Chapter 7. Approximation Methods – Part I
Very few problems in quantum mechanics can be solved exactly (analytically), therefore approximation methods have to be employed in almost all realistic situations. There are a small number
of situations where an exact solution of a simple equation gives a remarkably accurate account of
experimental data, but eventually refinement of experimental technique or previously overlooked
effects will require a more complicated theoretical description, which no longer yields to an exactly analytic solution. Indeed the majority of physicists dedicate their lives to either measure or
evaluate such small effects by making good approximations.
There is an enormous arsenal of approximation methods in quantum mechanics. In this class
we’ll study some major analytic approximation methods, which are still widely used today and
are the foundation for numerical computational methods. With the advent powerful computers,
these approximation methods can be vastly more precise than was imaginable originally, providing indispensible qualitative and even quantitative understanding to complex and realistic quantum systems.
In part I we will study the WKB method, the variational principle, and the time-independent
perturbation theory (Non-degenerate). In part II (Chapter 8), we will study and the degenerate
perturbation theory and the time-dependent perturbation theory.
7.1 WKB
This method is also called semiclassical approximation, developed by Wentzel, Krammers and
Brillouin in 1926. The basic idea is that the Schrödinger Equation can be solved easily for constant potential, so it can be solved approximately for a slowly-varying potential.
⎡ 2 d 2
⎤
⎢−
⎥ ψ(x) = E ψ(x) .
+V(x)
⎢ 2m dx 2
⎥
⎢⎣
⎥⎦
⎛i
⎞
If V(x) = 0 , ψ(x) = e ikx = exp ⎜⎜ 2mE x ⎟⎟⎟ .
⎟⎠
⎜⎝ 
⎛i
If V(x) =V0 , ψ(x) = e ikx = exp ⎜⎜⎜ 2m(E −V0 )
⎝
⎞
x ⎟⎟⎟ .
⎟⎠
Note: k = 2m(E −V0 ) /  can be real or imaginary for a piecewise potential.
We thus guess that for an arbitrary V(x)
102
(7.1)
⎛i
⎞
ψ(x) = e ik(x )x = exp ⎜⎜⎜ σ(x)⎟⎟⎟ ,
⎟⎠
⎝
(7.2)
where σ(x) is a complex function in general.
Plugging Eq. (7.2) into Eq. (7.1) converts the equation for ψ(x) to σ(x) :
i
σ′(x)ψ(x),

2
⎡
⎛ σ′(x)⎞⎟ ⎤⎥
i
⎢
⎜
⎟ ⎥ ψ(x) ,
ψ′′(x) = ⎢ σ′′(x)− ⎜
⎜⎝  ⎟⎟⎠ ⎥
⎢
⎣
⎦
ψ′(x) =
(7.3)
(7.4)
2
0 = (σ′) −iσ′′ + 2m[V(x)− E ] .
(7.5)
So far, no approximations are made, and Eq. (7.5) is mathematically identical to the Schrödinger
Equation [Eq. 6.1(7.1)].
The classical limit: Quantum mechanics reduces to classical mechanics when  → 0 . WKB
makes good use of this to approximate solutions of Eq. (7.5), therefore it is called the semiclassical approximation. In particular, WKB treats  as a small parameter, expanding σ(x) in
terms of  :
σ(x) = σ0(x) + (−i)σ1(x) + (−i)2 σ2(x) + 
2
0 = ⎡⎢σ0′ + (−i)σ1′ + (−i)2 σ2′ + ⎤⎥ −i ⎡⎢σ0′′ + (−i)σ1′′ + (−i)2 σ2′′ + ⎤⎥ + 2m[V(x)− E ] .
⎣
⎦
⎣
⎦
(7.6)
(7.7)
∞
Collecting the  n terms to rewrite Eq. (7.7) into the form of 0 = ∑  nFn (x) , so that we can
n=0
require Fn (x) = 0 for n = 0,1, 2 ; thus we obtain a set of equations:
0 = (σ0′ )2 + 2m[V(x)− E ],
(7.8)
0 = 2σ0′σ1′ + σ0′′ ,
(7.9)
0 = (σ1′)2 + 2σ0′σ2′ + σ1′′ ,
(7.10)
and so on. Usually, if V(x) doesn’t vary extremely fast, σ2 and higher-order terms can be safely
neglected, leading to the first-order WKB approximation:
σ(x) ≈ σ0(x)−iσ1(x) .
(7.11)
Here σ0(x) and σ1(x) can be easily found for an arbitrary potential V(x) by solving Eqs.
(7.8) and (7.9):
103
σ0′ (x) = ±p(x), where p(x) ≡ 2m[E −V(x)] .
(7.12)
Here p(x) is the classical momentum if E >V ; when E <V , it is an imaginary function. Then
x
σ0(x) = const. ± ∫ p(x ′)dx ′ ,
a
(7.13)
where a depends on the boundary conditions.
