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SJP QM 3220 3D 1
Angular Momentum (warm-up for H-atom)
Classically, angular momentum defined as (for a 1-particle system)
  
Lrp
y
xˆ
yˆ
zˆ
 x
y
z
px
py
pz
m


p  mv

r
x
O

Note: L defined w.r.t. an origin of coords.


L  xˆ ( yp z  zp y )  yˆ ( zp x  xpz )  z ( xpy  yp z )

 
, etc.
(In QM, the operator corresponding to Lx is Lx  yˆ pˆ z  zˆ pˆ y , p z 
i z



 
 dL
Classically, torque defined as   r  F , and  
(rotational version of F  ma )
dt

  
If the force is radial (central force), then   r  F  0  L  const .

H-atom:
electron


r
2
ˆF  - r̂ ke
r2

(Coulomb force)
proton at origin
In a multi-particle system, total average momentum:

Ltot   Lˆi is conserved for system isolated from external torques.
i
sum over particles
Internal torques can cause exchange of average momentum among particles, but

Ltot remains constant.
In classical and quantum mechanics, only 4 things are conserved:
 energy
 linear momentum
 angular momentum
 electric charge
Page H-1
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
̂
Back to QM. Define vector operator L
operator
unit vector
ˆ
L  Lˆ x xˆ  Lˆ y yˆ  Lˆ z zˆ
Recall

d Q

dt
d Lˆ

dt

Ĥ , Qˆ 
i

Hˆ , Lˆ 


i

d  Ly 
d  Lx 
xˆ 
yˆ  
dt
dt
Claim: for a central force such as in H-atom
V  V ( r )   ke
2
r
 
, then Ĥ , Lˆ  0
(will show this later)

dL
 0 (just like in classical mechanics)
This implies
dt
Angular momentum of electron is H-atom is constant, so long as it does not absorb
or emit photon. Throughout present discussion, we ignore interaction of H-atom
w/photons.
Will show that for H-atom or for any atom, molecule, solid – any collection of atoms

– the angular momentum is quantized in units of ħ. | L | can only change by integer
number of ħ's.
Units of L  L    
 
Note L   rp ,  p     (since p  k )
r 
 L  r      
r 


Claim: Lˆ x , Lˆ y  iLˆ z
and
Lˆ , Lˆ   iLˆ
i
Page H-2
j
k
(i, j, k cyclic:
x y z or
y z x or
z x y
)
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
To prove, need two very useful identities:

 
A  B, C   A, C   B, C 
AB, C   AB, C   A, C B

Proof: Lx , Ly  yp z  zp y , zp x  xpz 

yp z , zp x    yp z , xpz   zp y , zp x  zp y , xpz  
 

 

 


0
y p z , z  px


0
x
z
,
p
z py


i
i
all other terms
like [y, px] = 0
 i( xpy  yp x )  iLz


p , p   0,
(Have used x, p x   i, x, y   0, x, p y  0,
x
y
etc.
I'm dropping the ˆ over operators when no danger of confusion.
Since [Lx, Ly ] ≠ 0, cannot have simultaneous eigenstates of Lˆ x and Lˆ y .


 L2  L2  ( 1 Lˆ x , Lˆ y
2i 
x
x
i Lˆ z
2
)
2

   Lz
2
2
 
However, L2  L  L  L2x  L2y  L2z does commute with Lz.


Claim: L2 , Lz  0
L , L   0
2
i

 
 
, i = x, y, or z
 

Proof: L2 , Lz  Lx , Lz  Ly , Lz  Lz , Lz



0
2
2

 
2

 Lx Lx , Lz   Lx , Lz Lx  L y Ly , Lz  L y , Lz L y
 
 
 iL y
 iLy
iLx
 iLx
= 0 (Note cancellations)
[L2, Lz] = 0 => can have simultaneous eigenstates of Lˆ2 , Lˆ z (or Lˆ2 , Lˆi any i)
Page H-3
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Looking forward to H-atom:

Hˆ    2    V ( r )   
2m
Hˆ , Lˆ  0 , Hˆ , Lˆ  0
We will show that
2
z
=> simultaneous eigenstates of Hˆ , Lˆ2 , Lˆ z
energy q-nbr
 
n l m
Lz q-nbr
L2 q-nbr
When we solve the TISE Ĥ ψ = E ψ for the H-atom, the natural coordinates to use
will be spherical coordinates: r, θ, φ (not x, y, z)
z
r̂
θ
ˆ
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
̂
y
φ
x
2
2
2
in spherical coordinates is gawd-awful. But


x 2 y 2 z 2
separation of variables will give special solutions, energy eigenstates, of form
Just rewriting  2 
 ( r , , )  R( r ) Y ( , )  R( r ) ( ) ( )
The angular part of the solution Y (θ, φ) will turn out to be eigenstates of L2, Lz and
will have form completely independent of the potential V( r ).
*
Page H-4
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Given only [L2, Lz] = 0 and Lˆ2 , Lˆ z hermitean we know there must exist
simultaneous eigenstates f (which will turn out to be the Y (θ, φ) mentioned above)
such that
Lˆ2 f    f , Lˆ z f    f
(λ will be related to l, and μ will be related to m)
We will show that f will depend on quantum-numbers l, m, so we write it as flm, and
that
L2 f l m   2l (l  1)  f l m

Lz f l m  m  f l m
where l  0, 1
2
m  l ,  l  1 ,  l  1 , l
,1,3 , 
2
f l m  Yl m ( ,  ) will be determined later.
Notice max eigenvalue of Lz (= lħ ) is smaller than square root of eigenvalue of
L2   l l  1
So, in QM, Lz < | L |
Odd!
Also notice l = 0, m = 0 state has zero angular momentum (L2 = 0, Lz = 0) so, unlike
Bohr model, can have electron in state that is "just sitting there" rather than
revolving about proton in H-atom.
Proof of boxed  formulae: (This proof takes 2 ½ pages!)
Define L+ = Lx + i Ly = "raising operator"
L- = Lx - i Ly = "lowering operator"
(Note L+† = L- , L-† = L+ , A† = hermitean adjoint of A)
Neither L+ or L- are hermitean (self-adjoint).


