7. Angular Momentum
7A. Angular Momentum Commutation
Why it doesn’t commute
• The order in which you rotate things makes a difference, 12  21
• We can use this to work out commutation relations for the L’s
– It can be done more easily directly
• Recall: R  R  nˆ ,    exp   i nˆ  L 
• Also recall:
R  R1  R  R 2   R  R1R 2 
• We will calculate the following to second order in :
R  R  xˆ ,   R  R  yˆ ,   R  R  xˆ ,    R  R  yˆ ,   
 R  R  xˆ ,  R  yˆ ,  R  xˆ ,   R  yˆ ,   
• If rotations commuted, both sides would be the identity relation
Calculating the Left Side
R  R  xˆ ,   R  R  yˆ ,   R  R  xˆ ,    R  R  yˆ ,   
• To second order in :
R  R  xˆ ,   R  R  yˆ ,    exp  i  Lx  exp  i  Ly 
 1  i  Lx
 12  2 L2x
 1  i   Lx  Ly 
2
1  i  Ly
 12  2 L2y
 12  2  L2x  2 Lx Ly  L2y 
2

2
• Second half is same thing with   –
R  R  xˆ ,   R  R  yˆ ,   R  R  xˆ ,    R  R  yˆ ,   

1  i   L  L 
 1  i   Lx  Ly 
x
 1
2
y
 Lx  Ly 
2

L  
 12  2  L2x  2 Lx Ly  L2y 
2
 12  2  L2x  2 Lx Ly
2
2
2
y
  2  L2x  2Lx Ly  L2y 
2
 1   2  Lx , Ly 
R  R  xˆ ,   R  R  yˆ ,   R  R  xˆ ,    R  R  yˆ ,     1   2  Lx , Ly 
2
2
 O  3 
Calculating the Right Side
R  R  xˆ ,  R  yˆ ,  R  xˆ ,   R  yˆ ,   
• To second order in :
0
1

R  xˆ ,    0 cos 
 0 sin 

1  12  2

R  yˆ ,     0
 

0
0 
 1

   0 1 1  2


 sin   
2


1  12  2 
cos    0
0
0
 

1
0 
0 1  12  2 
1  12  2
0
 


R  xˆ ,   R  yˆ ,      2
1  12  2
 
2
 

1




• Second half is same thing with   –
1  12  2
0
 1  12  2
0
 
 R  xˆ ,  R  yˆ , 
 


2
2
1 2
1 2
1 2 
  
1 2 
 

 
R  xˆ ,   R  yˆ ,     
2 
2

1





1





Matching the Two Sides
• To second order in :
1

R  xˆ ,   R  yˆ ,   R  xˆ ,   R  yˆ ,      2
0

 2
1
0
0

0  = R  zˆ , 2 
1 

R  R  xˆ ,  R  yˆ ,  R  xˆ ,   R  yˆ ,     R R  zˆ , 2 
R  R  xˆ ,  R  yˆ ,  R  xˆ ,   R  yˆ ,     exp   i 2 Lz
R  R  xˆ ,  R  yˆ ,  R  xˆ ,   R  yˆ ,     1  i 2 Lz


 O  3 
R  R  xˆ ,   R  R  yˆ ,   R  R  xˆ ,    R  R  yˆ ,     1   2  Lx , Ly 
• Now match the two sides:
1   2  Lx , Ly 
2
 1  i 2 Lz
 Lx , Ly   i Lz
2
 O  3 
Levi-Civita Symbol
• Generalizing, we have
• Define the Levi-Civita symbol
 ijk
1 if ijk is an even permutation of xyz

 1 if ijk is an odd permutation of xyz
0
otherwise

• Then we write:
 Li , L j   i
 Lz , Lx   i
 Lx , Ly   i Lz ,  Ly , Lz   i Lx ,

ijk
 xyz   yzx   zxy  1
 xzy   zyx   yxz  1
Lk
k
• We will call any three Hermitian operators J
that work this way generalized angular momentum
 J i , J j   i

k
ijk
Jk
Ly
7B. Generalized Angular Momentum
J2 and the raising/lowering operators
 J i , J j   i

ijk
Jk
k
• What can we conclude just from the commutation relations?
• If J commutes with the Hamiltonian, than we can
 J, H   0
simultaneously diagonalize H and one component of J
– Normally pick Jz
2
2
2
2
J   J x  iJ y
J

