek δkk0
(1)
Ckk0 Vkk0 ,
←
k0
=
(1)
Ckk = 0
X |Vkk0 |2
Vkk0 Vk0 k
=
0
0
k 0 6=k ek − ek
k 0 6=k ek − ek
X
(2)
where we have assume V̂ is a Hermitian operator. The coefficient Ckk00 can be deter(2)
mined by choosing k 00 6= k, as before. Again, to determine Ek we only need to know
(1)
(2)
Ckk0 and Ckk00 is not necessary.
The 2nd-order result for the energy is then given by
(0)
(1)
(0)
(1)
(2)
Ek ≈ Ek + Ek + Ek = ek + Vkk +
Ψk ≈ Ψ k + Ψ k = Φ k +
with
0
Vk0 k ≡ hk | V̂ |ki =
3.1.5
|Vkk0 |2
0
k 0 6=k ek − ek
X
Vk 0 k
Φk 0
0
k 0 6=k ek − ek
X
Z
d3 r Φ∗k0 V̂ Φk .
Example
Consider an anharmonic oscillator with Hamiltonian
Ĥ = Ĥ0 + V̂
1
h̄2 ∂ 2
+ mωx2 ,
Ĥ0 = −
2
2m ∂x
2
where α is a small number (α h̄ω/x30 = h̄ω
been solved,
1
Ĥ0 |ni = h̄ω n +
|ni ,
2
1
Ĥ0 = h̄ω ↠â +
2
mω
2h̄
3/2
V̂ = αx3
). The eigenequation of Ĥ0 has
n = 0, 1, 2, · · ·, ∞
3.1. TIME-INDEPENDENT PT (NONDEGENERATE)
57
and we know the effect of ↠and â on the normalized state |ni
√
√
↠|ni = n + 1 |n + 1i , â |ni = n |n − 1i for n > 0 .
we hence need to express V̂ in terms of â and ↠. Using the definition
â ≡
r
we have
x̂ =
hence
V̂
ip̂
mω
x̂ + √
,
2h̄
2mh̄ω
â ≡
s
1
p̂ =
i
h̄ â + ↠,
2mω
h̄
= α
2mω
!3/2
h̄
= α
2mω
!3/2
h̄
= α
2mω
!3/2
r
†
3
â + â†
â + â†
â + â†
ip̂
mω
x̂ − √
2h̄
2mh̄ω
s
mωh̄ â − â†
2
â2 + â↠+ ↠â + â†2
â2 + 2↠â + 1 + â†2 .
Up to 2nd-order approximation, we have
En ≈ en + Vnn +
where
Vnn0 Vn0 n
e − e n0
n0 6=n n
X
1
en = h̄ω n +
2
and
Vnn = hn| V̂ |ni
h̄
= α
2mω
!3/2
hn| â + â†
h̄
= α
2mω
= 0
!3/2
hn| â + â†
i.e., the first-order energy correction is zero.