 2

2
2
Note L , L  0  L , L  L , Lx  i L , L y  0 




0
0


2
=> Consider f: L f    f , Lz f    f

2
Page H-5


 



M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Claim: g = L+ f is an eigenfunction of Lz with eigenvalue = (μ + ħ). So L+ operator
raises eigenvalue of Lz by 1 ħ.
L2 g  L2 ( L f )  L ( L2 f )    L f    g
Proof:

To prove Lz g = (μ + ħ) g, need to show that [Lz, L+] = ħ L+
L , L
z
x



 iL y  Lz , Lx   i L z , Ly  Lx  iL y  




iLy
iLx
L
Now Lz g  Lz ( L f )  L Lz f  L f      L f


μf
L Lz  L
   g 
So, operating on f with raising operator L+ raises eigenvalues of LZ by 1ħ but keeps
eigenvalue of L2 unchanged.
(Similarly, L- lowers eigenvalue of Lz by 1ħ.)
Operating repeatedly with L+ raises eigenvalue of Lz by ħ each time: L+ (L+ f ) has
(μ + 2ħ) etc.
But eigenvalue of Lz cannot increase without limit since Lz cannot exceed
L2
L2  L2x  L2y  L2z     2 ,





0
2
0
 
There is only one way out. There must be for a given λ a "top state" ft for which
L+ ft = 0.
Likewise, there must be for a given λ a "bottom state" fb for which L- fb = 0.
Lz
ft
LL+
L-
all with same λ =
eigenvalue of L2
L+
ħ
fb
Page H-6
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Write Lz f = m ħ ∙ f , m changes by integers only
Lz ft = ℓ ħ ∙ f t , ℓ = max value of m
L2 ft = ? Want to write L2 in terms of L+, Lz:


L L  ( Lx  iL y )( Lx  iL y )  L2x  L2y  i Lx , Ly
 
iLz
L2  L2z



Lz
=> L2  L L  L2z  Lz
(Also, L2  L L  L2z  Lz )
=> L2 f t  L L f t  L2z f t  Lz f t   2 (  1) f t



22 f t
0
2f t
So, L2 f   2 (  1) f where ℓ = max m, same λ for all m's.



Repeat for f b : Lz f b    f b ,
  min value of m.
L2 f b  L L f b  L2z f b  Lz f b   2 (   1) f b






0
 2 2 f b
2 f b
    (  1)  (   1)     (try it!)
So mmin = - mmax and m changes only in units of 1.
=> m = -ℓ, -ℓ+1, . . . ℓ-2, ℓ-1, ℓ
N integer steps
=> 2 ℓ = N , ℓ = N / 2 =>
ℓ = 0, 1/2, 1, 3/2, 2, 5/2, . . .
End of proof of 
Page H-7
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
m
2
0
1
1
0
0
-1
-1
-2
ℓ
0
2
1
3
We'll see later that there are 2 flavors of angular momentum:

1. Orbital
Ang. Mom.
(integer ℓ only)
Page H-8

2. Spin
Ang. Mom.
(integer or ½ integer OK)
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
The H- atom
mp >> me =>
proton (nearly)
stationary
me


r
mp ≈ 1840 me


pˆ 2
 V (r )
Hamiltonian of electron  Hˆ 
2m
 ke
 kZe2
, k 1
( or V ( r ) 
)
40
r
r

 
pˆ 2 pˆ 2  pˆ 2   2 2


  
2m
2m
2m
V (r ) 
2
TISE: Hˆ   En   special solutions (stationary states).
n
n
 ( x)  ( x, t )   ( x)e
n
n
 iEnt
n

General Solution to TDSE: ( x, t )   cn e
 iEnt

 ( x)
n
n
Spherical Coordinate System:
z
r̂
ˆ
θ
̂
z = r cos θ
x = r sin θ cos φ
y = r sin θ sin φ
r
φ
ψ = ψ (r, θ, φ)
volume
Normalization: dV  2  1



0
dr


0
d

2
0
d r 2 sin    1
2
Need  2 in spherical coordinates
Hard Way:  2 f    f 
Page H-9
2 f

x 2
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
f f r f  f 



g
x r x  x  x
 2 f g g r


  (nightmare !)
x 2 x r x
r r

Also need 9 derivatives:
,


x y
x
Easier Way: Curvilinear coordinates (See Boas)
path element:

ds  xˆdx  yˆdy  zˆdz
 rˆdr  ˆ r d  ˆ r sin d
 eˆ1h1 dx1  eˆ2 h2 dx2  eˆ3h3 dx3
3
  êi hi dxi (h i  "scale factor" )
i 1
f   eˆi
i 1
2 f 
1
h1h2 h3
1 f
hi xi
   h2 h3 f    h1h3 f 


 

