量子力学
Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal University
Chapter 6
Time-independent
Perturbation Theory
What
&
Why
Infinite square well with small perturbation
Problem
Suppose we have solved the (time - independen t)
Schr &o&dinger equation for some potential, now we
perturb the potential sligntly.
Difficulty & why
What
For this problem, we' d like to solve for
Perturbation theory is a sys -
the new eigenfunct ions and eigenvalues :
HΨn = En Ψn ,
tematic procedure for obtai ning approximate solutions
But unless we are very lucky, we' re
unlikely to be able to solve the Shr&o&ndi -
to the perturbed problem by
buiding on the known exact
nger equation exactly, for this more co mplicated potential.
solutions to the unperturbed
case.
OUTLINE
•
Non-degenerate perturbation theory
• Degenerate perturbation theory
• Some examples for application
Non-degenerate Perturbation Theory
• General formulation
• First-order theory
• Second-order energies
General formulation
The unperturbed case
Time - independent Shr&o&dinger equation :
0
0
n
0
n
0
n
H Ψ =E Ψ
Eigenvalues:
Eigenfunctions:
E
0
n
Requirement:
0
n
Ψ
Ψ
0
n
Ψ
0
m
= δ nm
Considering the perturbed case, we should solve
for the new eigenvalues and eigenfunctions.
We can write the Hamiltonian
of the new system into the
following two parts
0
H = H + H′
Hamiltonian of the
unperturbed system
Additional
Hamiltonian
of the
perturbed
system
Rewrite the new Hamiltonian as the sum of two terms
A small number ;later
we will crank it up to
1,and H will be the true,
exact Hamiltonian
H = H 0 + λH ′
Write the nth eigenfunction and eigenvalue as
power series in λ as follows
0
n
1
n
2
2
n
Ψn = Ψ + λΨ + λ Ψ + L
0
n
1
n
2
2
n
En = E + λE + λ E + L
The first-order
correction
The second-order
correction
Plugging the above
H,ψn and En into the
time-independent
equation and
Collecting like
powers of λ
HΨn = EnΨn
(
)
(
)
H 0 Ψn0 + λ H 0 Ψn1 + H ′Ψ + λ2 H 0 Ψn2 + H ′Ψn1 + L
(
)
(
)
= En0 Ψn0 + λ En0 Ψn1 + En1 Ψn0 + λ2 En0 Ψn2 + En1 Ψn1 + En2 Ψn0 + L
λ is a device to keep track of the different orders
To lowest order:
0
0
n
0
n
0
n
H Ψ =E Ψ
To first order:
0
1
n
0
n
0
n
1
n
1
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ
To second order：
0
2
n
1
n
0
n
2
n
1
n
1
n
2
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ + E Ψ
And so on
First-Order Theory
• The first-order correction to the
energy
• The first-order correction to the
wave function
The first-order
correction to the energy
0
1
n
0
n
0
n
1
n
1
n
0
n
H Ψ + H ′Ψ = E Ψ + E Ψ
0
n
0
1
n
Multiplying by (Ψn0 )∗
and integrating
0
n
0
0
0
1
1
0
0
′
Ψ H Ψ + Ψ H Ψn = En Ψn Ψn + En Ψn Ψn
And,
So,
Ψn0 Ψn0 = 1
0
n
0
1
n
0
n
0
n
1
n
Ψ H Ψ =E Ψ Ψ
1
n
0
n
0
n
E = Ψ H′ Ψ
Results: The first-order correction to the energy is the
expectation value of the perturbation in the unperturbed state.
Example
The first-order correction
to the wave function
Rewrite
equation
0
1
n
0
0 1
1
0
′
H Ψ + H Ψn = En Ψn + En Ψn
（H 0 − En0）Ψn1 = −
（H ′ − En1）Ψn0
The unperturbed wave functi-
Ψ = ∑C Ψ
ons constitute a complete set
m≠n
Additional,
H 0 Ψn0 = En0 Ψn0
1
n
(n)
m
0
m
0
0
(n)
0
1
0
′
(
E
−
E
)
C
Ψ
=
−
（
H
−
E
）
Ψ
∑ m n m m
n
n
m≠n
0
l
Take the inner product with Ψ
0
0
(n)
0
0
0
0
1
0
0
′
(
E
−
E
)
C
Ψ
Ψ
=
−
Ψ
H
Ψ
