Winter 2013
Chem 356: Introductory Quantum Mechanics
Chapter 7a – Hydrogen Atom and Angular Momentum ............................................................................ 88
Angular Momentum Theory ................................................................................................................... 88
The Radial Equation for the Hydrogen Atom .......................................................................................... 94
Chapter 7a – Hydrogen Atom and Angular Momentum
Angular Momentum Theory
Lˆx  yPz  zPy
Definitions:
Lˆ y  zPx  xPz
Lˆz  xPy  yPx
Lˆ2  Lˆx 2  Lˆ y 2  Lˆz 2
 r , Pb   i  
Derive fundamental commutation relations, for angular momentum
We know:
 Lˆx , Lˆ y   i Lˆz


 Lˆz , Lˆx   i Lˆ y


 Lˆ y , Lˆz   i Lˆx


Lˆ2  Lˆ 2  Lˆ 2  Lˆ 2
x
More definitions
y
z
Lˆ  Lˆx  iLˆ y
Lˆ  Lˆx  iLˆ y
Derive:
Lˆ2  Lˆ Lˆ  Lˆz 2  Lˆz
 Lˆ Lˆ  Lˆz 2  Lˆ z
 Lˆz , Lˆ   Lˆ
 Lˆz , Lˆ    Lˆ




Lˆz Lˆ  L Lz  Lz
 Lˆ2 , Lˆx    Lˆ2 , Lˆ y    Lˆ2 , Lˆz    Lˆ2 , Lˆ    Lˆ2 , Lˆ   0

 
 
 
 

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Winter 2013
Chem 356: Introductory Quantum Mechanics
Before we discuss further let me list a few useful tricks when dealing with commutators.
 Aˆ , Bˆ     Bˆ , Aˆ 




ˆ ˆ  ˆ ˆ
 ˆ ˆ
2) a. cA
 , B    A, cB   c  A, B  (pull c in front of everything)
1)
b. if ĉ commutes with  and B̂ then
ˆ ˆ ˆ  ˆ ˆ ˆ ˆ  ˆ ˆ
 AC
 , B    A, CB   C  A, B
ˆ ˆ ˆ ˆ  ˆ ˆ  ˆ ˆ ˆ
3) a.  AB
 , C   A  B, C    A, C  B
ˆ ˆ ˆ  ˆ ˆ ˆ ˆ  ˆ ˆ
b.  A
 , BC    A, B  C  B  A, C 
 Aˆ , Bˆ  Cˆ    Aˆ , Bˆ    Aˆ , Cˆ 
4)

 
 

Proofs are easy:
e.g.
ˆ ˆ ˆ  BCA
ˆ ˆ   ABC
ˆ ˆˆ
 Aˆ , BC


ˆ ˆ ˆ  BAC
ˆ ˆ ˆ  BAC  BCA
 ABC
  Aˆ , Bˆ  Cˆ  Bˆ  Aˆ , Cˆ 
Derive commutation relations easily:
 Lˆx , Lˆ y    yPz  zPy , zPx  xPz 


 
  yPz , zPx    yPz , xPz    zPy , zPx    zPy , xPz 
 yPx  Pz , z   yx  Pz , Pz   Py Px  z, z   xPy  z, Pz 
 xP
i
y
 yPx   i Lˆz
No deviatives, not acting on a function. Just use tricks!
 Lˆ2 , Lˆx    Lˆx 2 , Lˆx    Lˆ y 2 , Lˆx    Lˆz 2 , Lˆx 

 
 
 

 0  Lˆ y  Lˆ y , Lˆx    Lˆ y , Lˆx  Lˆ y  Lˆz  Lˆz , Lˆx    Lˆz , Lˆx  Lˆz
 i  Lˆ Lˆ  Lˆ Lˆ  Lˆ Lˆ  Lˆ Lˆ  0

y
z
z
y
z

y
y
z

 Lˆ  iLˆ   Lˆ  Lˆ
 Lˆ  Lˆ  i  Lˆ Lˆ  Lˆ Lˆ   Lˆ 
 Lˆ  i  i Lˆ   Lˆ  Lˆ
Lˆ Lˆ  Lˆzz2  Lˆz  Lˆx  iLˆ y
2
x
2
x
y
z
z
2
2
y
x
y
2
y
x
z
Lˆz
2
z
z
Chapter 7a – Hydrogen Atom and Angular Momentum
89
Winter 2013
Chem 356: Introductory Quantum Mechanics
 Lˆz , Lˆ    Lˆz , Lˆx  iLˆ y 

 