  

 x1  h1 x1  x2  h2 x2 

Spherical coordinates:
xi   r ,  , 
hi   1, r , r sin 
 2 f 
 2 f 
1   2 f 
1
 
f 
1


r

sin






r 2 r  r  r 2 sin   
  r 2 sin 2    2 
1
 (radial)
 2 (angular)
r
*
In Classical Mechanics (CM), KE = p2 /2m = KE =
(radial motion KE) + (angular, axial motion KE)
v

v
KE 

r
vr


Lˆ  r  mv  mrv
( v   L
p r2
1 2 m 2
mv  (vr  v2 ) 

2
2
2m

radial
O
Page H-10
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
mr
)
L2
2
2mr


1
 angular
2
r
Fall 2008
SJP QM 3220 3D 1
*
Same splitting in QM:
1  
 
1 2 
 

L̂2   r      2 
s



 2
2
i

 s      s   
2
(Notice L̂2 depends only on θ, φ and not r.)
 2
Hˆ 
   V ( r )  E 
2
2m
  2 1   2  
Lˆ2
r

  V ( r )  E


2m r 2 r  r 
2mr 2
Separation of Variables! (as usual)
Seek special solution of form:
 ( r , ,  )  R( r )  Y ( ,  )  R( r )  ( )  ( )
Normalization: ∫ dV | ψ |2 =


2
2
0 dr r R  0 d 0 d sin Y  1
 

1
1
2
2
(Convention: normalize radial, angular parts individually)
Plug ψ = R ∙ Y into TISE =>
  2 Y d  2 dR 
R ˆ2
L Y  V  R Y  E  R Y
r

2
2m r dr  dr  2mr 2
 2mr 2 1
Multiply thru by
:
2 R Y
 1 d  2 dR  2mr 2

1 ˆ2
LY
r
  2 (V  E ) 

2
dr 


Y
Rdr




g(

,

)
f (r )
=> f(r) = g (θ, φ) = constant C = ℓ(ℓ + 1)
Lˆ2Y   2 C  Y   2 (  1)Y
Page H-11
(Page H - 5)
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Have separated TISE into radial part f( r ) = ℓ(ℓ + 1), involving V( r ), and angular
part g (θ, φ) = ℓ(ℓ + 1) which is independent of V( r ).
=> All problems with spherically symmetric potential (V = V( r )) have exactly same
angular part of solution: Y = Y(θ, φ) called "spherical harmonics".
We'll look at angular part later. Now, let's examine


2
Radial SE:   
 R 
2mr 

  2 d  2 dR 
  2  rR
r

r

R
(
V

E
)

(  1)


2mr dr  dr 
2mr 2
Change of variable: u (r) = r ∙ R(r)


  dr u 2  1


0

Can show that
1 d  2 dR  d 2u
:
r
 
r dr  dr  dr 2
d 2u dR dR
d 2R



r
dr 2 dr dr
dr 2
du
dR
 Rr
,
dr
dr
same!
1 d  2 dR  1  dR 2 d R 
dR
d R
2
r
r 2
r
   2r
2 
r dr  dr  r  dr
dr
dr 
dr
2
2

  2 d 2u
 2 (  1) 

V


u  Eu
2
2m dr 2
2
m
r


Notice: identical to 1D TISE:
  2 d 2
 V   E  except
2m dr 2
r: 0 -> ∞ instead of x: - ∞ -> + ∞ and
V(x) replaced with
Veff  V ( r ) 
2
(  1)
2mr 2
Veff = "effective potential"
Boundary conditions:
Page H-12
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
u ( r = ∞ ) = 0 from normalization ∫ dr | u |2 = 1
u
u ( r = 0 ) = 0, otherwise R  blows up at r=0 (subtle!)
r
A
A B
V ( r )   , Veff    2
r
r r
Veff
2
ℓ=1
ℓ=2
r
E
ℓ=0
Notice that energy
eigenvalues given by
solution to radial
equation alone.
Seek bound
state
solutions E < 0
E > 0 solutions are
unbound states,
scattering solutions
Full solution of radial SE is very messy, even though it is effectively a 1D problem
(different problem for each ℓ )
Power series solution (see text for details). Solutions depend on 2 quantum
numbers: n and ℓ (for each effective potential ℓ = 0, 1, 2, … have a set of solutions
labeled by index n.)
Solutions: n = 1, 2, 3, …
ℓ = 0, 1, … (n - 1)
for given n
ℓmax = (n – 1)
n = "principal quantum number"
energy eigenvalues depend on n only (it turns out)
En 
E1
m( ke2 )2
,
E


(independent of ℓ)
1
n2
2 2
• same as Bohr model, agrees with experiment!
Page H-13
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
First few solutions: Rnℓ (r)
normalization
R10  A10e
r
a0
, a0  
"Bohr radius"
2
me
2