+
E
Ψ
Ψ
∑ m n m l m
l
n
n
l
n
m≠ n
1
n
If l = n, we get E ;
C
(n)
m
If l ≠ n, we get C .
Ψ =∑
1
n
m≠ n
(n)
m
0
n
Ψ H′ Ψ
=
0
m
0
′
Ψ H Ψn
0
n
0
m
E −E
0
m
0
n
E −E
0
m
0
m
Ψ
Notice: although perturbation theory often
yields surprising by accurate energies, the
wave functions are notoriously poor.
Second-order energies
0
2
n
1
0
2
1
1
2
0
′
H Ψ + H Ψn = E n Ψn + E n Ψn + E n Ψn
Take the inner product with Ψn0
Ψn0 H 0 Ψn2 + Ψn0 H ′Ψn1 = En0 Ψn0 Ψn2 + En1 Ψn0 Ψn1 + En2 Ψn0 Ψn0
Ψn0 H 0 Ψn2 = H 0 Ψn0 Ψn2 = En0 Ψn0 Ψn2
0
n
0
n
Meanwhile，Ψ Ψ
Ψ Ψ = ∑C
0
n
1
n
(n)
m
= 1；
0
n
0
m
Ψ Ψ
m≠n
2
E =∑
2
n
m≠ n
Ψm0 H ′ Ψn0
En0 − Em0
The Hermiticity of H0
We could proceed to
calculate the second-order
correction to the function,
the third-order correction
to the energy, and so on,
but in practice, the above
results is ordinarily as high
as it is useful to pursue this
method.
Results
Correction to the eigenvalue:
2
′ +∑
En = E + H nn
0
m
0
n
Ψ H′ Ψ
0
n
m ≠n
0
n
E −E
0
m
+L
(1)
Correction to the eigenfunction:
Ψn = Ψ + ∑
0
n
m≠ n
Ψm0 H ′ Ψn0
0
n
E −E
0
m
0
m
Ψ +L
(2)
Degenerate Perturbation Theory
• Two-ford degeneracy
• Higher-order degeneracy
Non-degenerate perturbation theory fail?
• In many cases, where two (or more) distinct
states share the same energy, non-degenerate
perturbation theory fail !
• Sometimes, two energy levels exists so near
that non-degenerate perturbation theory
cannot give us a satisfied answer.
So, we must look for some other way to
handle the problem.
Two-ford degeneracy
In order to see how the method generalizes,
we begin with the twofold degeneracy
Suppose ,
0
0
a
0
0
a
0
0
b
0
0
b
H Ψ =E Ψ ,
E0
H Ψ =E Ψ ,
λ
“Lifting of a degeneracy by a
paerturbation
0
a
0
b
Ψ Ψ
=0 .
The perturbation will
“break” the degeneracy
Any linear combination of the above states
0
0
a
0
b
Ψ = αΨ + β Ψ
is still an eigenstate of H0,with the same eigenvalue E0
0
0
0
H Ψ =E Ψ
Essential problem:
0
When we turn off the
perturbation, the “upper” state reduces down to one
0
0
linear combination of Ψa and Ψb , and the “lower”
state reduce to some other linear combination, but
we don’t know a priori what these good linear
combination will be. For this reason we can’t even
calculate the first-order energy (equation 1) because
we don’t know what unperturbed states to use.
As before, we write H, E, ψ in the following form
H = H
0
E = E
0
Ψ = Ψ
+ λH ′ ,
+ λE
0
1
+ λΨ
2
+ λ E
1
2
2
+ λ Ψ
+ L ,
2
+ L
Plug into the stationary equation
To lowest order, H 0 Ψn0 = En0 Ψn0
To first order, H 0 Ψ1 + H ′Ψ 0 = E 0 Ψ1 + E 1Ψ 0
0
1
0
0 1
1 0
′
H Ψ +HΨ =E Ψ +E Ψ
Take the inner
product wi th Ψa0
Ψa0 H 0 Ψ1 + Ψa0 H ′Ψ 0 = E 0 Ψa0 Ψ1 + E 1 Ψa0 Ψ 0
0
0
a
0
0
a
0
0
b
0
0
b
H Ψ =E Ψ ,
H Ψ =E Ψ ,
0
a
0
b
Ψ Ψ
0
=0 ,
0
a
0
b
Ψ = αΨ + β Ψ
0
a
0
a
Meanwhile, H 0 is
Hermitian, the first
term on the left
cancels the first
term on the right
0
a
0
b
α Ψ H′ Ψ + β Ψ H′ Ψ
= αE
1
Results
αWaa + βWab = αE
0
i
1
Where, Wij ≡ Ψ H ′ Ψ
0
j
The “matrix
elements” of H'
, (i, j = a, b)
0
b
Similarly, the inner product with Ψ yields
αWba + βWbb = βE
1
Eliminate βWab
α [WabWba − ( E −Waa )( E −Wbb )] = 0
1
1
If α is not zero, we can get the following equation
1 2
1
( E ) − E (Waa + Wbb ) + (WaaWbb − WabWba ) = 0
1
2
2
E = Waa + Wbb ± (Waa − Wbb ) + 4 Wab