â2 + 2↠â + 1 + â†2 |ni
â2 + 2n + 1 + â†2 |ni
58
CHAPTER 3. APPROXIMATION METHODS IN QM
To determine the 2nd-order energy correction, we need to evaluate Vn0 n . Consider
â + â†
=
=
â + â†
â + â
†
â2 + 2↠â + 1 + â†2 |ni
â2 + 2n + 1 + â†2 |ni
q
n (n − 1) |n − 2i + (2n + 1) |ni +
q
(n + 1) (n + 2) |n + 2i
q
√
n (n − 1) (n − 2) |n − 3i + (2n + 1) n |n − 1i + (n + 1) (n + 2)2 |n + 1i
q
q
√
+ n (n − 1)2 |n − 1i + (2n + 1) n + 1 |n + 1i + (n + 1) (n + 2) (n + 3) |n + 3i
q
q
√
=
n (n − 1) (n − 2) |n − 3i + (2n + 1) n + n (n − 1)2 |n − 1i +
q
q
√
+ (n + 1) (n + 2)2 + (2n + 1) n + 1 |n + 1i + (n + 1) (n + 2) (n + 3) |n + 3i
=
q
=
q
n (n − 1) (n − 2) |n − 3i + 3n3/2 |n − 1i + 3 (n + 1)3/2 |n + 1i
q
+ (n + 1) (n + 2) (n + 3) |n + 3i
therefore, there are only four nonzero contribution to Vn0 n
!3/2 q
Vn−3,n
h̄
= hn − 3| V̂ |ni = α
2mω
Vn−1,n
h̄
= hn − 1| V̂ |ni = α
2mω
!3/2
3n3/2
Vn+1,n
h̄
= hn + 1| V̂ |ni = α
2mω
!3/2
3 (n + 1)3/2
Vn+3,n
h̄
= hn + 3| V̂ |ni = α
2mω
!3/2 q
n (n − 1) (n − 2)
(n + 1) (n + 2) (n + 3)
therefore
2
2
2
2
Vn−1,n
Vn+1,n
Vn+3,n
Vn−3,n
Vnn0 Vn0 n
+
+
+
=
e − e n0
en − en−3 en − en−1 en − en+1 en − en+3
n0 6=n n
X
h̄
1 2
=
α
3h̄ω
2mω
!3
n (n − 1) (n − 2)
h̄
1
+ α2
h̄ω
2mω
!3
9n3
h̄
1
− α2
h̄ω
2mω
!3
9 (n + 1)3
3.2. DEGENERATE PT
59
h̄
1 2
α
−
3h̄ω
2mω
1 2
h̄
=
α
3h̄ω
2mω
!3
(n + 1) (n + 2) (n + 3)
!3
h
i
× n (n − 1) (n − 2) + 27n3 − 27 (n + 1)3 − (n + 1) (n + 2) (n + 3)
h̄
1
= − α2
h̄ω
2mω
!3
30n2 + 30n + 11
Hence, up to 2nd-order, the energy value is
1
1 2
h̄
En = h̄ω n +
−
α
2
h̄ω
2mω
!3
30n2 + 30n + 11
and the energy interval between adjacent energy levels
∆ = En − En−1
60 2
h̄
= h̄ω −
α
h̄ω
2mω
!3
n
See figure
3.2
Degenerate PT
In the above perturbation theory, the unperturbed states Φk are nondegenerate, ek0 6=
ek for all k 0 6= k. Now we extend the theory to the degenerate case. Our Hamiltonian
is still given by
Ĥ = Ĥ0 + V̂
where the perturbed term V̂ is assumed small and the unperturbed Hamiltonian Ĥ0
has been solved
Ĥ0 Φnd = en Φnd
with quantum number d = 1, 2, · · ·, M , labelling the degeneracy. In the case of
P
P
Hydrogen, we know for each n, we have degenerate qn d = (l, m), d = l,m and
M 2 . For convenience we only need one letter for the notation. We expect that the
60
CHAPTER 3. APPROXIMATION METHODS IN QM
introduction of perturbed term will lift the degeneracy in l, so the eigen equation for
Ĥ is
ĤΨnd = End Ψnd .
But if we still carry out the similar analysis as before, we expect to encounter a factor
such as Ĥ → Ĥ (λ) = Ĥ0 + λV̂ , and the expansion series
(0)
(1)
(0)
(1)
End = End + λEnd + · · ·
Ψnd = Ψnd + λΨnd + · · ·
substituting these into the original Schrödinger eq. and equating the coefficients of λ
on both sides of the eq. (consider λ as an arbitrary number, this must be true), we
have
(0)
(0) (0)
(0)
(0)
Ĥ0 Ψnd = End Ψn,d → Ψnd = Φnd , End = en
as zero-order result, and
(1)
(0)
(1)
(0)
(0)
(1)
V̂ Ψnd + Ĥ0 Ψnd = End Ψnd + End Ψnd
as first-order eq. etc. All those formulae remain the same, only replacing single qn by
double ones k → (n, d). However, there are divergence problem when we go beyond
zeroth-order. Consider the 1st-order correction to the wavefunction
(1)
Ψnd =
X
(1)
cnd,n0 d0 Φn0 d0
n 0 d0
=
X
d0 6=d
(1)
cnd,nd0 Φnd0 + other contributions from states with n0 6= n
with the diverging coefficient
(1)
cnd,nd0 =
Vnd,nd0
(0)
(0)
End − End0
=
Vnd,nd0
.
en − e n
Solution: Recall the basic assumption in PT
quantum number after perturbation = quantum number before perturbation.