Eq. (7.9) ⇒ σ1′(x) = −
σ0′′(x)
1 d
1 d
=−
ln[σ0′ (x)] = −
ln[±p(x)] , we obtain
′
2σ0(x)
2 dx
2 dx
1
σ1(x) = const. − ln[±p(x)].
2
(7.14)
Plugging Eqs. (7.14), (7.13) and (7.11) into Eq. (7.2), the (first-order) WKB approximation is
ψ(x) ≈
⎡i x
⎤
⎡ i x
⎤
A2
exp ⎢ ∫ p(x ′)dx ′⎥ +
exp ⎢− ∫ p(x ′)dx ′⎥ .
⎢ a
⎥
⎢  a
⎥
p(x)
p(x)
⎣
⎦
⎣
⎦
A1
(7.15)
Remarks: (1) Classically, the momentum p(x) = 2m[E −V(x)] , and the probability of find the
particle at position x is proportional to 1 / p(x) . Quantum mechanically, the WKB approxima2
tion leads to ψ(x) ∝ 1 / p(x) for E >V . (2) If V(x) =V0 , a constant potential, then Eq. (7.15)
reduces to ψ(x) = B1e ipx/ + B2e −ipx/ , with p = 2m(E −V0 ) , which are the solutions to the
Schrödinger equation for a constant potential. (3) A modern topic, the path integral method, has a
strong connection to the WKB approximation [Eq. (7.15)].
Turning Points
For an arbitrary potential V(x) ,
we will encounter the turningpoint problem when employing
WKB. Consider V(x) shown in
the figure on the right. Here
limx→−∞ V(x) → ∞ . The turning point is defined as x = b , where E =V(b) and the classical
momentum vanishes: p(b) = 0 .
In region I ( x > b ), E >V(x) , the phase integral starts from the turning point x = b :
104
x
σ0(x) = ∫ p(x ′)dx ′ = ∫
b
x
b
2m (E −V(x ′)) dx ′ .
(7.16)
Here p(x) is real, so σ0 is a pure quantum phase factor for wave function ψ(x) :
ψ(x) =
1
⎡A e iσ0 (x )/ + A e −iσ0 (x )/ ⎤ .
⎢ 1
⎥⎦
2
p(x) ⎣
(7.17)
A very useful trick is to use another two parameters, A and φ , so that
ψ(x) =
⎡ σ (x)
⎤
cos ⎢⎢ 0 + φ ⎥⎥ =
p(x)
⎢⎣ 
⎥⎦
A
⎡1 x
⎤
cos ⎢ ∫ p(x ′)dx ′ + φ ⎥ .
⎢ b
⎥
p(x)
⎣
⎦
A
(7.18)
In region II ( x < b ), E <V(x) , p(x) is purely imaginary. We set p(x) = iq(x) , where q(x)
= 2m[V(x)− E ] , and it is real.
b
b
σ0(x) = ∫ p(x ′)dx ′ = i ∫ q(x ′)dx ′ ,
x
ψ(x) =
x
⎛ 1 b
⎞
⎛ 1 b
⎞
A2
exp ⎜⎜⎜− ∫ q(x ′)dx ′⎟⎟⎟ +
exp ⎜⎜⎜+ ∫ q(x ′)dx ′⎟⎟⎟ .
⎟⎠
⎟⎠
⎝  x
⎝  x
q(x)
q(x)
A1
(7.19)
(7.20)
Here when x → −∞ , V(x) → ∞ therefore ψ(x) → 0 . But the second term in Eq. (7.20) will increase exponentially as x → −∞ , thus only the first term in Eq. (7.20) survives. Set B = A1 ,
ψ(x) =
⎛ 1 b
⎞
exp ⎜⎜− ∫ q(x ′)dx ′⎟⎟⎟ .
⎟⎠
⎜⎝  x
q(x)
B
(7.21)
Question: How to determine the relationship between A and B, and the phase shift φ ? Employing the boundary conditions at x = b :
ψI (x)
x=b
= ψII (x)
x=b
,
(7.22)
d
d
ψI (x) =
ψII (x) .
dx
dx
x=b
x=b
However, at x → b both p(x) and q(x) vanish, so ψ(x) and
(7.23)
d
ψ(x) diverge, we can’t apply
dx
the boundary conditions of Eqs. (7.22) and (7.23). Question: Why does it break down?
Every approximation has its limit beyond which the approximation method breaks down.
This is very important! Always keep in mind that you have to check if the approximation method
can be safely applied (i.e., the accuracy is acceptable); otherwise, your results are misleading. So
under what conditions can we apply the (first-order) WKB?