2 
1
±
The two roots
correspond to the
two perturbed
energies
(3)
But what if
α is zero?
How should we do in practice
Idea: It would be greatly to our advantage if
we could somehow guess the “good” states
right from the start.
M
O
R
A
L
1. Look around for some Hermitian operator
A that commutes with H' ;
2. Pick as your unperturbed states ones that
are simultaneously eigenfunctions of H0
and A ;
3. Use ordinary first-order perturbation theory.
If you can’t find such an operator, you’ll have to resort to
Equation 3, but in practice this is seldom necessary
Theorem
Proof
Let A be a Hermitian operator
Byassumption, [A, H′] = 0, so
that commutes with H ′.If Ψa0
and Ψb0 are eigenfunctions of A
0
′
Ψ [A, H ]Ψb = 0
0
a
with distinct eigenvalues,
= Ψa0 AH′Ψb0 − Ψa0 H′AΨb0
AΨa0 = µΨa0 , AΨb0 = νΨb0 , µ ≠ ν ,
= AΨa0 H′Ψb0 − Ψa0 H′νΨb0
Then Wab = 0 (and hence Ψa0 and
Ψb0 are the " good" states to use in
perturbation theory).
= (µ −ν ) Ψa0 H′Ψb0 = (µ −ν )Wab.
But µ ≠ν , soWab = 0. QED
Higher-order degeneracy
NEXT
PAGE
Higher-order degeneracy
0
E =E
0
1
0
n
，Ψ = ∑akΨ ,
0
Matrix equation
0
nk
0
0
1
1
H Ψ + H ′Ψ = E Ψ + E Ψ
0
1
′
(
H
−
E
∑ k′k δ k′k )ak = 0
k
Secular equation
1
det H k′′k − E δ k ′k = 0
Get E'
Get ak ,
then
get ψ0
Diagonalizes the perturbation H′
If you can think of an operator A that commutes with H',
and use the simultaneous eigenfunctions of A and H0, then
the W matrix will automatically be diagonal, and you
won’t have to fuss with solving the characteristic equation.
Example
Some Examples For Application
• The fine structure of hydrogen
• The zeeman effect
• Hyperfine splitting
The Fine Structure of Hydrogen
• The relativistic correction
• Spin-orbit coupling
The simplest
Hamiltonian of
the hydrogen
h2 2
e2 1
∇ −
H=−
2m
4πε 0 r
Replace m by the
reduced mass
Correct for the motion
of the nucleus
Relativistic correction
Spin-orbit coupling
Quantization of the Coulomb field
The magnetic interaction
between the dipole moments of
the electron and the proton
Fine structure
Lamb shift
Hyperfine structure
Hierarchy of correction to the Bohr
energies of hydrogen
Bohr energies:
Fine structure:
Lamb shift :
Hyperfine splitting:
2
2
of order
4
2
α mc
of order
5
2
α mc
of order
m  4 2
 m α mc
p