We need to modify this by introducing a new quantum number k as
before perturbation : (n, d)
after perturbation : (n, k) .
This is equivalent to introduce a new basic set {Ωnk , k = 1, 2, · · ·, M }as
Ωnk =
M
X
d=1
bkd Φnd ,
k = 1, 2, · · ·, M
3.2. DEGENERATE PT
61
with coefficients bkd to be determined by the condition to avoid the divergence mentioned above.
Now our new PT expansion series are
(0)
(1)
(0)
(1)
Enk = Enk + λEnk + · · ·
Ψnk = Ψnk + λΨnk + · · ·
and the zeroth-order equation does not change much
(0)
Ĥ0 Ψnk
=
(0) (0)
Enk Ψnk
→
(0)
Enk
(0)
Ψnk
= en ,
= Ωnk =
M
X
bkd Φnd
d=1
and the 1st-order eq.
(1)
(0)
(0)
(1)
V̂ − Enk Ψnk = Enk − Ĥ0 Ψnk
(1)
where we expand Ψnk in terms of Ωnk
(1)
Ψnk =
X
(1)
ckk0 Ωnk0 ,
k0
k = 1, 2, · · ·, M .
As before, we substitute this into the 1st-order eq. and take inner product with state
Ωnk00 on both sides, we have
Z
(1)
Enk
Vnk00 ,nk −
(1)
Enk δk00 k
d
3
r Ω∗nk00
or
V̂ −
Take k 00 = k, we have
Ωnk =
=
Z
Z
d3 r Ω∗nk00
d3 r Ω∗nk00
X
k0
X
k0
(1)
(0)
ckk0 Enk − Ĥ0 Ωnk0
(1)
ckk0 (en − en ) Ωnk0 = 0 .
(1)
Enk = Vnk,nk
and take k 00 6= k, we have
Vnk00 ,nk =
Z
d3 r Ω∗nk00 V̂ Ωnk = 0,
for all k 00 6= k .
This last equation means operator V̂ is diagonal in the new basis. Therefore, the
P
coefficients {bkd } in Ωnk = M
d=1 bkd Φnd is chosen so that V̂ is diagonal.
Example: Spin-orbit coupling (or LS-coupling). An electron moving in a
central potential V̂ = V (r) has Hamiltonian as
Ĥ0 = −
h̄2 2
∇ + V (r) .
2m
62
CHAPTER 3. APPROXIMATION METHODS IN QM
In general the eigenvalues Enl of Ĥ0 are 2 (2l + 1)-fold degenerate (for hydrogen,
Enl = En , and degeneracy is 2n2 -fold). One of relativistic effects (Dirac) is the
addition of a potential, the so-called spin-orbit coupling (or LS-coupling)
V̂LS = A (r) L̂ · Ŝ .
We can treat this V̂LS as perturbation potential and write
1
2
V̂LS = A (r) Ĵ2 − L̂2 −Ŝ ,
2
Ĵ ≡ L̂ + Ŝ
As we know, the eigenstates of Ĥ0 can be represented by the product of orbital wavefunctions and spin wavefunctions, or |nlmi |sms i = |nlml sms i. Within the first-order
degenerate PT, we need to use these |nlml sms i to diagonalize the perturbation operator V̂LS and to obtain the zero-order wavefunction and first-order energy correction.
We have already done so in the previous chapter. The zeroth-order wavefunction are
angular momentum states |nlsjmi which are eigenstates of Ĵ2 and Jˆz with angular
momentum quantum number, using s = 1/2
1
j = l ± , m = −j, −j + 1, · · ·, j − 1, j .
2
Hence, the energy level shift by LS-coupling is
(1)
ELS
3.3
3.3.1
3
h̄2
j (j + 1) − l (l + 1) −
hA (r)inl .