105
The central approximation we made in (first-order) WKB is Eq. (7.11), i.e.,
σ(x) = σ0(x) + (−i)σ1(x) + (−i)2 σ2(x) +  ≈ σ0(x)−iσ1(x) . This approximation is well held
if the recursive expansion of Eq. (7.7) can be cut off at the second order:
0 = ⎡⎢(σ0′ )2 + 2m(V − E)⎤⎥ −i (2σ0′σ1′ + σ0′′) +O( 2 ) ,
⎣
⎦
(7.24)
which requires that | (σ0′ )2 + 2m(V − E) |   | 2σ0′σ1′ + σ0′′ | . Since in the WKB solution, we
require σ0′ = p(x) , we choose (σ0′ )2 as a measure of the first term, and for the similar reasons we
choose σ0′′ as a measure of the second term. Then the validity condition for WKB is
| (σ0′ )2 |   | σ0′′ | ,
leading to 1  
(7.25)
σ0′′
d ⎛⎜ 1 ⎞⎟⎟
d ⎛⎜  ⎞⎟
1 dλ(x)
⎟
⎜
=

=
=
. We obtain the condition under
⎜
⎟
⎟
dx ⎜⎜⎝ σ0′ ⎟⎠⎟
dx ⎜⎝ p(x)⎟⎠ 2π dx
(σ0′ )2
which the WKB approximation can be applied,
dλ(x)
 1.
dx
(7.26)
Physically, it means that the de Broglie wavelength varies slowly over space.
Close to the turning point, i.e., when x → b , p → 0 , and because
dλ(x)
dλ(x)
dp −1(x)
1 dp(x)
1
→ ∞ , violating Eq. (7.26).
,
= 2π
= 2π 2
∝ 2
dx x→b
dx
dx
p (x) dx
p (x)
Question: then how to treat the region close to
x = b ? When an approximation method breaks
down, we can use (1) higher-order approximation
or (2) exact method if possible. Here we can solve
the Schrödinger equation near x = b , where
V(x) =V(b) +V ′(b)(x −b),
(7.27)
for | x −b |< δ → 0 .
We have solved ψ(x) using WKB in region I ( x > b ) and in region II (x < b) :
ψI (x) =
⎡1 x
⎤
cos ⎢ ∫ p(x ′)dx ′ + φ ⎥ , where p(x) = 2m[E −V(x)]
⎢ b
⎥
p(x)
⎣
⎦
A
106
(7.28)
ψII (x) =
⎛ 1 b
⎞
exp ⎜⎜⎜− ∫ q(x ′)dx ′⎟⎟⎟, where q(x) = 2m[V(x)− E ]
⎟⎠
⎝  x
q(x)
B
(7.29)
We need to find the relation between A and B, and the phase shift φ .
(1) Solving Schrödinger Eq. in | x −b |< δ for potential V(x) =V(b) +V ′(b)(x −b) to get ψ0(x) .
(2) Now we can enforce the boundary conditions:
matching ψ0(x) and ψI (x) at x = b + δ , and ψ0(x) and ψII (x) at x = b − δ .
(3) Take the limit of δ → 0 .
The procedure and math are quite involved. The final results are
A = 2B, φ = −π / 4
(7.30)
Applications
Example 1. Bound states in a potential well: V(x) → ∞
when x → ±∞ . The key is using the boundary conditions at x = x1 and x = x 2 , the two turning points.
Start the phase integral from the left,
⎡1 x
π⎤
cos ⎢ ∫ p(x ′)dx ′ − ⎥ .
⎢  x1
4 ⎥⎦
p(x)
⎣
A1
ψI (x) =
(7.31)
Or we can start the phase integral from the right,
ψI (x) =
⎡ 1 x2
π⎤
cos ⎢ ∫ p(x ′)dx ′ − ⎥ =
⎢ x
4 ⎥⎦
p(x)
⎣
A2
⎡1 x
π⎤
cos ⎢ ∫ p(x ′)dx ′ + ⎥ .
⎢  x2
4 ⎥⎦
p(x)
⎣
A2
(7.32)
But Eqs. (7.31) and (7.32) should be the same eigenfunction; therefore, A1 = A2 and
⎡1 x
π ⎤⎥ ⎡⎢ 1 x
π ⎤⎥
⎢
′
′
′
′
p(
x
)d
x
−
−
p(
x
)d
x
+
= nπ , for n = 0,1, 2 ⇒
⎢  ∫x1
4 ⎥⎦ ⎢⎣  ∫x 2
4 ⎥⎦
⎣
∫
x2
x1
1
p(x)dx = (n + )π
2
(7.33)
This is the WKB quantization condition for a potential well, very similar to the semi-empirical
Bohr-Sommerfeld quantization condition:
∫
x2
x1
p(x)dx = (n + 1)π, for n = 0,1, 2
107
(7.34)
The difference rises from the treatment of boundary conditions:
Bohr-Sommefeld: ψ(x) vanishes at x = x1 and x 2
ψ(x) starts to exponentially decay at x = x1 and x 2
WKB:
Now we consider a concrete example:
V(x) = α | x | , with α > 0 . First, we compute the position of turning points for eigenenergy En : x 2 = En / α
and x1 = −x 2 = −En / α .