40 2
me2

r   r 2 a0
e
R20  A20 1 
 2a0 
 r  r
R21  A21  e 2 a0
 a0 
NOTE:
• for ℓ = 0 (s states), R (r = 0) ≠ 0 => wavefunction ψ "touches" nucleus.
• for ℓ ≠ 0 , R (r = 0) = 0 => ψ does not touch nucleus.
ℓ ≠ 0 => electron has angular momentum. Same as classical behavior, particle with
  
non-zero L cannot pass thru origin ( L  r  p : r  0  p   )
Can also see this in QM: for ℓ ≠ 0, Veff has infinite barrier at origin = > u(r) must
decay to zero at r=0 exponentially.
Veff
r
Page H-14
=> exponential decay in
u( r )
R( r ) 
as well.
r
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Back to angular equation: Lˆ2Yl m   2 (  1)Ym Want to solve for the Ym ' s "spherical harmonics".
Before, started with commutation relations,
Lˆ , Lˆ   iLˆ , L , L   0
2
i
j
k
i
and, using operator algebra, solved for the eigenvalues of L2, Lz. We found
L2Ym   2(  1)Ym
L Y  mY
m
z 
m

where ℓ = 0, ½, 1, 3/2, …
m = - ℓ , - ℓ +1 … + ℓ
In the process, we defined raising and lowering operators:
Lˆ   Lˆ x  Lˆ y
Lˆ Ym  cmYm1
Lˆ Ym  cmYm1
(for m  m max  )
(for m  m min  )
(cm is some constant)
Lˆ  f top  Lˆ Y  0 and Lˆ Y  0
So, if we can find (for a given ℓ) a single eigenstate Ym , then we can generate all the
others (other m's) by repeated application of Lˆ or Lˆ .


Ym  Ym ( ,  ) .
It's easy to find the φ-dependence; don't need the L̂ business yet.
h 
Lˆ z 
(showed in HW)
i 
h Y
Lˆ zY 
 hmY (and you can cancel the h)
i 
Assume Y ( ,  )  ( ) ( ) 

d
 im  
d
 ( )  e  im
If we assume (postulate) that ψ is single-valued than
 (  2m)   ( )  eim2  1
=> m = 0, ± 1, ± 2, … But m = - ℓ, … + ℓ
So for orbital angular momentum, ℓ must be integer only: ℓ = 0, 1, 2, … (throw out
½ integer values)
Page H-15
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
*
(algebra!)
L  Lx  iL y

L  Lx  iL y

( L  Lt
 
 

e i 
 i cot 
 
 
 
 

e i  
 i cot 
 
 
f At g  Af g )
adjoint
Can deduce Ym from L̂Y  0
=>
Y
Y 
 i cot    0


Solution: (un-normalized)
Y ( , )  (sin )  e i
Checks: Plug back in.
(sin )1 cos ei  i cot  (sin ) (i)ei  0
(sin ) 
cos
 cot  (sin )     0 
sin



cot
Now, can get other Ym ' s by repeated application of L̂ . Somewhat messy (HW!)
2
Normalization from  d  d sin  Ym
Y00  const  1
Notice case ℓ = 0
(since

2
0
0
:
4
 d  d sin   d  4 )
Example:
Y11  
3
sine i
8
Y10 
3
cos
4
Y11  
3
sine i
8
Convention on ± sign: Y m  ( 1) m Ym 
*
Page H-16
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
The spherical harmonics form a complete, orthonormal set (since eigenfunctions of
hermitean operators)
 d (Y
) Ym' '   '  mm'
m *

Any function of angles f = f (θ, φ) can be written as linear combo of Ym ' s :
f ( , ) 

 0
Likewise:


0

 Y
m 
m

dr r 2 ( Rn )* Rn''   nn' '
=> H-atom energy eigenstates are
 nm ( r , , )  Rn ( r )Ym ( , )  Rn  m e im
n = 1, 2, … ; ℓ = 0, 1 … (n-1) ; m = -ℓ … + ℓ
Arbitrary (bound) state is
 

cnm  nm
(c's are any complex constants)
n , , m
energy of state (n, ℓ, m) depends only on n.
En = - constant/n2 (states ℓ, m with same n are degenerate)
ℓ=
0
n=
4
3
2
1
Page H-17
4
s
3
s
2
s
1
s
1
(1)
4p
3p
2
(3)
4d
3
(5)
4f
(7)
3d
2p
Degeneracy of nth level is
n2
(2•n2 if you include spin)
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Radial Probability Density
 dV 
2
1
 2

Prob (find particle in dV about r ) =  ( r ) dV
If ℓ = 0 , ψ = ψ(r) then  dV    dr 4r 2  (r )
2
Prob (find in r → r + dr) = P( r )dr  4r
2
2
 ( r ) dr
P(r) = radial probability density
Ground state:  100  Ae
P(r)  | A |2 4r 2 e
2 r
r
2
P(r)
a0
a0
Notice P(r) very different from ψ(r):
r
ao
ψ
r
If ℓ ≠ 0 , ψ = ψ (r, θ, φ) = R(r) Y(θ, φ), then
 dV 
2

r | R |  dΩ Y  1
dr
 


2
"solid angle"
2
2
1
1
Prob (find in r → r + dr) = r2 |R|2 dr
P(r) = r2 |R|2
Note:
even if ℓ ≠ 0
   (r )  R  Y  R 
1
4
so P( r )  r 2 R  4 r 2 
2
 R  4 
2
2
if
2
H-atom and emission/absorption of radiation:
If H-atom is in excited state (n = 2, ℓ = 1, m = 0) then it is in energy eigenstate =
stationary state. If atom is isolated, then atom should remain in state ψ210 forever,
since stationary state has simple time dependence:

(r , t )   210 (r )  e eE2t 
Page H-18
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
But, experimentally, we find that H-atom emits photon and de-excites: ψ210 -> ψ100
in ≈ 10-7 s -> 10-9 s
E
2p
Eγ = hf = ∆E
1s
The reason that the atom does not remain in stationary state is that it is not truly
isolated. The atom feels a fluctuating EM field due to "vacuum fluctuations".
Quantum Electrodynamics is a relativistic theory of the QM interaction of matter
and light. It predicts that the "vacuum" is not "empty" or "nothing" as previously
supposed, but is instead a seething foam of virtual photons and other particles.
These vacuum fluctuations interact with the electron in the H-atom and slightly alter
the potential V(r). So eigenstates of the coulomb potential are not eigenstates of the
actual potential: Vcoulomb + Vvacuum
Photons possess an intrinsic angular momentum (spin) of 1 ħ, meaning