= hnlsjm| V̂LS |nlsjmi =
2
4
Time-dependent perturbation theory
Newton’s equation
As we have seen before, the time-dependent Schrödinger eq. is
ih̄
∂
Ψ (r, t) = ĤΨ (r, t) ,
∂t
−ih̄
∂ ∗
Ψ (r, t) = ĤΨ∗ (r, t) .
∂t
The expectation value of an arbitrary operator  =  (t), assumed Ψ (r, t) is normalized,
Z
D E
 = hΨ|  |Ψi = d3 r Ψ∗ (r, t)  (t) Ψ (r, t)
has the following evolution equation
Z
i
d D E
∂ h ∗
 =
d3 r
Ψ (r, t) Â (t) Ψ (r, t)
dt
∂t
Z
Z
1
∂ Â
Ψ (r, t) +
=
d3 r Ψ∗ (r, t)
d3 r Ψ∗ (r, t) ÂĤ − Ĥ Â Ψ (r, t)
∂t
ih̄
3.3. TIME-DEPENDENT PERTURBATION THEORY
or
d D E
 =
dt
*
∂ Â
∂t
+
+
63
iE
1 Dh
Â, Ĥ .
ih̄
Using this formula, Newton’s equation becomes, using Ĥ = p̂2 /2m + V̂ (r)
D
E
d
hp̂i = − ∇V̂ .
dt
3.3.2
Time-dependent expansion
Consider now the following Hamiltonian
Ĥ = Ĥ0 + V̂ (t)
where Ĥ0 is time-independent with eigen eqs.
Ĥ0 Φn (r) = En Φn (r) ,
n = 1, 2, · · ·
and the time dependent part of the Hamiltonian is V̂ (t). The general time-dependent
wavefunction Ψ (r, t) can be expanded in the following general form as
Ψ (r, t) =
X
dn (t) Φn (r) =
n
X
cn (t) Φn (r) e−iEn t/h̄
n
and after substituting into the Schrödinger eq., we have
X
n
!
X
dcn
Φn (r) e−iEn t/h̄ =
En + V̂ cn Φn (r) e−iEn t/h̄
En cn + ih̄
dt
n
or
ih̄
X
n0
X
dcn0
Φn0 (r) e−iEn0 t/h̄ =
cn0 V̂ Φn0 (r) e−iEn0 t/h̄ .
dt
n0
Taking inner product with Φn (r) on both sides of eq., we have the time-evolution
equation for cn (t)
ih̄
dcn X
=
Vnn0 cn0 eiωnn0 t ,
dt
0
n
with
Vnn0 ≡ hΦn | V̂ |Φn0 i =
Z
d3 r Φ∗n V̂ Φn0 ,
n = 1, 2, · · ·
h̄ωnn0 = En − En0 .
these are a set of first-order, coupled equations for {cn , n = 1, 2, · · ·}. Normally difficult to solve. For some special case, exact solution can be found. See Mandl, Quantum
Mechanics, Section 9.2, P 198.
64
CHAPTER 3. APPROXIMATION METHODS IN QM
Consider a typical initial condition: at t = 0, the system is in one of the eigenstate,
say, |1i, namely
cn (0) = δ1n .
For V̂ (t) = 0, the amplitudes cn (t) retain this initial value at all times and
Ψ (r, t) =
X
cn (t) Φn (r) e−iEn t/h̄ = Φ1 (r) e−iE1 t/h̄
n
the time-dependent part is trivial.
3.3.3
Perturbation Theory
Now we treat V̂ as perturbation potential and write the Hamiltonian as before
Ĥ = Ĥ0 + λV̂ (t)
and correspondingly
(1)
2 (2)
cn = c(0)
n + λcn + λ cn + · · ·
the evolution equation for cn becomes
ih̄
X (0)
d (0)
(1)
(2)
2 (2)
cn + λc(1)
cn0 + λcn0 + λ2 cn0 + · · · λVnn0 eiωnn0 t .
n + λ cn + · · · =
dt
n0
Setting the coefficients of each power of λ equal on both sides we have a series successive eqs.,
d (0)
c
= 0, c(0)
n (t) = const.