J

J

J
x
y
z
• Define some new operators:
• Reverse these if we want:
J x  12  J   J   , J y  12 i  J   J  
• These satisfy the following properties
– Proof by homework problem
J †  J ,  J 2 , J   0   J 2 , J   ,
 J z , J   
J  , J 2  J z2  J J   J z
Eigenstates
 J 2 , J   0   J 2 , J   ,
 J z , J   
J
•
•
•
•
j, m
Since J2 commutes with Jz, we can diagonalize them simultaneously
We will (arbitrarily for now) choose an odd way to write the eigenvalues
Note that j and m are dimensionless
J 2 j, m  2  j 2  j  j, m
Note that J2 has positive eigenvalues
J z j, m  m j, m
– We can choose j to be non-negative
• We can let J act on any state |j,m to produce a new state J |j,m:
J 2  J  j , m   J 2 J  j , m  J  J 2 j, m  J 
J z  J  j, m
  J

2
j
2
 j  j, m 
J z   J z , J   j , m   J  m  J   j, m 
• This new state must be proportional to:
J  j, m  j, m  1
2
j
2
 j   J  j, m
 m  1  J 
j, m


Eigenstates (2)
j
 j  j, m
J †  J , J 2  J z2  J J   J z
J 2 j, m 
J  j, m  j, m  1
J z j, m  m j, m
2
2
• To find proportionality, consider
J  j, m
2
 j, m J J  j, m  j , m  J 2  J z2
J z  j, m 
2
j
2
 j  m2
• This expression must not be negative:
2
2
2
2
2
2
j m
j

j

m
m
j  j m m
j  j m m 0
• When it is positive, then we have
2
2
J
j
,
m

j

j

m
m j, m  1

– Choose the phase positive
• Conclusion: given a state |j,m, we can produce a series of other states
j , m  1 , j, m  2 , j, m  3
• Problem: if you raise or lower enough times, you eventually get |m| > j
• Resolution: You must have:
m j
J  j, mmax  0  J  j, mmin
max
mmin   j
m
Summary
• Eigenstates look like j, m
• The values of m are
m  j , j  1, j  2, ,  j
2
j
  j  j, m
mmax   j
J 2 j, m 
mmin   j
J z j, m  m j, m
– There are 2j + 1 of them
• Since 2j+1 is an integer: j  0, 12 ,1, 23 ,
J  j, m 
2
j 2  j  m2 m j , m  1
J x  12  J   J   , J y  12 i  J   J  
We can use these expressions to write out J’s as matrices of size (2j + 1)  (2j + 1):
• First, pick an order for your eigenstates, traditionally  j , j , j, j  1 , , j,  j 
• The matrix Jz is trivial to write down, and is diagonal
 J z mm'  j, m J z j, m  m j, m j, m  m mm
• The matrix J+ is a little harder, and is just above the diagonal
 J  mm'  j, m J  j, m  j 2  j  m2  m j, m j, m  1  j 2  j  m2  m m,m1
• You then get J- = J+† and can find Jx and Jy
Sample Problem
• Basis states
 1,1 , 1, 0 , 1, 1 
• Jz is diagonal:
 J z mm 
m mm
1 0 0 


Jz  0 0 0 
 0 0 1 


 J  mm 
0

J   0

0

j  j  m  m m,m1
22
0
0
2
0 

22 

0 

J  j, m 
j 2  j  m2 m j , m  1
12  1  02  0  2
• J+ just above the diagonal:
2
J z j, m  m j, m
Write out the matrix
form for J for j = 1
1  1   1   1  2
2
2
 0