Then, Eq. (7.33) ⇒
1
(n + )π =
2
∫
x2
x1
= 2∫
x2
0
2m (En − α | x |) dx
4 2m 3/2
2m(En − αx) dx =
En
3α
(7.35)
We find the eigenenergies
⎡⎛
1 ⎞ 3απ ⎤⎥
En = ⎢⎢⎜⎜n + ⎟⎟⎟
2 ⎟⎠ 4 2m ⎥⎥⎦
⎢⎣⎜⎝
2/3
.
(7.36)
How accurate is the WKB? We can also solve this problem numerically using a computer. If
we define rn ≡
En (WKB)
En (exact)
, then r0 = 0.7162 , r1 = 0.9923 , r2 = 0.9942 , r3 = 0.9985 …
For n ≥ 10 , the WKB energies are essentially the same as the exact values.
Example 2. Tunneling rate of an energy barrier.
We have studied a tunneling rate for a square barrier; however, in reality, the energy barrier can
have a rather complicated profile, as shown here.
Then how can we evaluate the tunneling rate
T=
ψ(x 2 )
ψ(x1 )
2
2
using the WKB approximation?
108
For x1 < x < x 2 , E <V(x) , and p(x) = 2m(E −V ) = iq(x) , where p(x) is imaginary
while q(x) is real. The WKB wavefunction:
ψ(x) ≈
⎛ 1 x
⎞
exp ⎜⎜− ∫ q(x ′)dx ′⎟⎟⎟ .
⎟⎠
⎜⎝  x1
q(x)
A
(7.37)
Here we omit the exponentially increasing component. The tunneling rate
2
⎡ q(x )
⎛ 1 x2
⎞⎤
1
T ≈ ⎢⎢
exp ⎜⎜⎜− ∫ q(x)dx ⎟⎟⎟⎥⎥ .
⎟⎠⎥
⎝  x1
⎢ q(x 2 )
⎣
⎦
(7.38)
Normally the tunneling rate T  1 and q(x1 ) / q(x 2 )  1 , then
T ≈ e −2γ , with γ =
1 x2
q(x)dx .
 ∫x1
(7.39)
Here γ is Gamow’s factor. An example is alpha decay, which is your homework problem.
7.2 Variational Method
Perturbation theory requires exact solutions to H 0 ; however, it is often very difficult to solve the
eigenstate problem. The variational theory offers a powerful method to estimate the ground-state
energy E 0 and wavefunctions ψ0(r) .
The variational theorem: For an arbitrary wavefunction φ(r)
φH φ
φφ
≥ E0 .
(7.40)
Proof: The eigenstates of H , ψn , form a basis to expand φ :
φ = ∑ cn ψn ,
(7.41)
n
where cn = ψn φ . Since H ψn = En ψn ,
φ H φ = ∑ |cn |2 En ≥ E 0 ∑ |cn |2 ,
n
n
and
109
(7.42)
φ φ = ∑ |cn |2 ,
(7.43)
n
therefore, the theorem (7.40) is proved.
The variational method is based on the variational theorem. The left hand side of (7.40),
E[φ(r)] ≡
∫ φ (r)Hφ(r)d r ,
=
∫ | φ(r) | d r
*
φH φ
φφ
3
2
3
(7.44)
is a functional of φ(r) . Since E[φ(r)] ≥ E 0 , one can make E[φ] as small as possible by ‘varying’ φ to get close to E 0 [and φ(r) also becomes close to ψ0(r) ]. φ(r) is called the trial wavefunction, which depends on m parameters λ1 , λ2, , λm , and then E[φ] also depends on these
parameters. The minimum of E[φ] [constrained by the expression of φ(r) ] is found by solving
the following set of m equations,
∂E
= 0,
∂λ1
∂E
= 0, ,
∂λ2
∂E
= 0.
∂λm
(7.45)
A good format of φ(r) will result in a good estimation of the ground-state energy. If φ(r)
happens to be exact, then the exact ground-state energy is obtained. But how to construct a trial
wavefunction? We appeal to physical intuition — the asymptotic behaviors at boundary or large
distance. Different forms of φ(r) will give different estimation of E 0 ; in practice, we can use a
crude method to get crude φ(r) , then we use the variational method to refine φ(r) and E 0 .
Example: the ground-state energy of the one-dimensional infinite potential well
⎧ 0, for | x |< a
⎪
⎪
V(x) = ⎨
⎪
∞, for | x |≥ a.
⎪
⎪
⎩
(7.46)
⎛ πx ⎞
cos ⎜⎜ ⎟⎟⎟ ,
⎜⎝ 2a ⎟⎠
a
(7.47)
We have found the exact solution
ψ0(x) =
E0 =
1
 2π 2
.