dt n
X (0)
d
ih̄ c(1)
=
cn0 Vnn0 eiωnn0 t
n
dt
n0
X
d
(1)
=
cn0 Vnn0 eiωnn0 t
ih̄ c(2)
n
dt
n0
··· .
ih̄
Consider the typical initial condition: applying the perturbation V̂ (t) after t > 0,
namely
(
= 0,
t≤0
V̂ (t)
6= 0,
t>0
and assume for t ≤ 0, the system is in one of the eigenstates of Ĥ0 , say |1i,
cn (0) = c(0)
n (t) = δ1n
3.3. TIME-DEPENDENT PERTURBATION THEORY
65
Hence the first-order equation becomes
ih̄
d (1) X (0)
cn =
cn0 Vnn0 eiωnn0 t = Vn1 eiωn1 t
dt
n0
Integrating these equations from t = 0 to t > 0, subject to the condition that cn (0) =
δ1n , we have
(1)
c1 (t) = 1 +
c(1)
n (t) =
(1)
1
ih̄
Z
t
0
V̂11 (t0 ) dt0
1 Zt
0
V̂n1 (t0 ) eiωn1 t dt0 ,
ih̄ 0
2
ωn1 =
En − E 1
,
h̄
n 6= 1
where c1 (t) is the probability (in 1st-order approximation) that the system at
2
t > 0 stays in the original state |1i, and c(1)
n (t) is the probability (in 1st-order
approximation) that the system at t > 0 is in another state |ni (n 6= 1).
Example. A hydrogen atom is placed in a spatially homogeneous time-dependent
electric field in the z-axis
(
0,
t≤0
ε=
−t/τ
ε0 e
,
t>0
with τ > 0. At time t = 0, the atom is in the (1s) ground state with wavefunction
and energy
1 −r/a0
4π0h̄2
e2
φ1s = q
, a0 =
.
, E1s = −
e
8π0 a0
me2
πa30
Find the probability, in the 1st order perturbation theory, that after a sufficiently
long time the atom is in the (2p) excited state with wavefunction and energy
φ2p = q
1
32πa50
r cos θ e−r/2a0 ,
E2p =
E1s
.
4
Solution: According to the 1st-order perturbation theory, the probability amplitude for the transition (1s) → (2p) is
1
c (t) =
ih̄
with
ω2p,1s =
Z
t
0
0
V̂2p,1s (t0 ) eiω2p,1s t dt0
E2p − E1s
3E1s
=−
h̄
4h̄
and
0
0
V̂ (t0 ) = eε0 e−t /τ z = eε0 e−t /τ r cos θ .
66
CHAPTER 3. APPROXIMATION METHODS IN QM
The matrix element V̂2p,1s (t0 ) is
0
V̂2p,1s (t0 ) = h2p| V̂ (t0 ) |1si = eε0 e−t /τ h2p| r cos θ |1si
= eε0 e
Z
−t0 /τ

1
q
r cos θ e−r/2a0 · q
32πa50
Z ∞
1
πa30
Z π

e−r/a0  r 2 sin θdrdθdφ
2π
1
2
4 −3r/2a0
√
dφ
cos
θ
sin
θdθ
r
e
dr
= eε0 e
0
0
4 2πa40 0
1
4!25 a50 2
215/2
−t0 /τ
−t/τ
√
= eε0 e
·
· · 2π = Ce
, C ≡ 5 eε0 a0
35
3
3
4 2πa40
−t0 /τ
Z
and the amplitude
C
c (t) =
ih̄
Z
t
0
0
0
eiω2p,1s t −t /τ dt0 =
C eiω2p,1s t−t/τ − 1
ih̄ iω2p,1s − 1/τ
and the transition probability is
|c(t)|2 =
C 2 1 + e−2t/τ − 2e−t/τ cos(ω2p,1s t)
,
2
ω2p,1s
+ 1/τ 2
h̄2
and after a sufficiently long time, t = ∞, the probability is
|c (∞)|2 =
3.3.4
(eε0 a0 )2
215
.