of order
α mc
An application of time-independent
perturbation theory
The relativistic correction
General discussion
2
Classical formula
2
e 1
h
2
H=−
∇ −
2m
4πε 0 r
Represent
kinetic energy
T
=
2 m
The relativistic formula
T=
The total
relativistic
energy
p
2
mc
2
( c)
1− υ
2
− mc
The rest energy
2
Express T in terms of the relativistic momentum
The relativistic momentum：p =
mυ
( c)
1− υ
( )
2
2


υ
m υ c + m c 1−
c 

2 2
2 4
2
p c +m c =
=
T
+
mc
2
υ
1−
c
2
2 2
2 4
( )
(
)
2
The nonrelativistic
T=
p 2 c 2 + m 2 c 4 − mc 2
limit p << mc
2
4
2
4


1
p
1
p
p
p




T = mc 2 1 + 
− 3 2 +L
 − 
 L − 1 =
 2  mc  8  mc 
 2m 8m c
2
4
p
p
T=
− 3 2 +L
2m 8m c
The lowest-order
relativistic contribution to
the Hamiltonian
The classical result
In first-order perturbation theory, the correction to En is given
by the expectation value of H' in the unperturbed state
1
1
4
2
2
E = − 3 2 ψ pˆ ψ = − 3 2 pˆ ψ pˆ ψ
8m c
8m c
1
r
 pˆ 2

And, 
+ V ψ = Eψ
 2m

ˆp 2ψ = 2m(E − V )ψ
[
1
2
2
2
E =−
E
−
E
V
+
V
2
2mc
1
r
]
In the case of hydrogen
2
2
2






1
e
1
e
1
2





Er1 = −
E
+ 
n + 2 En 
2
2
2mc 
4πε 0  r  4πε 0  r 



Where En is the Bohr energy of the state in question
And,
1
1
= 2 ,
r
na
1
1
=
2
r
l + 1 n3a 2
2
(
)
2
2
2






1
e
1
e
1
2






Er1 = −
E
+
2
E
+
n
n
 4πε  n 2 a  4πε  l + 1 n 3a 2 
2mc 2 
0 
0 


2


(
)
4πε 0 h
a≡
2
me
 m  e2 2  1
  2
En = −  2 
 2h  4πε 0   n
2
P l u g g i n g
2
2
2






1
e
1
e
1
1
2






+
+
Er = −
E
2
E
n
n
 4πε  n 2 a  4πε  l + 1 n3a 2 
2mc 2 
0 
0 


2


(


E
4n
1


Er = −
−
3
2
2mc  l + 1

2 

2
n
)
Spin-Orbit Coupling
The orbiting positive charge
sets up a magnetic field B
in the electron frame, which
exerts a torque on the
spinning electron, tending
to align its magnetic
mo m e n t ( µ ) a l o n g t h e
d i r e cti on o f t h e fi e l d .
Hydrogen atom, from the
electron’s perspective.
H = −μ⋅Β
The Magnetic Field of the Proton
According to the Biot - Savart law : B =
e
An effective current I =
T
µ0 I
2r
The charge of proton
The period of the orbit
2πmr 2
The orbital angular momentum of the electron : L = rmυ =
T
Additional, B and L point in the same direction
Β=
1
e
2
4 πε 0 mc r
3
L
The Magnetic Dipole Moment of the Electron
Classical
electrodynamics
2
A ring of charge, rotating about its axis
qπr
µ=
T
2
2πmr
S=
T
 q 
μ= 
S
 2m 
However, as it turns
out, the electron’s
magnetic moment is
twice the classical
answer
Classical
result
?
e
μe = − S
m
The “extra” factor of 2 was explained by Dirac in his
relativistic theory of the electron. For a related
discussion, see V.Namias,Am.J.Phys.,57,171(1989).
e
Β=
L
2 3
4πε 0 mc r
The Spin-Obit Interaction
1
 e2  1
 2 2 3 S ⋅ L
H = 
 4πε 0  m c r
e
μe = −
S
m
FR We did the analysis in the rest frame of
A the electron, but that’s not an inertial
U
system——it
accelerates
as
the
electron
D
orbits around the nucleus.
S
O
L
U
T
I
O
N
Make an appropriate kinematic correction,
known as the Thomas precession.
The Spin-Obit Interaction
Exactly
The modified
gyromagnetic ratio cancel one
another
for the electron
the Thomas
precession
factor
D I S R E G A R D
Safely, on the basis of a naive classical model
 e  1
′ = 
 2 2 3 S ⋅ L
H SO
8
πε
m
c
r
0 