  1  L   (  1)  2
and Lzmax  
So when an atom absorbs or emits a single photon, its angular momentum must
change by 1 ħ, by Conservation of Angular Momentum, so the orbital angular
momentum quantum number ℓ must change by 1.
"Selection Rule": ∆ℓ = ± 1 in any process involving emission or absorption of 1
photon => allowed transitions are:
s
p
d
5
4
3
2
n=1
If an H-atom is in state 2s (n = 2, ℓ = 0) then it cannot de-excite to ground state by
emission of a photon. (since this would violate the selection rule). It can only lose
its energy (de-excite) by collision with another atom or via a rare 2-photon process.
Page H-19
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Matrix Formulation of QM
  cn
n


1
 
0
u1   
0
 
M
complete orthonormal set
c1 
 
c 2 
c 
n ,
  c n    3 
M
 
c n 
 
 M
0
 
1
u2   
0
 
M
 c1 
1
 0
 
 
 
 c2 
 0
1
   cn u n     c1    c2    
c
0
0
n
 3
 
 



 
 
 
If ket's are represented by column vectors, then bra's are represented by the
transpose conjugate of column = row, complex conjugate.
   cn un     cn* un
n
n
  c1* c2* c3* 
 c1 
 
2
2
c 
*
*
*
   c1 c2 c3   2   c1  c2   
c
 3

 
Operators can be represented by matrices:


c
2
n
n
no hat on matrix element
 A11 A12 


Aˆ  Amn   m Aˆ n   A21 A22

 



where {| n >} is some complete orthonormal set.

Page H-20

M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Why is that? Where does that matrix come from?
Consider the operator  and 2 state vectors  ,  related by
  Â 
(  )
In basis {| n >} ,
  cn n

n
 n 1n23
n
   dn n

n
cn
 n 1n23
n
dn
Now project equation  onto | m > by acting with bra:

m

m Aˆ 

c
n
m Aˆ n
n
dn   Amn c n
n


But, this is simply the rule for multiplication of matrix  column.
d1 
A11 A12
K  c1 
 

  
d2   A21 A22
 c 2 
d3 

  
 

  
 M
 M
  M
So there you have it, that's why the operator is defined as this matrix, in this basis!
Now, suppose Aˆ  Ĥ and  n
Hˆ n  En n ,
Ĥ 2
E
 1

 E2 2  



0
 are energy eigenstates, then
Hˆ mn  m Hˆ n  En mn
0 
E2
E3




0
 0
 
 
1
1

E
2
0
 0
 
 


 
 
A matrix operator m Aˆ n is diagonal when represented in the basis of its own
eigenstates, and the diagonal elements are the eigenvalues.
Notice that in general operators don't commute Aˆ Bˆ  Bˆ Aˆ . Same goes for Matrix
Multiplication:
≠
Page H-21
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Claim: The matrix of a hermitian operator is equal to its transpose conjugate:
*
Aˆ hermitian  Amn  Anm
Proof:
m Aˆ n  Aˆ m n  n Aˆ m
*
*
 Amn  Anm


t
*
Similarly, adjoint (or "Hermitian conjugate") Aˆ t : Amn
 Anm
Proof:
*
Aˆ m n  m Aˆ t n  n Aˆ m
Of course, it's difficult to do calculations if the matrices and columns are infinite
dimensional. But there are Hilbert subspaces that are finite dimensional. For
instance, in the H-atom, the full space of bound states is spanned by the full set {n, ℓ,
m} (= | nℓm>). The sub-set {n=2, ℓ=1, m = +1, 0, -1} forms a vector space called a
subspace.
Subspace? In ordinary Euclidean space, any plane is a subspace of the full volume.

If we consider just the xy components of a vector Rxy  xˆRx  yˆ Ry , then we have a
perfectly valid 2D vector space, even though the "true" vector is 3D.
Likewise, in Hilbert space, we can restrict our attention to a subspace spanned by a
small number of basis states.
Example: H-atom subspace {n=2, ℓ=1, m = +1, 0, -1}
Basis states are

m
  1 ,
0 , 1 (can drop n=2, ℓ=1 in label since they are
fixed.)
Lˆ z m  h m m
(l  1)
Lˆ m  h l (l  1) m
2

 Lz mn
2

2h2 m (for all m)
1 0 0 


 m Lˆ z n  h 0 0 0 


0 0 1

Page H-22
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
1 0 0


L mn  m Lˆ n  2h 0 1 0


0 0 1
(What about Lx? Ly?)
2

2
Before seeing what all this matrix stuff is good for, let's examine spin because it's
very important physically and because it will lead to 2D Hilbert space with simple 2
x 2 matrices.
Review of Dirac Bra - Ket Notation
bracket or inner-product:
f g   dx f * ( x ) g ( x ) or
 dV f
*
 
( r ) g ( r ) or dΩ ( ,) 
Which integral you do depends on the configuration space of problem.
Key defining properties of bracket:

