5.73 Lecture #24
24 - 1
J Matrices
Last time:
starting with [ J i , J j ] = ih Σ ε ijk J k
DEFINITION !
k
J 2 jm = h 2 j( j + 1) jm
J z jm = hm jm
J ± = J x ± iJ y
J ± jm = h[ j( j + 1) − m(m ± 1)]
1/ 2
jm ± 1
nonzero matrix elements and “Condon Shortley” phase choice
j′ m′ J 2 jm = h 2 j( j + 1)δ j′ jδ m ′m
j′ m′ J z jm = hmδ j′ jδ m ′m
j′ m′ J ± jm = h[ j( j + 1) − mm′ ] δ j′ jδ m ′m ±1
1/ 2
(J , J , J , J , J , J ) all stay within j
2
z
x
y
+
−
all matrix elements of J 2 , J z, J x , J ± are real and positive (only those of J y are imaginary)
TODAY:
1. What do the matrices look like for J = 0, 1 / 2, 1?
2. many operators are expressed as an angular momentum times a
constant — Zeeman example — density matrix
r
3. other operators involve things like q or products of
two angular momenta
Stark effect
Wigner-Eckart Theorem
* classify operators by commutation rule
* matrix elements in convenient basis sets
* transform between inconvenient and
convenient basis sets.
revised November 15, 2001
5.73 Lecture #24
24 - 2
[p , p ] = 0
x
y
A student in 1999 suggested that he could find f(x,y) such that
∂2 f
∂2 f
≠
∂x∂y ∂y∂x
Thus [p x , p y ] ≠ 0!
This is possible, but f(x,y) would have to have a form that excludes it as an
acceptable ψ(x,y). Typically, the f(x,y) will have to be discontinuous or have
discontinuous first derivatives. For all well behaved V(x,y), ψ(x,y) will have
continuous first derivatives. The f(x,y) used to prove a commutation rule must
be acceptable as a quantum mechanical wavefunction, ψ(x,y). This is a good
thing because (see Angular Momentum Handout)
e − ia p x
e − ia p x
h
h
x1 = x1 + a
generates a linear translation of + a in x direction.
linear translations commute (but rotations do not)
This is the basis for (or a consequence of )
[p , p ] = 0
[J , J ] = ih Σ ε
i
j
i
j
k
ijk
Jk
revised November 15, 2001
5.73 Lecture #24
24 - 3
Nonlecture
prepare (excite)
evolve
detect
e.g.
excite:
E
e − iH t
h
D
basis set 0 , 1 , 2
0
Ε 0 = 1

1
1
0
0
The “excitation matrix” E
creates equal amplitudes in
two excited eigenstates:
1   1   0
0  0 =  1
   
0  0  1
2
1
ρ(0) = E 0 0 E†
evolve:
0
If we are in the eigenbasis of H
e − iH t
h
− iE t
 a  ae a
 b =  be − iE b t
   − iE c t
 c  ce
h
h
h




(translation in time)
but otherwise, need [Te − iT
ρ(t ) = Te − iT
†
HT t
h
†
HTt / h
T† ] = U(t, 0)
T†E 0 0 E†Te + iT
= U(t, 0)ρ(0)U† (t, 0)
detect:
D
†
HT t
h
T†
ρ(0) in eigenbasis
of H
the “detection matrix”
D t = Trace(ρD)
Building Blocks
revised November 15, 2001
5.73 Lecture #24
24 - 4
r
Many QM operators have the form f ( J)
e.g. Zeeman effect
r
Others have the form f (q)
e.g. Stark effect
r
H Zeeman = − γB ⋅ J
(B is magnetic field)
r
r r
HStark = eE ⋅ q
(E is electric field)
r
f ( J1 , J 2 )
Others have the form of

e.g. spin - orbit

H SO = aL ⋅ S
We are going to want to be able to write matrix representations of
these operators.
Let us begin by writing matrices for J2, Jz, Jx, Jy, J+, J– .
jm = 00
only basis state is
1 × 1 matrix
j=0

J 2 00 = 






0
same for all components
11
1 1
and −
22
2 2
j = 1/2
J
2 (1 / 2 )
3 1
= h2 
4 0
0