†
J  J   2

 0
0

0 0

2 0
0
Sample Problem (2)
1 0 0 


Jz  0 0 0 
 0 0 1


Write out the matrix
form for J for j = 1
• Now work out Jx and Jy:
0

J x   12

0

1
2
0
1
2
• As a check, find
0

1

2

0 
 J  J 
J y  12 i  J   J  
Jx 
1
2
 0  1 i
0 
2


1
1
Jy   2 i
0
 2 i


1
 0
i
0 
2

J2
 12 0 12 

2
2
2
2
2

0
1
0
J  Jx  J y  Jz


1 0 1
2
2
 12 0  12 

2
0
1
0


 1 0 1 
 2
2 
0
2

J   0 0

0 0

0
 0

J   2 0

2
 0
J2   j2  j 
1 0 0

2
0
0
0

 
0 0 1


0 

2

0 

0

0

0
2
1
2 0 0

2
0
2
0


0 0 2


Special Cases and Pauli Matrices
• The matrices for j = 0 are really simple:
– We sometimes call this the scalar representation
J x  J y  J z   0
The j = ½ is called the spinor representation, and is important
• There are only two states
 12 ,  12 , 12 ,  12 
• Often these states abbreviated

,

• The corresponding 22 matrices are written in terms of the Pauli matrices
J  12 σ
0 1
 0 i 
1 0 
x  
,  y  
 , z  

1
0
i
0
0

1






• The Pauli matrices are given by:
• Useful formulas:
J z    12
 , J x   12
, J y    12 i
7C. Spherically Symmetric Problems
Spherical Coordinates


1 2
1 2
P V  R  
P  V R2
2
2
• All components of L commute with H, because they commute with R2
• It makes sense to choose eigenstates of H, L2 and Lz n, l , m
• Consider this Hamiltonian:
H
H n, l , m  En n, l , m , L2 n, l , m 
2
l
2
 l  n, l , m , Lz n, l , m  m n, l , m
• It seems sensible to switch to spherical coordinates:
x  r sin  cos 
• We write Schrödinger’s equation in spherical coordinates
y  r sin  sin 
2
E  
2  V  r 
z  r cos 
2
 1 2
1
 
 
1
 2 
E  
r   2
 V  r 
 sin 
 2 2

2 
2 
2  r r
r sin   
  r sin   
2
L in Spherical coordinates (1)
• We need to write L in spherical coordinates
x  r sin  cos 
• Start by writing angular derivatives out:
y  r sin  sin 
  x  y  z
 
 z  r cos 




 r sin   
sin  
cos  

x  y  z 
y
 x

  x  y  z
 






 r cos  
cos  
sin    r
sin 

x  y  z 
y
z
 x

• It’s not hard to get Lz from these equations:

Lz  i


 

 
 
Lz   XPy  YPx   i  x
 sin 
y

  i r sin   cos 

y

x

y

x




L in Spherical coordinates (2)
 



 
 


 r cos  
cos  
sin    r
sin 
 r sin   
sin  
cos  
y
z

y
 x

 x
 
x  r sin  cos  ,
y  r sin  sin  ,
z  r cos 
• Now it’s time to get clever: consider






2
2
sin  cot 
 cos 
x
z
 r sin  cos 
 r cos   cos   sin  


z
x
z
x
 
 
Ly  i  z
x


x

z




Ly  i sin  cot 
 i cos 


• And we get clever once more:




 
2
2
 r sin  sin 
z
y
cos  cot 

sin   r cos   sin   cos  
y
z
y
z
 


 
 
Lx  i cos  cot 
 i sin 
Lx  i  y
z



x 
 z
Other Operators in Spherical coordinates


Lx  i cos  cot 
 i sin 




Ly  i sin  cot 
 i cos 


• It will help to get the raising and lowering operators:


L  Lx  iLy  cot   i cos  sin  
  i sin   cos  


2







1



1




2
2
L  e i  i cot 

L  
sin 
 2



2 
  
sin





sin








• And we need L2: L2  L2x  L2y  L2z 
• Compare to Schrödinger:
 1 2
1
 
 
1
 2 
E  
r   2
 V  r 
 sin 
 2 2

2 
2 
2  r r
r sin   
  r sin   
2
Solving Spherically Symmetric Problems:
• Rewrite Schrödinger’s equation:
• Our eigenstates will be n, l , m
H n, l , m  En n, l , m , L2 n, l , m 
2
1
2
E  
r