<f | g >* = < g | f >
c = constant

<f | c • g > = c < f | g > , < c • f | g > = c* < f | g >

<α | ( b| β > + c | γ >) = b < α | β > + c < α | γ >
Dirac proclaims: < g | f > = < g | next to | f >
bracket = "bra" and "ket"
Ket | f > represents vector in H-space (Hilbert Space)
"ket"
"wavefunction"

 is to  ( x) as R is to Rx , Ry , Rz 
Both ψ and ψ(x) describe same state, but | ψ > is more general:
g x0   ( x  x0 )
 ( x0 )  x0 
fp 
1
e ipx 
( p )  p 
cn   
Page H-23
Different
"representations"
of same
H=space vector
|ψ>
Ĥ u n  En u n
un  
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
-representation, momentum-rep, energy-rep.
Page H-24
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
What is a "bra"? < g | is a new kind of mathematical object, called a "functional".
g _ 
 dx g
*
( x) _
insert state function here
input
output
function:
number
number
operator:
function
function
functional: function
numbe
< g | wants to bind with | f > to produce inner product < g | f >
For every ket | f > there is a corresponding bra < f |. Like the kets, the bra's form a
vector space.

| c f > → < c f | = c* < f |

(?)
| α f + β g > → < α f + β g | = α* < f | + β * < g |

< α f + β g | h > = α* < f | h > + β * < g | h >
Complex number  bra = another bra
any linear combo of bra's = another bra

=> bra's form
vector space
The vector space of bras is called a "dual space". It's the dual of the ket vector space.
Aˆ f  Aˆ f is a ket. What is the corresponding bra? Aˆ f  Aˆ f  f At
Def'n of A†
ˆ ( Aˆ t  "A - dagger" )
Definition: hermitean conjugate or adjoint At of operator A
Aˆ f g  f At g
for all f, g.
ˆ  At , then Aˆ is hermitean or self-adjoint.)
(If A
Some properties:


Aˆ Bˆ   Bˆ Aˆ

Aˆ   Aˆ
t
t
t
t
t

Proof :

 g Aˆ t f

t

f Aˆ t g  Aˆ t f g 

*
*
 Aˆ g f  f Aˆ f 

 


Page H-25
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
The adjoint of an operator is analogous to complex conjugate of a complex number:
c **  c, Aˆ tt  Aˆ
c *  c  c real, Aˆ t  Aˆ  Aˆ hermitean.
The "ket-bra" | f > < g | is an operator. It turns a ket (function) into another ket
(function):

f g

h  f
gh
Projection Operators

Hˆ un ( x)  Enun ( x)  Hˆ n  En n
 (x)   c n un (x)   un  un (x) 
n
n
  cn n   n  n   n n 
n

n
n
n
n  1̂ =>
n
Pˆn  n n
"Completeness relation"
(discrete spectrum case)
= "projection operator"
P̂n picks out portion of vector | ψ > that lies along | n >
Pˆn   n

n   cn n
u2 = |2>
|ψ
>
n n is like xˆ ( xˆ  _ )

xˆ xˆ  R  xˆ Rx
 
|2> <2|ψ>
u1 = |1>
u1<u1|ψ> = |1> <1|ψ>
Page H-26
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
r
r r
r
   n n  like R  xˆ x  R  yˆ yˆ  R
n
ˆ   n n like 1  xˆ xr  _
1
 yˆ yˆ  _ 
n



Anywhere there is a vertical bar in the bracket, or a ket or a bra, we can replace the
bar with 1   n n
n
Example:    1
=>

n n   1   cn*cn   cn  1
2
n
n
n
If eigenvalue spectrum is continuous (as for xˆ or pˆ ) then must use integral, rather
than sum, over states.
 dx
Example:
x x  1̂
Completeness Relation
(continuous spectrum)
( p)  f p  
 dx
fp x x  
1
dx eipx h(x)

2h
The Measurement Postulates 3 and 4 can be restated in terms of the projection
operator:

Starting with state    cn n   n n  ,
n
n
where sum {n} is over any complete set of states, if we measure observable
associated with n, then we will find value n0 with probability
2
P( n0 )  cn   n0 n0    Pˆn0 
P( n0 )  Pˆn0
  Pˆn0 
Probability of finding eigenvalue n0 = expectation value of projection operator P̂n0 .
And as result of measurement state | ψ > collapses to state n0  Pˆn0  .
(apart from normalization)
Page H-27
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
We can now generalize to case of states described by more than one eigenvalue,
such as H-atom.
   cnm nm     nm nm 
nm
nm

If we measure energy (but not also L , Lz ), find n0, then we are projecting onto
subspace spanned by {ℓ, m } with some n0.
Pˆn0   n0 m n0 m
  0, 1  n0  1
m
P( n0 )   Pˆn0    cn0m
2
m    
m
State collapses to Pˆn0    n0 m n0 m 
m
must renormalize
Page H-28
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Spin ½
Recall that in the H-atom solution, we showed that the fact that the wavefunction
(r) is single-valued requires that the angular momentum quantum # be integer: l
= 0, 1, 2.. However, operator algebra allowed solutions l = 0, 1/2, 1, 3/2, 2…
Experiment shows that the electron possesses an intrinsic angular momentum
called spin with l = ½. By convention, we use the letter s instead of l for the spin
angular momentum quantum number : s = ½.
The existence of spin is not derivable from non-relativistic QM. It is not a form of
orbital angular momentum; it cannot be derived from L  r  p .
(The electron is a point particle with radius r = 0.)
Electrons, protons, neutrons, and quarks all possess spin s = ½. Electrons and
quarks are elementary point particles (as far as we can tell) and have no internal
structure. However, protons and neutrons are made of 3 quarks each. The 3 halfspins of the quarks add to produce a total spin of ½ for the composite particle (in a
sense,  makes a single ). Photons have spin 1, mesons have spin 0, the deltaparticle has spin 3/2. The graviton has spin 2. (Gravitons have not been detected
experimentally, so this last statement is a theoretical prediction.)
Page H-29
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Spin and Magnetic Moment
We can detect and measure spin experimentally because the spin of a