1
1 1
h
2 0
0

−1
J(z1/ 2 ) =
(1 / 2 )
J+
0
= h
0
1

0
2 × 2 matrices
J2
11
11  1 1
= h2
+1
22
22  2 2
m = +1/2
e.g.
 0 1  1  0

  =  
 0 0  0  0
J(+1/ 2 )
11
=0
22
 0 1  0  1

  =  
 0 0  1  0
J(+1/ 2 )
1 1
11
−
=h
2 2
22
m = –1/2
revised November 15, 2001
5.73 Lecture #24
0
J(−1/ 2 ) = h
1
24 - 5
0

0
J(x1/ 2 ) =
1
1 0
=
J
J
+
( + − ) h 1
2
2 
J(y1/ 2 ) =
1
i 0
=
−
J
J
h
−
( + −)
2i
2  −1
verify that J = J + J + J
2
2
x
0
J = 
4 1
1  0

0  1
0
J = 
4 i
− i  0

0  i
2
x
2
y
h
h
2
2
2
y
1

0
1 h  0
= 
0 2  i
 1 0


4  0 1
(J(1/ 2) ) 2 = 3h
2
z
1 h 2  1
= 
0 4  0
− i h 2  1
= 
0  4 0
− i

0
2
0

1
0

1
An amazing amount of insight gained from this complete set of 2 × 2 matrices
 1 0
I=

 0 1

0
σ x = 
1


0
σ y = 
i


σ =  1
 z  0

CTDL, pages 417-454
1. Pauli Matrices
2. Diagonalization of 2 × 2
1
(1/ 2)
 → Jx
0
−i
(1/ 2)
 → Jy
0
0
(1/ 2)
 → Jz
−1
3. Geometric interpretation of 2 × 2
ρ in terms of fictitious spin 1/2
4. spin 1/2 ρ
5. magnetic resonance
σ x, σ y ] = ?
What is [σ
3 matrices with eigenvalues ±1
revised November 15, 2001
5.73 Lecture #24
 m11
arbitrary M = 
 m21
24 - 6
m12  m11 + m22
m − m22
m + m21
m − m21
I + 11
σ z + 12
σ x + i 12
σy
=
m22 
2
2
2
2
r r
M = a 0I + a ⋅ σ
scalar
part of
M
vector
part of
M
1
Tr (M)
2
r 1
a = Tr(Mσ )
2
1
a x = Tr (Mσ x )
2
1
a y = Tr (Mσ y )
2
1
a z = Tr (Mσ z )
2
a0 =
Center of Gravity
M↔ρ
Information in 2 ↔2 ρ is
repackaged into a 3
component vector.
Visualization of dynamics!
This provides a basis for taking apart the dynamics of an arbitrary 2 × 2 ρ into
dynamics of x, y, z fictitious spin-1/2 components. Beat the S = 1/2 Zeeman
problem to death and use it as basis for understanding dynamics of any 2 × 2 space.
revised November 15, 2001
5.73 Lecture #24
24 - 7
A set of 3 × 3 matrices
J=1
J2
(1)
1
= 2 h2  0