  Lˆz , Lˆx   i  Lˆz , Lˆ y   i Lˆ y  i i Lˆx
  Lˆ  iLˆ   Lˆ


x
y



Next we will derive the spectrum (eigenvalues and eigenfunctions) using only commutation relations.
 Lˆ2 , Lˆz   0  Lˆ2 , Lˆz have a set of common eigenfunctions


Lˆ2 g  ag
a ,b
a ,b
Lˆz ga ,b  bga ,b
L2  Lx 2  Ly 2  Lz 2
Because
a  b2
Proof:
 Lˆx 2   0 for any  (use Hermiticity Lˆ x )
Next: it g a ,b is eigenfunction of L̂2 , Lˆ z , then so is Lˆ g a ,b
 
Lˆ  Lˆ g   Lˆ Lˆ g
2
2
Proof: Lˆ Lˆ g a ,b  Lˆ Lˆ g a ,b  aLˆ aa ,b
z

a ,b

z
 b 
a ,b
 Lˆ g a.b
 Lˆ ga,b
Lˆ g a ,b : same eigenvalue of L̂2 , raised eigenvalue b 
 
Lˆ  Lˆ g   L L g
of Lˆ z (or: Lˆ ga ,b  0 )
Lˆ2 Lˆ g a ,b  Lˆ Lˆ2 g a ,b  aLˆ g a ,b
z

a ,b

z
a ,b
 L g a ,b
 (b  ) L g a ,b
Lˆ , Lˆ : ladder operators
Same eigenvalue a
Chapter 7a – Hydrogen Atom and Angular Momentum
90
Winter 2013
Chem 356: Introductory Quantum Mechanics
When do the ladder operators terminate?
Lˆ ga ,bmax  0
use a  b 2 ; b 2  a


Lˆ2 g a ,bmax  Lˆ Lˆ  Lˆz 2  Lˆz g a ,bm
 0  bmax 2  bmax  a
 bmax  l
At the other end:

a
2
l (l  1)
Lˆ g a ,bmin  0


Lˆ2 ga ,bmin  Lˆ Lˆ  Lˆz 2  Lz g a ,bmin
  bmin 2  bmin   a 
2
l (l  1)
 bmin  l
Hence from purely the commutation relations we find
a  l (l  1)
b
2
 l, l  1, l  2,....l  2, l  1, l 
2l  1 states associated with a  l (l  1)
2
There are 2 possibilities for such a structure:
A: l is integer
B: l is a half integer
l
l (l  1)
m  l...l
Degen
Spatial
name
Spin
name
0
1
2
3
1
2
3
2
5
2
0
2
6
12
3
4
15
4
35
4
0
-1,0,1
-2,-1,0,1,2
-3,-2,-1,0,1,2,3
1 1
 ,
2 2
3 1 1 3
 , , ,
2 2 2 2
5 3 1 1 3 5
 , , , , ,
2 2 2 2 2 2
1
3
5
7
s
p
d
f
Singlet
Triplet
Quintet
Septet
2
-
Doublet
4
-
Quartet
6
-
Sextet
Chapter 7a – Hydrogen Atom and Angular Momentum
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Winter 2013
Chem 356: Introductory Quantum Mechanics
When do spin-multiples show up?
1
particle
2
-
Electron in spin
-
Nuclear spin; many nuclear spin states are possible
Many electron states: triplet, quartet, doublet excited states.
We have used only the commutation relations  Lx , Ly   i Lz and Lˆ2  Lˆx 2  Lˆ y 2  Lˆz 2 to derive the
eigenvalues for L̂2 and Lˆ z operators. Any set of 3 operators that satisfy the commutation relations will
yield the same ‘spectrum’
Lˆ2 
2
l (l  1)
Lˆz  m
m  l , l  1,.....l  1, l
We can do more, if we go back to our original problem, using the  ,  representation of operators.
Lˆz  i
Lˆ2 
2


 1  
 
1 2 
sin



 sin   
  sin 2   2 


One can also derive expressions for L̂ and L̂

cos   
Lˆ  ei   i
sin   
 
 
cos   
Lˆ  ei   i
sin   
 
It is easy to find eigenfunctions for Lˆ z
i

f ( )  m f ( )


Boundary Condition f ( )  eim
f ( )  eim

m is an integer
The periodic boundary condition is the reason that we do not know how to represent spin functions:
 i 2 1 i 2
i
e 
e