charged particle is always associated with a magnetic moment.
Classically, a magnetic moment is defined as a vector  associated with
i
r
a loop of current. The direction of  is perpendicular to the plane of
i
the current loop (right-hand-rule), and the magnitude is
m, q
r
  iA  ir .
2
The connection between orbital angular momentum (not spin) and magnetic
moment can be seen in the following classical model: Consider a particle with mass
m, charge q in circular orbit of radius r, speed v, period T.
i
q
,
T
v
2r
T

i
 qv 
2
  iA  
 r
2

r



qv
2r
| angular momentum | = L = p r = m v r , so v r = L/m , and  

qvr
2
qvr
q

L.
2
2m
So for a classical system, the magnetic moment is proportional to the orbital
angular momentum:
 
q
L
2m
(orbital)
.
The same relation holds in a quantum system.
In a magnetic field B, the energy of a magnetic moment is given by
E    B   z B (assuming B  B zˆ ).
In QM, Lz  m .
Writing electron mass as me (to avoid confusion with the magnetic quantum number
m) and q = –e we have  z  
Page H-30
e
m , where m =  l .. +l. The quantity
2 me
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
B 
e
is called the Bohr magneton. The possible energies of the magnetic
2 me
moment in B  B zˆ is given by Eorb   z B   B Bm .
For spin angular momentum, it is found experimentally that the associated magnetic
moment is twice as big as for the orbital case:
 
q
S
m
(spin)
(We use S instead of L when referring to spin angular momentum.)
This can be written  z  
e
m   2 B m .
me
The energy of a spin in a field is E spin   2 B B m (m = 1/2) a fact which has been
verified experimentally.
The existence of spin (s = ½) and the strange factor of 2 in the gyromagnetic ratio
(ratio of  to S ) was first deduced from spectrographic evidence by Goudsmit and
Uhlenbeck in 1925.
Another, even more direct way to experimentally determine spin is with a SternGerlach device, next page
Page H-31
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
(This page from QM notes of Prof. Roger Tobin, Physics Dept, Tufts U.)
Stern-Gerlach Experiment (W. Gerlach & O. Stern, Z. Physik 9, 349-252 (1922).
B
z
y
x
r r r r
r r r
F    B  ( ) B (in current free regions),


r
B
or here, F  zˆ (z z ) (this is a little
z
crude - see Griffiths Example 4.4 for a better treatment, but this gives the main idea)

Deflection of atoms in z-direction is proportional to z-component of magnetic
moment z, which in turn is proportional to Lz. The fact that there are two beams is
proof that l = s = ½. The two beams correspond to m = +1/2 and m = –1/2. If l = 1,
then there would be three beams, corresponding to m = –1, 0, 1. The separation of
the beams is a direct measure of z, which provides proof that z   2 B m
The extra factor of 2 in the expression for the magnetic moment of the electron is
often called the "g-factor" and the magnetic moment is often written as
z   g  B m . As mentioned before, this cannot be deduced from non-relativistic
QM; it is known from experiment and is inserted "by hand" into the theory.
Page H-32
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
However, a relativistic version of QM due to Dirac (1928, the "Dirac Equation")
predicts the existence of spin (s = ½) and furthermore the theory predicts the value
g = 2. A later, better version of relativistic QM, called Quantum Electrodynamics
(QED) predicts that g is a little larger than 2. The g-factor has been carefully
measured with fantastic precision and the latest experiments give g =
2.0023193043718(76 in the last two places). Computing g in QED requires
computation of ab infinite series of terms that involve progressively more messy
integrals, that can only be solved with approximate numerical methods. The
computed value of g is not known quite as precisely as experiment, nevertheless the
agreement is good to about 12 places. QED is one of our most well-verified
theories.
Spin Math
Recall that the angular momentum commutation relations
[L2 , Lz ]  0 ,
[Li , L j ]  i Lk
(i j k cyclic)
were derived from the definition of the orbital angular momentum operator:
L  r  p.
The spin operator S does not exist in Euclidean space (it doesn't have a position or
momentum vector associated with it), so we cannot derive its commutation
relations in a similar way. Instead we boldly postulate that the same commutation
relations hold for spin angular momentum:
[S2 ,Sz ]  0 ,
Page H-33
[Si , Sj ]  i Sk . From these, we derive, just a before, that
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
S2 s ms 
2
s (s  1) s ms
Sz s ms 
ms s ms

1
2
 
3
4
2
s ms
s ms
( since s = ½ )
( since ms = s ,+s = 1/2, +1/2 )
Notation: since s = ½ always, we can drop this quantum number, and specify the
eigenstates of L2 , Lz by giving only the ms quantum number. There are various ways
 12 ,  12
to write this: s ms  ms