0
0
0

1
0
1
0
J (z1)
1
= h 0

0
J (+1)
0
= 21/ 2 h 0

0
1
0
0
0
1

0
(1)
−
0
= 2 h 1

0
0
0
1
0
0

0
J
0
0
0
1/ 2
0
0

−1
J (x1)
0
= 2 −1/ 2 h 1

0
1
0
1
J (y1)
0
= 2 −1/ 2 h i

0
−i
0
i
 0 1 0  1  0
 0 0 1  0 =  0

   
 0 0 0  0  0
J(+1) 11 = 0
0
1

0
0
− i

0
For 2 × 2 problem (e.g. J = 1/2), needed 4 independent 2 × 2 matrices (because there
are 4 elements in a 2 × 2 matrix) to represent arbitrary 2 × 2 matrix.
for 3 × 3 problem, need 9 independent 3 × 3 matrices
(x, y,z,
x 2 , y 2 , z 2 , xy, xz, yz) (because there are 9 elements in a 3 × 3 matrix)
actually scalar (I), vector, tensor
s+p+d
[for 2 × 2 it was s + p = 4].
Can you write out each of the J(3/2) matrices (16 4 ↔4 matrices)?
revised November 15, 2001
5.73 Lecture #24
24 - 8
9 3 × 3 basis matrices is not nearly so nice as the 4 basis matrices for 2 × 2 problem.
But this turns out to be what is needed to “understand” and picture spin = 1 systems.
similarly for j = 3/2, 2, etc.
There are 2 lovely consequences of being able to take an arbitrary matrix and
rewrite it as sum of J matrices.
1. If M is the matrix of an operator – a term in the Hamiltonian – then it is clear
that this operator may be re-expressed as a sum of operators, each of which
behaves exactly like a (combination of) component(s) of J – evaluated in the
jm⟩ basis set.
M ( j ) = a0 I + ∑ ali J i J j +
i
∑
c3ijk J i J j J k
i, j , k
+…
basis for classification of operators into Tm( k )
and Wigner– Eckart Theorem for evaluation of matrix elements.
r
2. especially for 2 level systems, if M = ρ and a is defined from M as on page 24-6,
then we have a vector picture to understand preparation, evolution, detection
z
z
y
y
r
a
π/2 pulse
evolution of vector, fictitious B-fields
z
y
π pulse
revised November 15, 2001
5.73 Lecture #24
24 - 9
Now let’s do some J = 1 examples
Zeeman effect for an
q
l
= 1, (p orbital) state
r r r
L=q×p
r
r r
µ L ∝ −q × p
r
r
µ L ∝ −L
e-
( up out of page)
page)
(down into
r
r
µ L ≡ −γL
field
strength
current on a
circular wire
r
r
ˆ
E = B ⋅ µ L = Bzk ⋅ (− γL) = − γBzL z
classical energy
field exclusively along z
for L = 1 system:
H Zeeman
1
= −γBz h 0

0
0
0

−1
0
0
0
case (1) Let Ψ(0) = LML = 11
 1
ρ = Ψ Ψ =  0 (1
 
 0
0
1
0) =  0

0
0
0
0
0
0

0
ELM L = E11 = Trace(ρH)
 1
= − hγBz Tr  0

 0
= − hγBz
0
0
0
0  1
0  0

0  0
0
0
0
0 
0 

−1 
revised November 15, 2001
 0 0 0
ρ =  0 1 0


 0 0 0
E10 = Trace (ρρH) = 0
5.73 Lecture #24
What about?
E1−1 = + γBz h
24 - 10
no motion of E
( ) −1/ 2
case (2) Let Ψ 0 = 2 ( 11 + 10 )
Ψ (t ) = 2 −1/ 2 [ 11 e − iE t + 10 e − i 0 t
11
h
h
]
1
( 11 11 + 10 10 + 11 10 e − iE11 t + 10 11 e + iE11 t
2
 1
e − iω11 t 0
1
1
0
ρ(t ) =  e iω11 t

2 
0
0
0


ρ(t ) =
h
h
)
nothing at location of coherence in ρ
1 0 0 
H(B || +z) = − γBz h 0 0 0 


 0 0 −1
1
E (t ) = H = Trace(Hρ) = − γBz h(1)
2
no motion of E
Looked at 2 cases:
1. pure state
11 , B || z
2. mixed state
1
E = − γBz h
2
E = − γBz h
2 -1/2 ( 11 + 10 ), B || z
always mixed state gives time independent ⟨E⟩
NMR: oscillating Bx, By, cw Bz
revised November 15, 2001
5.73 Lecture #24
24 - 11
Stark Effect: Electric field
r r
r
classical
E ∝ ε ⋅ (qe − − q p + ) ≈ ε z z
so we will need matrix elements of x, y, z in jm basis set. How?
Based on
[ z, L ] = −ih Σ ε
j
k
zjk
qk
vector operator definition — later
Other angular momenta
1. l electron orbital ang. mom.
2. s electron spin
3. I nuclear spin
These separate angular momenta interact with each other
spin - orbit: ζ (r )λ ⋅ s
Zeeman:
- γBz (L z + g sSz + g II z )
hyperfine:
a I•S
coupled and uncoupled basis sets:
m sms ↔ jlsm j
l
l
revised November 15, 2001
5.73 Lecture #24
case (3)
24 - 12
same Ψ(0) = 2 −1/ 2 ( 11 + 10
)
r
but H is for B || x
one at coherence in ρ
 0 1 0
H = − γBx h2 −1/ 2  1 0 1
 0 1 0
Something subtle is intentionally wrong here. Can you find it?
1
E(t ) = Tr(Hρ) = − γBx h[e + iω 11t + e − iω 11t + 0]
2
= −γBx h cos ω11t
E (t )
− γBx h
t
revised November 15, 2001