8ma 2
110
(7.48)
Because wavefunction must vanish at x = ±a , a trial wavefunction must satisfy φ(±a) = 0 , and
we have many choices. The simplest one is φ(x) = A(a 2 − x 2 ) , where A is the normalization parameter (NOT a variational parameter). Therefore, we don’t need to solve Eq. (7.45), instead, total energy is evaluated directly from Eq. (7.44):
d2 2
2
∫−a (a − x ) x 2 (a − x )dx 10 π 2! 2
E=
= 2
= 1.0132E 0 .
a
2
2
2 2
π
8ma
∫ (a − x ) dx
a
2
2
(7.49)
−a
Next let’s try a bit more sophisticated one, φ(x) = A(| a |λ − | x |λ ) , with a variational parameter λ , which is determined by evaluating E(λ) ,
E(λ) =
and then solving the equation
λ=
(λ + 1)(2λ + 1)  2
,
2λ −1
4ma 2
(7.50)
dE(λ)
= 0 . We find that E(λ) has a minimum at
dλ
⎛ 5 + 2 6 ⎞⎟
1+ 6
⎜
⎟⎟ E ≈ 1.00298 E .
≈ 1.7247 , with E min = ⎜⎜
0
⎜⎜⎝ π 2 ⎟⎟⎠ 0
2
7.3 Time-Independent Perturbation (Non-Degenerate)
Very few problems can be solved analytically; however, practically in all realistic cases the effects of different factors are not equipollent. Thus one can first single out the most significant
factors, which normally can be solved either analytically or approximately. The influence of neglected factors can then be taken into account approximately as a perturbation.
Usually (not always), perturbations only slightly modify the wavefunctions and energy levels, so that the qualitative picture of the perturbed system retains the main features of the unperturbed one, and the changes in wavefunctions and energy levels due to perturbation can be expressed in terms of the unperturbed wavefunctions and energy levels. In this class we will focus
on these well-behaved perturbations. However, in certain cases an infinitesimal perturbation can
generate significant effects on the unperturbed system, and the associated quantum chaos is currently among the frontiers in research on the foundation of quantum mechanics.
7.3.1 Formal Theory (Non-Degenerate)
111
Let’s start with Hamiltonian H = H 0 + H ′ , where H ′ is much weaker than H 0 .
What we know: (1) the eigenstates n (0) and eigenvalues En(0) for H 0 : H 0 n (0) = En(0) n (0)
′ = n (0) H ′ m (0)
(2) H nm
Goal: the eigenstates n and eigenvalues En for H : H n = En n , from perturbation theory
′ , and n (0) . We can write H = H 0 + λH ′ , where 0 ≤ λ ≤ 1 , then
in terms of En(0) , H nm
En = En(0) + λEn(1) + λ 2En(2) + 
(7.51)
n = n (0) + λ n (1) + λ 2 n (2) + 
(7.52)
We want to express the j-th order corrections to energy, En(j ) , and to eigenstate, n (j ) , in
′ , and n (0) . In principle, one can obtain the exact solutions to En and n by
terms of En(0) , H nm
summing all orders (from 0 to infinity) of corrections; however, in practice, we normally do firstand second-order corrections to eigenenergies, while only first-order corrections to eigenstates.
Assume all En(0) are significantly different, i.e.,
′
En(0) − Em(0)  H nm
(7.53)
which is the condition for applying time-independent non-degenerate perturbation theory.
H n = En n
⇒
(H
0
(
)
+ λH ′) n (0) + λ n (1) + λ 2 n (2) + 
)(
(
)
= En(0) + λEn(1) + λ 2En(1) +  n (0) + λ n (1) + λ 2 n (2) + 
From the above we derive one equation for each order. We plug the zeroth-order solutions to the
first-order equation to get the first-order corrections, and then plug the first-order solutions to the
second-order equation to get the second-order corrections, and so on.
Zeroth-order: H 0 n (0) = En(0) n (0)
First-order:
(H
(
0
)
(
)
(7.54)
)
(
)
(7.55)
− En(0) n (1) = En(1) − H ′ n (0)
Second-order: H 0 − En(0) n (2) = En(1) − H ′ n (1) + En(2) n (0)
Multiply n (0) on both sides of Eq. 6.43, using the zeroth-order solution H 0 n (0) = En(0) n (0)
112
(
)
(
)
n (0) H 0 − En(0) n (1) = 0 = n (0) En(1) − H ′ n (0) = En(1) − n (0) H ′ n (0) ⇒
′
En(1) = n (0) H ′ n (0) = H nn
(7.56)
Multiply m (0) on both sides of Eq. (7.54), where m ≠ n
(
)
(
)
m (0) H 0 − En(0) n (1) = m (0) En(1) − H ′ n (0)
′ = −H mn
′ , since m ≠ n
⇒ (Em(0) − En(0) ) m (0) n (1) = En(1)δn,m − H mn
(1)
m (0) , a linear combination of the basis
We can express n (1) = ∑C nm
m
(1)
C nm
= m (0) n (1) =
′
H mn
E
(0)
n
−E
(0)
m
{ n } , and then
(0)
(1)
= 0 . Thus we obtain
, for m ≠ n ; while C nn
(1)
n (1) = ∑C nm
m (0) = ∑
m
m≠n
′
H mn
E
(0)
n
−E
(0)
m
m (0)
(7.57)
Here Eqs. (7.56) and (7.57) are the first-order non-degenerate time-independent perturbations
to energy levels and eigenstates, respectively. Notes: We can obtain this form of the first-order
correction to eigenstates only if En(0) − Em(0) ≠ 0 , i.e., non-degeneracy. When
′ , the first-order corrections are much smaller than their corresponding unperEn(0) − Em(0)  H nm
turbed results.