 , 
 , 
These states exist in a 2D subset of the full Hilbert Space called spin space. Since
these two states are eigenstates of a hermitian operator, they form a complete
orthonormal set (within their part of Hilbert space) and any, arbitrary state in spin
a
space can always be written as   a   b    
b
(Griffiths' notation is
  a   b  )
Matrix notation: 
1
  ,
0
 0
    . Note that       1 ,
1
 0
If we were working in the full Hilbert Space of, say, the H-atom problem, then our
basis states would be n
m ms . Spin is another degree of freedom, so that the
full specification of a basis state requires 4 quantum numbers. (More on the
connection between spin and space parts of the state later.)
Page H-34
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
[Note on language: throughout this section I will use the symbol Sz (and Sx , etc) to
refer to both the observable ("the measured value of Sz is  / 2 ") and its associated
operator ("the eigenvalue of Sz is  / 2 ").]
The matrix form of S2 and Sz in the m (z) basis can be worked out element by
ˆ A  mA
ˆ n .)
element. (Recall that for any operator A,
mn
3
4
 S2  
3
4
S2 
2
 S2 
,
1 0


0 1
2
Sz 
 0 , etc.
 Sz   
1
,
2
 Sz 
 0 , etc.
1 1 0 


2  0 1
Operator equations can be written in matrix form, for instance,
Sz 
 
2

 1 0  1 
1

     
2  0 1 0 
2  0

We are going ask what happens when we make measurements of Sz , as well as Sx
and Sy , (using a Stern-Gerlach apparatus). Will need to know: What are the
matrices for the operators Sx and Sy ? These are derived from the raising and
lowering operators:
S  Sx  iSy
S  Sx  iSy

S  S 
 2i1 S  S 
Sx 
Sy
1
2
To get the matrix forms of S+ , S , we need a result from the homework:
S s, ms

s (s  1)  m(m  1) s, ms  1
S s, ms

s (s  1)  m(m  1) s, ms  1
Page H-35
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
For the case s = ½, the square root factors are always 1 or 0. For instance, s = ½,
m = 1/2 gives s(s  1)  m(m  1) 
S 

 ,
 S 
 0,
Sx 
1
2

Sy 
Often written: S 
 , S 
 0 , leading to
Notice that S+ , S are not hermitian.
 S  and Sy 
0 1


2 1 0

 1 . Consequently,
, etc. and
 0 0
S  

1 0
S
 23     12  12 
 0 and S 
 S 
0 1
S  

0 0
Using Sx 
S 
1
2
1
2i
S
 S  yields
 0 i 


2i 0 
These are hermitian, of course.
0 1
 0 i
1 0 
 , where x  
 , y  
 , z  
 are
2
1
0

i
0
0

1






called the Pauli spin matrices.
a
Now let's make some measurements on the state   a   b     .
b
Normalization:
  1 
a  b  1.
2
2
a
Suppose we measure Sz on a system in some state     .
b
Postulate 2 says that the possible results of this measurement are one of the Sz
eigenvalues:  / 2 or  / 2 . Postulate 3 says the probability of finding, say  / 2 ,
is Prob(find  /2) =
Page H-36
| 
2
a
  0 1  
b
2
 b .
2
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Postulate 4 says that, as a result of this measurement, which found  / 2 , the initial
state  collapses to  .
But suppose we measure Sx ? (Which we can do by rotating the SG apparatus.)
What will we find? Answer: one of the eigenvalues of Sx, which we show below are
the same as the eigenvalues of Sz :  / 2 or  / 2 . (Not surprising, since there is
nothing special about the z-axis.) What is the probability that we find, say, Sx =
 / 2 ? To answer this we need to know the eigenstates of the Sx operator. Let's
call these (so far unknown) eigenstates (x ) and (x )
(Griffiths calls them
(x ) and (x ) ). How do we find these? We must solve the eigenvalue equation:
Sx     , where are the unknown eigenvalues. In matrix form this is,
 0

 /2
/ 2  a 
a
 
      which can be rewritten 
0  b 
b
 /2
/ 2  a 
   0 . In
  b 
linear algebra, this last equation is called the characteristic equation.
This system of linear equations only has a solution if
 
Det 
 /2
/ 2

 
 
/2
/2

 0 . So  2   / 2   0
2

   /2
As expected, the eigenvalues of Sx are the same as those of Sz (or Sy).
Now we can plug in each eigenvalue and solve for the eigenstates:
 0 1  a 
a

      a  b ;
2  1 0  b 
2 b
So we have (x) 
Page H-37
1  1
 
2  1
and
 0 1  a 
a

       a   b .
2  1 0  b 
2 b
(x) 
1 1
 
2  1
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
1
Now back to our question: Suppose the system in the state (z)    , and we
0
measure Sx. What is the probability that we find, say, Sx =  / 2 ? Postulate 3 gives
the recipe for the answer:
Prob(find Sx   /2) =
 |
(x)
(z)
2

1
2
1
1 1  
0
2

1
2
2
 1/ 2
a
Question for the student: Suppose the initial state is an arbitrary state    
b
and we measure Sx. What are the probabilities that we find Sx =  / 2 and  / 2 ?
Page H-38
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
SJP QM 3220 3D 1
Let's review the strangeness of Quantum Mechanics.
Suppose an electron is in the Sx =  / 2 eigenstate (x ) 
1
2
 1
  . If we ask: What
 1
is the value of Sx? Then there is a definite answer:  / 2 . But if we ask: What is the
value of Sz , then this is no answer. The system does not possess a value of Sz. If we
measure Sz, then the act of measurement will produce a definite result and will force
the state of the system to collapse into an eigenstate of Sz, but that very act of
measurement will destroy the definiteness of the value of Sx. The system can be in
an eigenstate of either Sx or Sz, but not both.
Page H-39
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008