L
  V  r 


2
2
2 r r
2 r
2
2
l
2
 l  n, l , m , Lz n, l , m  m n, l , m
• The angular properties are governed by l and m
  r   R  r  Y  , 
• This suggests factoring  into angular and radial parts:
L2  RY   2  l 2  l   RY 
L2Y  2  l 2  l  Y
L2  2  l 2  l 
Lz  RY   m  RY 
Lz  m
LzY  mY
2
• Substitute into Schrodinger:
2
1
2
ERY


rRY

L
RY  V  r  RY


2
2
• Cancel Y:
2 r r
2 r
2
2
d2
2
ER  
rR

l
 l  R V r  R


2
2 
2 r dr
2 r
The Problem Divided:
  r   R  r  Y  , 
2
2
d2
2
ER  
rR

l
 l  R V r  R


2
2 
2 r dr
2 r
L2Y 
2
l
2
 l Y
• It remains to find and normalize R and Y
LzY  mY
• Note that Y problem is independent of the potential V
• Note that the radial problem is a 1D problem
– Easily solved numerically
• Normalization:



2
2 2
2
2
2 3
2
1     r  d r   r dr  sin  d  d R  r  Y  ,     R  r  r dr  Y  ,   d 
0
0
 f  ,  d   
0
0

0
sin  d 
2
0
f  ,   d   d  cos
1
1
• Split this up how you want, but normally:
 Y  ,  d   1
2


0
R  r  r 2 dr  1
2
2
 0
f  ,   d
7D. The Spherical Harmonics
Dependence on , and m restrictions
• We will call our angular functions spherical harmonics and label them Yl m  ,  
• We previously found:

Lz  i

• For general angular momentum we know:
J z j, m  m j, m
m  j, j 1,
, j
LzYl m  mYl m , m  l , l  1,
• We can easily determine the  dependence of the spherical harmonics
 m
m
im
m
im
Y

,


f

e
i
Yl  mYl m




Yl
e
l

• Also, recall that  = 0 is the same as  = 2
f    f   e2 im
• It follows that m (and therefore l) is an integer l  0,1, 2,
, l
Finding one Spherical Harmonic:
Yl m  ,   f   eim
• We previously found:
• For general angular momentum:
J  j, m 
j 2  j  m2 m j , m  1
L  e
 i
LYl m 

  

 i cot 







l 2  l  m2
m Yl m1
• If we lower m = – l, we must get zero:
d 
 i  il 

 
l
 i 
 il  e
e  l cot  
f  
0  LYl  e  i cot 


f

e



d 







d
f    f   l cot 
d
• Normalize it:
2
l
1   Yl  ,   d   N  sin  e
l
2
 il 2
f    N sinl 

2
0
0
d   N 2  sin 2l  sin  d 
1
Yl  ,    l
2 l!
l
ln f  l ln sin    k
df
 l cot  d
f
 2l  1! sin l  eil
4
2l
4

2
l !

2
d  N
 2l  1!
2
Finding All Spherical Harmonics:
1
Yl  ,    l
2 l!
l
 2l  1! sin l  eil
4
• To get the others, just raise this repeatedly:
Yl
m 1
 ,   
LYl m  ,  
l 2  l  m2  m
1
Yl  ,    l
2 l!
m

L  e
 i
LYl m 
1
 l  m  l  m  1

  

 i cot 







l 2  l  m2
m Yl m1


  m
e  i cot 

 Yl  ,  
  

i
 2l  1 l  m ! ei  i cot     
 

4  l  m !  
   
l m
sin  e 
l
• Sane people, or those who wish to remain so, do not use this formula
• Many sources list them
– P. 124 for l = 0 to 4
• Computer programs can calculate them for you
– Hydrogen on my website
 il
Properties of Spherical Harmonics:
• They are eigenstates of L2 and Lz
• They are orthonormal:
m * m
 Yl  Yl d   ll mm
L2Yl m 
2
l
2
 l  Yl m
LzYl m  mYl m
• They are complete; any angular function can be written in terms of them:

l
f  ,      clmYl m  ,  
l  0 m  l
clm   Yl m  ,   f  ,   d 
*
• This helps us write the completion relation

l
*
m



f  ,       Yl  ,   f  ,    d  Yl m  ,  


l  0 m  l
 l m

f  ,       Yl  ,    Yl m  ,    f  ,    d 
 l 0 ml


l
1
Yl  ,    Yl  ,      cos   cos          
           


sin 
l  0 m  l
m
m
More Properties of Spherical Harmonics:
• Recall: parity commutes with L
• It follows that L2  Yl m   L2Yl m 
2
L2Yl m 
2
m
l

l

Y
  l 
•
•
•
•
L2
 Yl ~ Yl
m
Recall: is real but Lz is pure imaginary
Take the complex conjugate
of our relations above:
*
m
m
This implies Yl  ~ Yl
m
m *
m
Y