310 (3E1s /4)2 + (h̄/τ )2
Fermi golden rule
Now we consider the following simpler case, a step function perturbation potential
V̂ (t) =
(
0,
V̂ ,
t≤0
t>0
where V̂ is independent of time. What is the transition probability per unit time
for the system originally in state |1i at t ≤ 0 to the state |2i at t > 0?
Solution: From the above 1sr-order perturbation theory, the time-integration in
the transition probability amplitude can be carried out,
(1)
c2←1 (t) =
1
ih̄
Z
t
0
V2,1 1 − eiω2,1 t
h̄ω2,1
E2 − E 1
=
h̄
0
V2,1 (t0 ) eiω2,1 t dt0 =
V2,1 = h2| V̂ |1i ,
ω2,1
3.4. VARIATIONAL METHOD
67
and the probability is, using sin y = (eiy − e−iy )/2,
4 |V2,1 |2 sin2 (ω2,1 t/2)
2πt
= 2 |V2,1 |2 D (ω2,1 , t)
2
2
ω2,1
h̄
h̄
2 sin2 (xt/2)
D (x, t) =
, x = ω2,1
πt
x2
2
(1)
P2←1 (t) = c2←1 (t) =
where function D (x, t) has a simple integral
Z
2
D (x, t) dx =
πt
−∞
∞
Z
∞
−∞
sin2 (xt/2)
dx = 1,
x2
using
Z
∞
−∞
sin2 (ax)
dx = πa,
x2
similar to the continuous delta function δ (x). In the limit t → ∞, the two become
identical. So we have the transition probability given by, for large t
P2←1 (t) =
2πt
2πt
2
|V2,1 |2 δ (E2 − E1 )
2 |V2,1 | δ (ω2,1 ) =
h̄
h̄
and the probability per unit time
Ṗ2←1 (t) =
2π
|V2,1 |2 δ (E2 − E1 ) .
h̄
This is the famous Fermi Golden rule.
3.4
3.4.1
Variational method
Variational principle
Variational principle (Rayleigh-Ritz): The expectation value of a given Hamiltonian
Ĥ in any state Ψ is always greater or equal to the exact ground-state energy E 0 ,
hΨ| Ĥ |Ψi
≥ E0
hΨ |Ψi
the equality holds if |Ψi is the exact ground state.
Proof: Suppose
Ĥ |Φn i = En |Φn i , n = 0, 1, 2, · · ·
E0 < En , for n > 0
68
CHAPTER 3. APPROXIMATION METHODS IN QM
Write |Ψi as a linear combination of |Φn i
∞
X
|Ψi =
n=0
Cn |Φn i = C0 |Φ0 i +
2
hΨ |Ψi = |C0 | +
∞
X
n=1
∞
X
n=1
Cn |Φn i
|Cn |2
Therefore
Ĥ |Ψi = Ĥ C0 |Φ0 i +
= E0 C0 |Φ0 i +
∞
X
n=1
∞
X
n=1
!
Cn |Φn i
Cn En |Φn i
and
2
hΨ| Ĥ |Ψi = E0 |C0 | +
2
∞
X
n=1
> E0 |C0 | + E0
so
En |Cn |2
∞
X
n=1
2
2
|Cn | = E0 |C0 | +
∞
X
n=1
2
|Cn |
!
= E0 hΨ |Ψi
hΨ| Ĥ |Ψi
≥ E0 .
hΨ |Ψi
In general, we do not know the exact ground state energy E0 . We wish to find
an approximation for its value. By the variational principle, our approximation hĤi
using any trial wavefunction |Ψi will always be greater than the exact value E0 .
Hence, the lower the expectation value hĤi (i.e., the closer to the exact value), the
better the trial wavefunction. A typical varitaional calculation is as follows: we first
choose a trial wavefunction Ψ with a few variational parameters α1 , α2 , · · ·
Ψ (α1 , α2 , · · ·)
and calculate the expectation value
E (α1 , α2 , · · ·) =
hΨ| Ĥ |Ψi
.
hΨ |Ψi
Next, we minimize this E (α1 , α2 , · · ·) through variational equations
∂
E (α1 , α2 , · · ·) = 0
∂α1
∂
E (α1 , α2 , · · ·) = 0
∂α2
···
3.4. VARIATIONAL METHOD
69
The solution for (α1 , α2 , · · ·) of these equations then put back to E (α1 , α2 , · · ·) and
this minimum E represents the best estimate for the trial function of the form
Ψ (α1 , α2 , · · ·).