2
1
 eigenvalue
1
2
1
1
i 
, but rotating  over 2 , the function e 2 changes ‘sign’!
Chapter 7a – Hydrogen Atom and Angular Momentum
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Winter 2013
Chem 356: Introductory Quantum Mechanics
Let us return to eigenfunctions of L̂2 and Lˆ z . These eigenfunctions are denoted
Yl m ( ,  )  Pl m ( )eim
Pl m ( ) are called the ‘associated Legendre’ polynomials. They will be seen to be polynomials in sin  ,
cos .We can easily generate them using the ladder operators.
The highest m  l function in the multiplet has to satisfy
LˆYl l ( ,  )  0
Yl l ( ,  )  Pl l ( )eil
 
cos    l
Lˆ Pl l ( )eil  ei   i
Pl ( )eil

sin   
 
 P l ( )
cos  l 
0  ei (l 1)  l
l
Pl ( ) 


sin



The solution to this equation is
Pl l ( )   sin    sin l ( )
l
cos 
l
 sin    0
sin 
Hence we have found the highest m  l function Yl l ( ) , for any l
since
l (sin  )l 1 cos   l
Yl l ( )   sin   eil
l  0,1, 2,3, 4,......
l
All the other functions can be generated by differentiation: acting with

cos   
Lˆ  ei   i
 yields other functions in multiplet
sin   
 
Eg. To generate the p-functions ( l  1 )
Y11  sin  ei
cos 


Y10 ~ e  i  cos  
sin   ei ~ cos 
sin 


Y01 : ei   sin   0 ~ sin  ei
This procedure works for all l . I did not worry about normalization.
In general one finds that Pl m ( ) only depends on m , hence the  -part of Yl m ( ) and Yl  m ( ) are the
same. Then one can combine the functions
1 m
eim  eim
m
m
Y ( ,  )  Yl ( ,  )  Pl ( )
2 l
2


 Pel ( ) cos 
m
Chapter 7a – Hydrogen Atom and Angular Momentum
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Winter 2013
Chem 356: Introductory Quantum Mechanics
1 m
Yl ( , )  Yl  m ( )   ....  Pl m ( ) sin 
2i
And
Using the new linear combination we get
sin  cos  ~
x
 Px
r
z
 Pz
r
y
sin  sin  ~  Py
r
cos  ~
The Yl m ( ,  ) are then related to the usual Cartesian angular momentum functions.
In the lecture notes on angular momentum I work out the d-functions in this way. You are asked to do it
on the assignment.
The Radial Equation for the Hydrogen Atom
In spherical coordinates:
 2 1   2  
Lˆ2
e2
Hˆ 
r




2  r 2 r  r  2  r 2 4 0 r
Or

try solution f (r )Yl m ( ,  )

 2 1   2 f  l (l  1)
r

2 r 2 r  r 
2 r 2
2
 2 2 f
 2 f  l (l  1)


2 r 2 r 2 r 2  2 r 2
2
Multiply through by

e
 4 0

e2
4 0 r

  f (r )
4 0 r 
e2
1
a0
Where a0 is the Bohr radius, also define


2
2 2
Define
2
2
2
2
E 
2 df d 2 f  l (l  1) 2 



 f (r )   f (r )
r dr dr 2  r 2
a0 r 
These type of 1d differential equations are easy to solve on a computer. I give a few examples of
solutions in my lecture notes on the Hydrogen atom.
Chapter 7a – Hydrogen Atom and Angular Momentum
94
Chem 356: Introductory Quantum Mechanics
Winter 2013
The general solution to the Hydrogen atom is:
 n,l ,m (r, , )  Pn,l (r )er / na Yl m ( ,  )
0
where Pn,l (r )  ar
n 1
 br n2  ...r l is a polynomial in r having n  l  1 radial nodes
En,l  En 
2
2 a0 2 n 2
The energies are exactly the same as for the Bohr atom. The theory is very different. This strange
coincidence may have held back physics for 10 years!
The Yl m ( ,  ) themselves have l angular nodes, and so the total number of nodes associated with
wavefunctions with En is always n  l 1  l  n 1
1s
 E0
No nodes
2s
 E0
4
1 radial, 0 angular
2p
 E0
4
No radial, 1
angular
3s
 E0
9
2 radial nodes
3p
 E0
9
1 radial, 1 angular
3d
 E0
9
0 radial, 2 angular
To understand radial distribution


0

2
0
0
2
dr  d  d   n ,l ,m (r , ,  ) r 2 sin 
You have been shown before
Y
l
m
2
( ,  )  constant, no  ,  dependence
m
 spherically symmetric
 justification to look at radial part only, multiplied by r 2 volume element.
 usual radial distribution, see figure 7.2 in McQuarrie
Chapter 7a – Hydrogen Atom and Angular Momentum
95