1
Y
It works out to


l 
l
l
2
 l  Yl m
LzYl m  mYl m
Lz  Yl m   LzYl m  m  Yl m 
• Hence when you let parity act, you
must be getting essentially the same state
2
Yl m   1 Yl m
l
m
L Yl
2

m *
 Lz Yl


m *
2
l
2
 l Yl
 m Yl

m *

m *
The Spherical Harmonics
Y  ,   
0
0
1
Y  ,   
0
1
2 
3
2 
Y11  ,   
Y30  ,   
1
3
Y
 ,   
5cos   3cos  


7
4
21 e  i
sin   5cos 2   1
8 
105 e
Y  ,   
4 2
3
3
Y
 ,   
3 e i
sin 
2 2
3
2 i
2
3
cos 
35 e
8 
5
4 
2
3cos
  1

15 e  i
sin  cos 
2 2
sin  cos 
Y21  ,   
sin 3 
15 e 2i
Y  ,   
sin 2 
4 2
2
3i
Y  ,   
0
2
2
2
7E. The Hydrogen Atom
Changing Operators
• Hamiltonian for hydrogen (SI units):
• These operators have commutators:
•
•
•
•
H tot
ke e 2
1 2
1 2

P1 
P2 
2m1
2m2
R1  R 2
 R1i , P1 j   i  ij   R2i , P2 j  , all others vanish
m1R1  m2 R 2
Classically, what we do:
R cm 
, Pcm  P1  P2
m1  m2
– Total momentum is conserved
m2 P1  m1P2
– Center of mass moves uniformly
R  R1  R 2 , P 
– Work in terms of relative position
m1  m2
Quantum mechanically: Let’s try
 Rcm,i , Pcm, j   i  ij   Ri , Pj  ,
Find commutation operators for these
all others vanish
M  m1  m2
– Proof by homework problem
Find the new Hamiltonian
2
1
1
1
k
e
1
1
2
2
e
– Proof by homework problem H 


Pcm 
P 
tot
 m1 m2
2M
2
R
Reducing the problem: 6D to 3D
H tot
1 2
1 2 ke e 2

Pcm 
P 
2M
2
R
 Rcm,i , Pcm , j   i  ij   Ri , Pj 
M  m1  m2
1
1
1


 m1 m2
• Note that for actual hydrogen,  is essentially the electron mass
• Split the Hamiltonian into two pieces
1 2 ke e 2
1 2
H
P 
H cm 
Pcm
H tot  H cm  H
2
R
2M
• These two pieces have nothing to do with each other
1 2
– It is essentially two problems
Ecm 
p cm
• Hcm is basically a free particle of mass M,
2M
– It is trivial to solve
Etot  Ecm  E
• The remaining problem is effectively a single particle of mass 
in a spherically symmetric potential
Reducing the problem: 3D to 1D
1 2 ke e 2
H
P 
2
R
• Because the problem is spherically symmetric, we will have states
• These will have wave functions
n, l , m
 nlm  r   R  r  Yl m  , 
• The radial wave function will satisfy:
2
2
k
e
d2
2
e
ER  
rR

l

l
R

R




2
2
2 r dr
2 r
r
2
• Note that m does not appear in this equation, so R won’t depend on it
 nlm  r   Rnl  r  Yl m  , 
• We will focus on bound states E < 0
Rnl  r 
The Radial Equation: For r Large and Small
2
2
k
e
d2
2
e
ER  
rR

l

l
R

R




2
2
2 r dr
2 r
r
2
Let’s try to approximate behavior at r = 0 and r = :
• Large r: keep dominant terms, ignore those with negative powers of r:
2
2
2
k
e
d2
d
2
e
ER  
R