Clearly, the success of this nonperturbative method will depend crucially on the
form of the trial wave functions we choose to start with.
3.4.2
Examples
Hydrogen atom. The trial function for the ground state of Hydrogen is choose as
Φα = Ce−αr ,
C=
s
α3
π
the Hamiltonian is
h̄2 2
e2
Ĥ = −
∇ −
,
2m
4π0 r
∂
1 ∂
r2
∇ = 2
r ∂r
∂r
2
!
+
1
 (θφ)
r2
and the expectation value is
Eα = hΦα | Ĥ |Φα i =
e2 α
h̄2 α2
−
2m
4π0
hence
∂Eα
e2 m
1
=0 → α=
2 =
∂α
a0
4π0h̄
where a0 is the Bohr radius. So the best estimate for the ground state energy is
h̄2 1
1 e2
e2 α 1
Eα =
=
−
−
2m a20 4π0 a0
2 4π0 a0
and for the ground state is
Φ= q
1
a30 π
e−r/a0
both are actually exact. This is not surprising because the trial function has the
correct form.
Helium atom. In this case we do not know the exact result and we are dealing
with two identical particles. Consider in general Helium-like ions with the following Hamiltonian describing two electrons interacting each other and with a nucleus
containing Z protons
h̄2 2
h̄2 2
Ze2
Ĥ = −
∇1 −
∇2 −
2m
2m
4π0
1
1
e2 1
+
+
.
r1 r2
4π0 r12
70
CHAPTER 3. APPROXIMATION METHODS IN QM
We want to approximate the ground state of this system by a simple trial wavefunction
(actually, PT), using the knowledge from hydrogen-like ions.
Ze2 1
h̄2
∇2 − 4π
are
We know the eigenstates φnlm (r) for a single electron Ĥ = − 2m
0 r
s
Z 3 −Zr/a0
e
πa30
s
Zr −Zr/a0
Z3
1
−
φ2s (r) = R20 (r) Y00 (θ, φ) =
e
8πa30
2a0
...
φ1s (r) = R10 (r) Y00 (θ, φ) =
with φ1s (r) as the ground state, and eigenvalues
Enl = En = −
e2 Z 2
,
8π0 a0 n2
n = 1, 2, · · · .
We also know electrons have spin degree of freedom, indicated by two states |↑i or
|↓i. We want to construct the simplest two-body wavefunction from these single-body
state. There is a (anti)symmetry requirement we need to impose. The following is a
trial wavefunction satisfying this requirement,
Ψ0 (1, 2) = φ1s (r1 ) φ1s (r2 ) χ12 ,
1
χ12 = √ (|↑i1 |↓i2 − |↓i1 |↑i2 )
2
hence Ψ (1, 2) = −Ψ (2, 1). This is a spin-singlet state. Another possible trial state
is the spin-triplet,
1
Ψ00 (1, 2) = √ [φ1s (r1 ) φ2s (r2 ) − φ2s (r1 ) φ1s (r2 )] τ12 ,
2
1
τ12 = |↑i1 |↑i2 , √ (|↑i1 |↓i2 + |↓i1 |↑i2 ) , |↓i1 |↓i2
2
but this will give higher energy due to the present of the higher level state φ2s (r1 ).
(It is quite easy to test experimentally if the ground state Helium is singlet or triplet.)
Using the singlet state Ψt (1, 2), we evaluate the energy expectation value
D
Ĥ
E
=
=
=
=
e2
1
Ze2 1
h̄2 2
+
∇1 −
2·
2m
4π0 r1
4π0 r12
!
2
2 2
Z e
5 Ze
2· −
+
8π0 a0
4 8π0 a0
5
e2
−2Z 2 + Z Ry, Ry =
4
8π0 a0
−5.50 Ry, Z = 2
*
+
which is only 6% above the experimental result of −5.81 Ry.