2
′ , L] ≠ 0
[H SO
′ , S] ≠ 0
[H SO
The spin and orbital
angular momenta
are not conserved
Put it another way
The total angular momentum J ≡ L + S
[H ′
SO
2
2
2
]
,L = 0
[H ′
SO
2
]
,S = 0
2
L , S , J , J z are conserved, and the eigenstates of these
quantities are good states to use in perturbation theory.
J ≡L+S
1
L ⋅S = (J − L
2
2
2
2
−S
2
Meanwhile,
)
 e2  1
′ = 
 2 2 3 S ⋅ L
H SO
 8πε 0  m c r
1
1
=
3
r
l l + 1 (l + 1)n 3 a 3
2
h2
[ j ( j + 1) − l (l + 1) − s (s + 1)]
2
E
1
SO
 e  1
 2 2
= 
 8πε 0  m c
Express in
terms of En
2
J = L + S + 2L ⋅ S
The eigenvalue s of L ⋅ S :
2
2
h
2
1
ESO
(
)
[
j ( j + 1) − l (l + 1) − 3 ]
2
4
(
)(l + 1)n a
2
 n[ j ( j + 1) − l (l + 1) − 3 ]

4
ll+ 1
En2
=
2 
mc 

3
(
ll+ 1
3
2
)(l + 1)


Fine structure
The relativistic correction
2
 4n

E
n

E r1 = −
− 3
2
2 mc  l + 1

2


The spin-orbit coupling
+
The correction to
the Bohr formula
The complete finestructure formula
1
ESO
En2
=
mc 2
[
 n j ( j + 1) − l (l + 1) − 3

4

l l + 1 (l + 1)

2
(
)


E
4
n
3 −

E f1s =
2 mc 2 
j+ 1 
2

2
n
Combine
with the
Bohr
formula


13.6eV  α 2  n
3 
Enj = −
1+ 2
−
2

1
 n j+
n
4 
2



•Fine structure breaks the degeneracy in
l
•The energies are determined by n and
j
]


The Fine Structure of Hydrogen
Energy levels of hydrogen, including fine structure (not to scale)
The Zeeman Effect
• Weak-field zeeman effect
• Strong-field zeeman effect
• Intermediate-field zeeman effect
General discussion
Consider an atom placed in a
uniform external magnetic field Bext
What
&
How
For a single electron
H Z′ = −(μl +μs ) ⋅Βext
Associated with
orbital motion
Associated with
electron spin
e
(L + 2S ) ⋅Βext
H Z′ =
2m
New phenomenon
Zeeman Effect
The energy levels are shift when an atom is
placed in a uniform external magnetic field
H
Depend on the strength of the external
field in comparison with the internal field
that gives rise to spin-orbit coupling
O
Weak-field Intermediate Strong-field
W
Bext<<Bint
Comparable
Bext<<Bint
Weak-field Zeeman Effect
Bext<<Bint
Fine structure
dominates
Weak-field Zeeman splitting of the
ground state; the upper line (mj=1/2)
has slope 1, the lower line (mi=-1/2 )
has slope -1
HZ as a small
perturbation
′ , L] ≠ 0
[H SO
′ , S] ≠ 0
[H SO
The spin and orbital
angular momenta
are not conserved
Put it another way
The total angular momentum J ≡ L + S
[
]
2
′
H SO , L = 0
2
2
[
]
2
′
H SO , S = 0
2
L , S , J , J z are conserved, and the eigenstates of these
quantities are good states to use in perturbation theory.
n, l , j and m j are the " good" quantum numbers
The zeeman correction to the energy
E
1
Z
= nljm
j
H Z′ nljm
j
e
=
Β ext ⋅ L + 2 S
2m
The expectation
value of S
?
In the presence of spin-orbit coupling,
L and S are not separately conserved;
they precess about the fixed total
angular momentum, J.
S ave =
(S ⋅ J ) J
J
2
L= J −S
Meanwhile,
S ave
(
S ⋅ J)
=
J
J
2
h2
1 2
2
2
S ⋅ J = J + S − L = [ j ( j + 1) + s(s + 1) − l (l + 1)]
2
2
(
)
 (S ⋅ J ) 
L + 2S = 1 + 2 J
J 