R

l

l
R

R


2
2
2 dr
 r dr
2 r
r
2
•
•
•
•
•
2
Define a such that:
r a
E
2
R
~
e
Then we have
2 a
Want convergent

R
~
r
Now, guess that for small r we have
Substitute in, keeping smallest powers of r
Er   
2
    1 r  2 
2
• Want it convergent
2
l

2
2
R ~ er a , for large r
 2    l2  l
 l  r  2  kee2 r  1
  l or  =  l  1
R ~ r l , for small r
The Radial Equation: Removing Asymptotic
2
r a
ke e2
d2
2
R
~
e
, for large r

R


rR

l

l
R

R




2 a 2
2 r dr 2
2 r 2
r
• Factor out the expected asymptotic behavior:
R  r   f  r  er a
– This is just a definition of f(r)
• Substitute in, multiply by 2/2
2
2
2

k
e
1 r a
1 d2
l

l
r a
r a
r a
e
 2 e f 
rfe

fe

fe

2 
a
r dr
r2
r 2
2
2
1 r a
1 d2
2 d
1  r a l 2  l  r a 2  ke e 2  r a
r a
r a
 2 fe  
rf  e 
 rf  e  2 fe  2 fe  2 fe
2 
a
r dr
ar dr
a
r
r
2  ke e 2
1 d2
2 d
l2  l
rf  
 rf   2 f  2 f  0
2 
r dr
ar dr
r
r
• Define the Bohr radius:
a0 
2
 ke e 2
The Radial Equation: Taylor Expansion
1 d2
2 d
l2  l
2
rf  
 rf   2 f  f  0
2 
r dr
ar dr
r
ra0
R ~ r l , for small r

• Write f as a power series around the origin
i
f

f
r

i
– Recall that at small r it goes like rl
i l
• Substitute in

2 
2 
2
i 2
i 1
2
i 2
i 1
f
i

i
r

f
i

1
r

l

l
f
r

f
r
0








i
i
i
i
a i l
a0 i l
i l
i l

• Gather like powers of r: 
2
2  i 1
2
2
i 2
fi  i  i  l  l  r   fi   i  1   r

a0 
i l
i l
a
• On right side, replace i  i – 1 

2
2
2
2
i 2
• On left side, first term vanishes  fi  i  i  l  l  r   fi 1  i   r i 2
i l  1
i 1l
 a a0 
• These must be identical expressions, so:
2i a  2 a0
fi  2
fi 1
2
i i l l
Are We Done?

2
2
2
i
a

2
a
0
f   fi r
R r   f r  e
fi  2
fi 1 E  
a0 
2
2
i l
i i l l
 ke e 2
2 a
• It looks like we have a solution for any E:
– Pick fl to be anything
– Deduce the rest by recursion
• Now just normalize everything
i
2
12
2i a
• Problem: No guarantee it is normalizable

f
f
fi  2 fi 1
i 1
i
 
• Study the behavior at large i:
ia
i
!
i
a
i
1  2r 
f  r      e 2r a
R  r  e r a e2 r a e r a
i i! a 
r a
i
• Only way to avoid this catastrophe is to make sure some f vanishes, say fn
2n a  2 a0
0  fn  2
f n 1
2
n  nl l
2n 2
0

a a0
a  na0
  ke e 
E

2 
2
2 n 

2
2
2
Summarizing Everything
 ke2 e4
• Because the exponential beats the polynomial,
a0 
E 2 2
these functions are now all normalizable
 ke e 2
2 n
– Arbitrarily pick fl > 0
n 1
2
2

r
na


 Rnl  r  r dr  1 Rnl  r   e 0  fi r i
• Online “Hydrogen” or p. 124
i l
• Note that n > l, n is positive integer
n  1, 2,3,
2  i  n  fi 1
• Include the angular wave functions
fi 
l  0,1, , n  1
m
n  i 2  i  l 2  l  a0
 nlm  r   Rnl  r  Yl  , 
m  l , l  1, , l
• Restrictions on the quantum numbers:
• Another way of writing the energy:
2
E
 ke2e4
2
2 n
2