3.4. VARIATIONAL METHOD
Math Note: Evaluate
1
r12
D
1
r12
1
= 2
π
E
71
:
Z
a0
6 Z
d 3 r1 d 3 r1
1
e−2Z(r1 +r2 )/a0 .
|r1 − r2 |
First we use the expression (Fourier transformation)
1
=
|r1 − r2 |
hence
1
r12
1
= 2
π
Z
a0
6 Z
d3 k ik·(r1 −r2 ) 4π
e
k2
(2π)3
Z
d3 k 4π
(2π)3 k 2
now
Z
and finally
d3 r1 eik·r1 −2Zr1 /a0 = h
1
r12
4Z
=
πa0
Z
∞
0
(x2
Z
3
d r1 e
2
ik·r1 −2Zr1 /a0 16πZ/a0
k 2 + (2Z/a0 )2
i2
dx
5Z
.
4 =
8 a0
+ 1)
72
CHAPTER 3. APPROXIMATION METHODS IN QM
We can improve our wavefunction by introducing a variational parameter Z 0
s
φ1s (r) =
Z 03 −Z 0 r/a0
e
,
πa30
Ψ0 (1, 2) = φ1s (r1 ) φ1s (r2 ) χ12
and repeat our calculation as before, see example one,
D
Ĥ
E
Z0
Ze2 1
h̄2 2
∇1 −
= 2·
2m
4π0 r1
*
0 2
02 2
+
!Z
0
e2
+
4π0
1
r12
Z0
0 2
ZZ e
5 Ze
Z e
−
+
8π0 a0 4π0 a0
4 8π0 a0
5
e2
= 2 Z 02 − 2Z 0 Z + Z 0 Ry, Ry =
8
8π0 a0
= 2·
and variational eq.
∂ D E
5
Ĥ 0 = 0 → Z 0 = Z −
0
Z
∂Z
16
and the estimated ground-state energy
D
Ĥ
E
Z 0 =Z−5/16
= −2 Z −
= −
5
16
2
Ry,
Z=2
2 · 272
Ry = −5.70 Ry
162
which is closer to the experimental result of −5.81 Ry. (higher no more than 2%).
Physically, the variational parameter Z 0 represents the mutual screening by the two
electrons.
Now how does one go from here? Can we further improve our estimate and how?
A more sophisticated variational wavefunction (Hylleraas) is
Ψ (1, 2) = P (s, t, u) Ψ0 (1, 2)
Z0
Z0
s ≡
(r1 + r2 ) , t ≡
(r1 − r2 ) ,
a0
a0
where P (s, t, u) is expanded in power series of s, t and u
P (s, t, u) =
X
cn,2l,m sn t2l um
l,n,m
and variational eqs. are
∂
∂cn,2l,m
D
E
Ĥ = 0,
∂ D E
Ĥ = 0
∂Z 0
u≡
Z0
r12
a0
3.4. VARIATIONAL METHOD
73
and by including 10 cn,2l,m parameters, the accuracy can reach to 6-7 digits. See
Intermediate QM, 1964, H.A. Bethe.
Variational method and diagonalization. For a given basis set (truncated or
untruncated), {|ni , n = 1, 2, · · ·, N }, the trial wavefunction can be written as
|Ψi =
N
X
n=1
cn |ni ,
hΨ| =
N
X
n=1
hn| c∗n
where {c∗n , cn ; n = 1, 2, · · ·, N } are variational parameters determined by
δ D E
δ D E
Ĥ
=
Ĥ = 0,
δc∗n
δcn
or
hΨ |Ψi
or
PN
n0 =1
N
X
n0 =1
D
E
Ĥ =
Hnn0 cn0 − hΨ| Ĥ |Ψi cn
= 0,
hΨ |Ψi2
D
E
Hnn0 cn0 = Ĥ cn ,
hΨ| Ĥ |Ψi
hΨ |Ψi
Hnn0 ≡ hn| Ĥ |n0 i
n = 1, 2, · · ·, N .
This is precisely
D E the eigenequation for the Hamiltonian matrix {Hnn0 } with the energy
eigenvalue Ĥ . Therefore, the variational method and diagonalization is equivalent.