Now we derive the second-order energy corrections En(2) from Eq. (7.55). Using the similar
approach, multiply n (0) on both sides, we get
(
)
(
)
n (0) H 0 − En(0) n (2) = n (0) En(1) − H ′ n (1) + n (0) En(2) n (0)
⇒
⇒
0 = En(1) n (0) n (1) − n (0) H ′ n (1) + En(2)
En(2) = n (0) H ′ n (1) − En(1) n (0) n (1)
Using the first-order results: n (1) = ∑
m≠n
′
H mn
E
(0)
n
−E
(0)
m
m (0) ⇒ n (0) n (1) = 0 , therefore
(1)
En(2) = n (0) H ′ n (1) = ∑C nm
n (0) H ′ m (0) = ∑
m≠n
m≠n
113
′
H nm
2
En(0) − Em(0)
.
(7.58)
You can derive n (2) , En(3) , and so on; but in practice, they are rarely used. A summary:
n (1) = ∑
′;
En(1) = H nn
E
=∑
(2)
n
m≠n
m≠n
′
H nm
′
H mn
En(0) − Em(0)
λ = 1, H = H 0 + H ′
m (0)
⇒
2
En ≈ E
En(0) − Em(0)
(0)
n
′ +∑
+ H n,n
m≠n
n ≈ n (0) + ∑
m≠n
′
H nm
En(0) − Em(0)
′
H mn
E
(0)
n
2
−E
(0)
m
m (0)
7.3.2 Applications of the Non-Degenerate Time-Independent Perturbation Theory
Example 1: Charged harmonic oscillator in a weak electric field. The Hamiltonian is
H =−
2 d 2
1
+ mω 2x 2 + Fx = H 0 + H ′ ,
2
2m dx
2
(7.59)
where H ′ = Fx and F = qε , with ε the magnitude of electric field and q the charge.
1
We know that H 0 n (0) = En(0) n (0) with En(0) = (n + )ω .
2
We need to work out H nk′ = F n (0) x k (0) , using the ladder operators
x = x 0(a + a † ) , where x 0 =

; then
2mω
⎧
⎪
⎪
Fx 0 n
⎪
⎪
⎪
H nk′ = ⎨ Fx n + 1
⎪
0
⎪
⎪
⎪
0
⎪
⎩
k = n −1
k = n +1
Otherwise
′ =0,
En(1) = H nn
En(2) = ∑
k≠n
=
| H nk′ |2
En(0) − Ek(0)
(Fx 0 )2 n
ω
⇒ En ≈ E
(0)
n
+
+E
=
′ |2
| H n,n−1
(0)
En(0) − En−1
(Fx 0 )2(n + 1)
(1)
n
−ω
+E
(2)
n
+
=−
′ |2
| H n,n+1
(0)
En(0) − En+1
F 2x 02
ω
=−
F2
2mω 2
1
F2
= (n + )ω −
.
2
2mω 2
114
(7.60)
It is interesting to compare the exact solution, which is straightforward to obtain, since a
charged simple harmonic oscillator in electric filed is still a SHO (Why? Consider the classical
SHO under a constant force). Set y = x +
H =−
2 d 2
1
F2
2 2
+
mω
y
−
2m dy 2 2
2mω 2
⇒ En +
F
, the Hamiltonian
mω 2
2
⎛ 2 d 2
⎞
1
2 2⎟
⎟⎟ ψ(y) = (En + F )ψ(y)
⇒ ⎜⎜⎜−
+
mω
y
⎟⎠
⎜⎝ 2m dy 2 2
2mω 2
F2
1
1
F2
=
(n
+
)ω
⇒
E
=
(n
+
)ω
−
.
n
2
2
2mω 2
2mω 2
Question: the exact solution is the same as that using the second-order perturbation theory, why?
Next, we employ the perturbation theory to find the wavefunctions:
H′
(1) (0)
(1)
ψn(1)(x) = ∑C nk
ψk (x), where C nk
= (0) kn (0) .
En − Ek
k≠n
(7.61)
(1)
≠ 0 only for k = n ± 1.
In this case, C nk
(1)
Eq. (7.60) ⇒ C n,n−1
=
Fx 0 n
(1)
and C n,n+1
=−
ω
ψn (x) ≈ ψn(0)(x) +
Fx 0 n + 1
ω
, we obtain
Fx 0 ⎡
⎤
(0)
(0)
⎢ n ψn−1(x)− n + 1 ψn+1(x)⎥ .
⎦
!ω ⎣
(7.62)
Let’s compare with the exact results for the ground state.