 2 c2
2n 2
ke e 2
1


c 137.04
• For an electron orbiting a nucleus,  is almost
exactly the electron mass, c2 = 511 keV
E
13.6 eV
n2
Radial Wave Functions
R10  r  
2e r a0
a03
e  r 2 a0 
r 
R20  r  
1


3
2
a
2a0 
0 
R21  r  
re
 r 2 a0
2 6a05
2e  r 3a0 
2r
2r 2 
R30  r  

1 
2 
3
3 3a0  3a0 27 a0 
4 2 re  r 3a0 
r 
R31  r  
1 

5
6
a
27 3a0 
0 
e r 4 a0 
3r
r2
r3 
R40  r  
 2
1 
3 
3
4 a0  4a0 8a0 192a0 
5 re  r 4 a0 
r
r2 
R41  r  
1


2 
5
4
a
80
a
16 3a0 
0
0 
r 2 e  r 4 a0 
r 
R42  r  
1


7
12
a
64 5a0 
0 
R43  r  
R32  r  
r 3e r 4 a0
768 35a09
2 2 r 2 e  r 3a0
81 15a07
Sample Problem
What is the expectation value for R for a
hydrogen atom in the state |n,l,m = |4,2,-1?

r 
R42  r  
1 

7
64 5a0  12a0 
2  r 4 a0
r e
 4,2,1  r   R42  r  Y21  , 


R   R    r   r  d r   r dr  d    r    R
0
0
2
3
3
2
2
42
 r  r dr  d  Y  , 
3
• The spherical harmonics are orthonormal over angles
2
1
2
2
1


1
1
r   r 2 a0 4 3
 r 2 a0
7 7
9 9
1 8 8
1

a
r

a
r

a
r
e
dr
R 
1

e
r
r
dr



0
6 0
144 0

7 0 
0
20480
20480a0  12a0 
1
1

a0  28  7! 16  29  8! 144
 210  9!  21a0
20480

> integrate(radial(4,2)^2*r^3,r=0..infinity);
Degeneracy and Other Issues
n  1, 2,3,
n, l , m
Note energy depends only on n, not l or m
l  0,1, , n  1
• Not on m because states related by rotation
 ke2 e4
• Not on l is an accident – accidental degeneracy E   2 2 m  l , l  1, , l
2 n
How many states with the same energy En?
n 1
• 2l + 1 values of m
D  En     2l  1  n 2
• l runs from 0 to n – 1
l 0
• Later we will learn about spin, and realize there are actually twice as many states
Are our results truly exact?
• We did include nuclear recoil, the fact that the nucleus has finite mass
• Relativistic effects
– Small for hydrogen, can show v/c ~  ~ 1/137
• Finite nuclear size
– Nucleus is 104 to 105 times smaller
– Very small effect
• Nuclear magnetic field interacting with the electron
7E. Hydrogen-Like Atoms
Other Nuclei
Can we apply our formulas to any other systems?
2
k
e
Z
1
2
e
• Other atoms if they have only one electron in them
H
P 
2
R
• The charge on the nucleus multiplies potential by Z:
• Reduced mass essentially still the electron mass
2
2

c

m
c
 511 keV
e
2
2
• Just replace e by e Z
a0  a0 Z
• Atom gets smaller
– But still much larger than the nucleus
• Relativistic effects get bigger
 ke2e4 Z 2
 2 c2 Z 2
13.6Z 2 eV
– Now v/c ~ Z
E


2 2
2
2 n
2n
n2
Bizarre “atoms”
We can replace the nucleus or the electron with other things
Anti-muon plus electron
• Anti-muon has same charge as proton, and much more mass than electron
• Essentially identical with hydrogen
Positronium = anti-electron (positron) plus electron
• Same charge as proton
• Positron’s mass = electron’s mass
• Reduced mass and energy states reduced by half
Nucleus plus muon
• Muon 207 times heavier than electron
• Atom is 207 times smaller
• Even inside a complex atom, muon sees essentially bare nucleus
• Atom small enough that for large Z, muon is partly inside nucleus
Anti-hydrogen = anti-proton plus anti-electron
• Identical to hydrogen