 j ( j + 1) − l (l + 1) + 3 
4 J
= 1 +
2 j ( j + 1)




Choose the z-axis to lie along Bext
1
Z
E = µ B g J Bext m j
Where, µ B
The
landé
g-factor,
gJ
Bohr
magneton
eh
≡
= 5 . 788 × 10
2m
−5
eV
Τ
Strong-field Zeeman Effect
From csep10.phys.utk.edu/.../light/zeeman-split.html
Strong-field Zeeman Effect
Bext>>Bint
The zeeman
effect dominates
The splitting of spectrum
Fine structure
as the
perturbation
Strong-field Zeeman Effect
n, l , ml and ms are now the " good" quantum numbers
Choose the z-axis to lie along Bext
The zeeman Hamiltonian
H Z′ =
Proof!
The " unperturbe d" energies
13 .6 eV
E nm l m s = −
+ µ B Bext (ml + 2 m s )
2
n
e
Bext (Lz + 2S z )
2m
The fine structure correction:
E 1fs = nlm l m s H r′ + H so′ nlm l m s
H'r : The same as before
S ⋅ J = S x Lx + S y Ly + S z Lz = h 2 ml ms

 l (l + 1) − m m  
13
.
6
3
eV

2
l s

−
E 1fs =
α


1
n3
4
n
l l +
(l + 1)  

2

(
)
What
if l =0?
Intermediate-field Zeeman Effect
Neither H'Z nor
H'fs dominates
Treat them on
an equal footing
As perturbation
to Bohr
Hamiltonian
H ′ = H Z′ + H fs′
Intermediate-field Zeeman Effect
My
choice
The case : n = 2； The basis : the
states characteri zed by l , j , and m j
 Ψ1 ≡
l = 0
 Ψ 2 ≡
Using the ClebschGordan coefficients
Ψ3

Ψ4

Ψ5

l = 1
Ψ6

Ψ7

Ψ
 8
≡
3
2
3
2
≡
3
2
−3
2
≡
3
2
1
2
=
≡
1
2
1
2
= −
≡
3
2
−1
2
=
≡
1
2
−1
2
= −
1
2
= 11
1
2
2
3
1
−1
2
10
1
2
1
3
2
10
3
1 −1
2 2
,
1 −1
2 2
= 00
,
,
1
2
1
2
1 −1
3
1 1
2 2
= 00
,
1
2
= 1 −1
1 1
2 2
1 −1
1
2
1
2
1
+
2
+
1
2
1
2
2
+
1
2
3
+
3
3
1
−1
2
1
2
11
11
3
−1
2
1
2
1 0
10
,
1
2
−1
2
1
2
,
,
−1
2
.
Matrix of H'
 5γ −β