From perturbation theory,
ψ0(x) ≈ ψ0(0)(x)−
since ψ1(0)(x) =
⎛
Fx ⎞
ψ1(0)(x) = ⎜⎜⎜1 − ⎟⎟⎟ ψ0(0)(x) ,
!ω
!ω ⎟⎠
⎝
Fx 0
(7.63)
x (0)
ψ (x) . (How would you prove that?)
x0 0
The exact solution:
ψ0(y) = A0e
⎛⎜ y ⎟⎞2
⎟⎟
−⎜⎜
⎜⎝ 2x ⎟⎟⎠
0
, with A0 =
1
F
.
, y =x+
2 1/4
(2πx 0 )
mω 2
(7.64)
The first-order Taylor expansion assuming small F :
ψ0(x) ≈ A0e
2
⎜⎛ x ⎟⎞⎟ − 1 xF
−⎜⎜
⎟
⎜⎝ 2x ⎟⎟⎠
2x 02 mω 2
0
e
⎛
1 xF ⎞⎟⎟ ⎛⎜
Fx ⎞⎟ (0)
⎟⎟ ψ (x).
≈ ψ0(0)(x)⎜⎜⎜1 − 2
=
1
−
⎜
⎟
⎜⎝
ω ⎟⎠ 0
2x 0 mω 2 ⎟⎟⎠ ⎜⎝
115
(7.65)
Example 2: Fine structure of the hydrogen atom
The above figure shows the hydrogen optical absorption spectral lines on a logarithmic scale,
which can be accurately explained by solving the Schrödinger equation for Coulomb potential.
Specifically,
1
2
En = − 2 E 0, where the Rydberg energy E 0 =
= 13.60 eV.
n
2mea 02
116
(7.66)
The eigenfunctions ψnlm (r, θ,φ) can be written as
r, θ,φ n,lm = Rn (r)Yl m (θ,φ) .
(7.67)
However, when resolution is increased, one finds small deviations from theory. For example, for the 3s to 2p transition, using the
reduced mass, you’ll get 656.47 nm for hydrogen and 656.29 nm for
deuterium, respectively. But experimentally, physicists measured a
small shift of 0.15 nm (0.00042 eV) with a tiny splitting of 0.016 nm
(0.000045 eV). These effects are called the fine structure of H spectral lines, and the physical origins are rather complicated.
If you look even closer, you will find more complicated
and even weaker effects, including the Lamb shifts and the
hyperfine structure. Lamb shifts are due to electron and
vacuum interaction, which are significant only for the swave, while the hyperfine structure is due to nuclear spin.
The energy hierarchy is summarized on the right.
Here we will not discuss the Lamb shift and the hyperfine
structure. The Hamiltonian at the fine-structure level:
H = H 0 + H KE + H DW + H so .
(7.68)
The perturbation H ′ = H KE + H DW + H so , accounting for the relativistic kinetic energy, the
Darwin and the spin-orbit coupling corrections, respectively.
(1) The relativistic kinetic energy corrections.
The kinetic energy:
p2
p4
K.E. = p c + m c −mc ≈
−
,
2m 8m 3c 2
2 2
2 4
2
(7.69)
for low-energy electrons. Then
H KE = −
(1)
E KE,nlm
=−
p̂ 4
,
8m 3c 2
1
n,lm p̂ 4 n,lm .
3 2
8m c
117
(7.70)
(7.71)
You can imagine that evaluating Eq. (7.71) in real-space integrals is very difficult. Here is
the trick. We note that
2
2
⎛ p2 ⎞
⎛
e2 ⎞
p = 4m ⎜⎜⎜ ⎟⎟⎟ = 4m 2 ⎜⎜⎜H 0 + ⎟⎟⎟
⎜⎝ 2m ⎟⎠
⎜⎝
r ⎟⎠
4
2
(7.72)
so that
⎡
1
1 ⎤
n,lm p̂ 4 n,lm = 4m 2 ⎢⎢(En(0) )2 + 2En(0)e 2
+e 4 2 ⎥⎥ ,
r
r ⎥⎦
⎢⎣
where
(7.73)
1
1
1
1
= n,lm n,lm and 2 = n,lm 2 n,lm . We find that (Homework problem)
r
r
r
r
e2
1
= −2En(0) ,
r
(7.74)
2
e
4
4nE 0
1
.
=
2
l +1/ 2
r
The first-order relativistic kinetic energy corrections are
⎡ 3
⎤
α4
1
(1)
⎥
E KE,nlm
= − mc 2 ⎢⎢− 4 + 3
⎥
2
4n
n
(l
+
1
/
2)
⎣
⎦
(7.75)
(7.76)
(2) The Darwin correction.
This is the correction due to the relativistic Dirac equation. Physically, when an electron approaches the nuclear, it will be “trembling”, a phenomenon called “zitterbewegung”. It is a consequence of the interference of the electron with its antiparticle, which appears as a virtual particle when the energy of the electron comes close to its rest energy. The consequence of the zitterbewegung is that the electron sees a smeared-out Coulomb potential of the nucleus. Mathematically, the Darwin correction is found by acting with the Laplacian on the potential.
In the case of the Coulomb potential, this yields
∇2V(r)  ∇2(1 / r)  δ(r) ,
(7.77)
so that
H DW = gδ(r), where g =
 2π 2e 2
.
2m 2c 2
(1)
E DW,nlm
= nlm H DW nlm = g | ψnlm (r = 0) |2 .
118
(7.78)
(7.79)
l
⎛ r ⎞⎟
⎛ Zr ⎞⎟
⎛
⎞
⎟⎟ L(2l+1) ⎜⎜ 2Zr ⎟⎟⎟ [See Eq. (4.66) in Chapter 4],
Because unl (r) = rRnl (r) = rAnl ⎜⎜⎜ ⎟⎟ exp ⎜⎜⎜−
⎜⎝ na 0 ⎟⎟⎠ n−l−1 ⎜⎜⎝ na 0 ⎟⎟⎠
⎝⎜a 0 ⎟⎟⎠
we obtain
⎪⎧⎪ 0
(l > 0)
⎪⎪
3/2
ψnlm (r = 0) = ⎪⎨ 1 ⎛⎜ 1 ⎞⎟
(7.80)
⎪⎪
⎟⎟
⎜⎜
(l = 0),
⎪⎪ π ⎜⎝ na ⎟⎟⎠
0
⎪⎩
E
and
(1)
DW,nlm
⎧⎪ 0
⎪⎪
= ⎪⎨ α 4
⎪⎪
mc 2
⎪⎪⎩ 2n 3
(l > 0)
(l = 0),
(7.81)
which indicates that the Darwin correction is positive and only applies to the s-wave.
(3) The spin-orbit interaction.
The spin-orbit coupling Hamiltonian is
H so =
e2
S⋅L
2m 2c 2r 3
(7.82)
The total angular momentum is defined as
J = S+ L ,
(7.83)
1
S ⋅ L = (J 2 − L2 −S 2 ) .
2
(7.84)
therefore
We need to construct the common eigenstates for J2 , L2 and S2 , designated as j,m;l,s ,
where j = | l −s |, | l −s | +1, , j + s −1, j + s and m = m j = −j, − j + 1, , j −1, j .
In this basis,
j ′, m ′;l ′,s S ⋅ L j,m;l,s = δj, j ′δm,m′δl,l ′
2
2
⎡ j(j + 1)−l(l + 1)−s(s + 1)⎤ .
⎢⎣
⎥⎦
(7.85)
Here s = 1 / 2 , so that j = l ± 1 / 2 ( l ≠ 0 ) and j = s = 1 / 2 ( l = 0 ). For l = 0 , there is no
spin-orbit coupling and E so(1) vanishes. For l ≠ 0 , we obtain
119
E
(1)
so,nlm
 2e 2 1
=
4m 2c 2 r 3
⎧
⎪
⎪ l
⎨
⎪
−(l + 1)
⎪
⎪
⎩
(j = l + 1 / 2)
(7.86)
(j = l −1 / 2)
Using the result of (Again your homework problem)
1
1
1
= 3 3
,
3
r
a 0 n l(l + 1 / 2)(l + 1)
one finds the first-order spin-orbit interaction corrections
⎧
⎪
α4
1
⎪ l
(1)
2
E so,nlm =
mc 3
⎨
4
n l(l + 1 / 2)(l + 1) ⎪
−(l + 1)
⎪
⎪
⎩
(7.87)
(j = l + 1 / 2)
(j = l −1 / 2)
(7.88)
All these three corrections are of the α 4 effects, and we combine them to write down the total fine-structure energy shifts
(1)
E f.s.,nj
=−
⎛ 1
α4
3 ⎞⎟
2⎜
⎟
mc
−
⎜
⎜⎝ j + 1 / 2 4n ⎟⎟⎠ .
2n 3
(7.89)
Some remarks:
1) Here j = l ± 1 / 2 for l > 0 , while j = 1 / 2 for l = 0 .
(1)
2) For l > 0 , ΔE f.s.
comes from the spin-orbit interaction and relativistic kinetic energy;
(1)
however, when l = 0 , i.e., for the s-wave, ΔE f.s.
comes from the Darwin correction and
relativistic kinetic energy.
3) The exact fine-structure energy levels obtained from the Dirac Equation are
2
⎡ ⎛
⎞⎟
⎢
⎜
α
⎟⎟
Enj = mc 2 ⎢1 + ⎜⎜
⎟
⎢ ⎜⎜
2
2 ⎟
⎢ ⎝ n −(j + 1 / 2) + (j + 1 / 2) − α ⎟⎠
⎣
⎤
⎥
⎥
⎥
⎥
⎦
−1
2
−mc 2 .
(7.90)
Expand Eq. 6.79 to order α 4 , we recover the perturbative results (Eq. 6.78), amazing!
4) Evaluating the correction due to the spin-orbit interaction requires the degenerate perturbation theory, which will be discussed in next Chapter.
120