 0

 0
 0

 0
 0

 0
 0

Where
0 

5γ +β
0
0
0
0
0
0 

0
0
0
0
0
0 
γ −2β
0
0
γ −2β
0
0
0
0 

2
2
0
0
0
γ −3 β 3 β
0
0 
2
1

β
γ
β
0
0
0
5
−
0
0
3
3

2
2
0
0
0
0
0
γ +3 β 3 β 
2
1 
0
0
0
0
0
β
5
γ
+
3
3 β
0
,
0
0
0
( ) 13 . 6 eV
γ ≡ α 8
2
0
and
0
β ≡ µ B B ext .
Energy levels for the n=2 states of hydrogen,
with fine structure and zeeman splitting
ε1
ε2
ε3
ε4
= E 2 − 5γ + β
= E 2 − 5γ − β
= E2 − γ + 2β
= E2 − γ − 2β
ε 5 = E 2 − 3γ + β 2 +
4γ
2
+ (2 3 )γβ + β
2
4
ε 6 = E 2 − 3γ + β 2 −
4γ
2
+ (2 3 )γβ + β
2
4
ε 7 = E 2 − 3γ − β 2 +
4γ
2
− (2 3 )γβ + β
2
4
ε 8 = E 2 − 3γ − β 2 −
4γ
2
− (2 3 )γβ + β
2
4
Hyperfine splitting
Brief introduction to proton
Made by Arpad Horvath.
commons.wikimedia.org/wiki/
Image:Quark_struct...
The quark structure of
the proton. There are
two up quark in it and
one down quark. The
strong force is
mediated by gluons
(wavey). The strong
force has three types
of charges, the so
called red, green and
the blue. cut is a
douche bagg
Brief introduction to proton
magnetic
field itself constitutes a magnetic dipole,
The proton
smaller than the electron’s
ge
S p ; g = 5 .59 ( measured value ）
μp =
2m p
According to classical electrodynamics, a dipole
µ sets up a magnetic field.
µ0
2µ0
3
[3(μ⋅ r )r -μ] + μδ (r )
Β=
3
4πr
3
For a related discussion, see D.J.Griffiths,Am.J.Phys.,50,698(1982).
The electron near proton
The Hamiltonian of the electron,due to
the proton’s magnetic dipole moment
H = −μ⋅Β
µ ge [3(S ⋅ r )(S ⋅ r ) − S
=
2
H hf′
0
8πm p me
p
e
r3
p ⋅ Se
] + µ ge
2
0
3m p me
S p ⋅ S eδ 3 (r )
According to perturbation theory, we can
get the first-order correction to the energy
2
2
(
)
(
)
3
S
⋅
r
S
⋅
r
−
S
⋅
S
µ
ge
µ
ge
2
p
e
p
e
1
0
0
Ehf =
+
S p ⋅ S e Ψ (0 )
3
8πm p me
r
3m p me
For the ground state hydrogen
The wave function is
spherically symmetrical
3(S p ⋅ r )(S e ⋅ r ) − S p ⋅ S e
r
Meanwhile , Ψ100 (0 )
2
3
1
= 3
πa
2
E
1
hf
=0
Spin-spin
coupling
µ 0 ge
=
S p ⋅Se
3
3π m p m e a
Spin-spin coupling
The individual spin angular momenta are no longer conserved;
the “good” states are eigenvectors of the total spin.
1 2
S p ⋅S e = S − S e2 − S p2
2
(
S ≡ Se + S p
The electron and proton
both have spin 1/2 .
In the
triplet
In the
singlet
( )
)
2
3
S =S =
h
4
2
e
state,
state,
2
p
S
2
= 2 h
S
2
= 0 .
+ 1 , ( triplet);
4 gh
 4
1
Ehf =
2 2 4 
3m p me c a − 3 , (singlet).
 4
2
2
Effect of spin-spin coupling
Hyperfine
splitting in
the ground
state of
hydrogen
•Lifting the triplet configuration
•Depressing the singlet configuration
The energy gap as a result of spin-spin coupling
The frequency of the
4 gh 2
−6
∆E =
=
5
.
88
×
10
eV photon emitted in a
2 2 4
3m p me c a
transition from the
∆E
ν=
= 1420ΜΗ z
h
triplet to the singlet
state
Corresponding wavelength : c = 21cm
ν
Fall
in
the
microwave
region
This famous “21-centimeter line” is among
the most pervasive and ubiquitous forms of
radiation in the universe.
21centimeter
line