Download Quantum Physics 2005 Notes-8 Three-dimensional Schrodinger Equation Notes 8

Quantum Physics 2005
Notes-8
Three-dimensional Schrodinger Equation
Notes 8
Quantum Physics F2005
1
The Schrodinger Equation
Here’s what we have been working with:
Hˆ ! = Eˆ!
h2 " 2
"#
$
# + V ( x, t ) # = i h
2
"t
2m "x
Here’s what the 3D equation is:
Hˆ ! = Eˆ!
h2 r 2
"#
r
$
% # + V ( r , t ) # = ih
"t
2m
Notes 8
Quantum Physics F2005
2
A quick review of coordinate systems
Cartesian coordinates: iˆ, ˆj , kˆ form an orthogonal (rectangular) system
with each axis at right angles to the other two.
r
! Line element: dl = dxiˆ + dyˆj + dzkˆ
! Volume element: dV = dxdydz
rˆ "
r
"
"
% = iˆ + ˆj + k
"x
"y
"z
z
2
2
2
"
"
"
%2 = 2 + 2 + 2
"x "y "z
y
x
Notes 8
Quantum Physics F2005
3
Cylindrical
Converting to Cartesian
coordinates:
! Line element:
! Length ds:
x = ' cos& ,
y = ' sin & , z = z
' = ( x 2 + y 2 )1 / 2 , tan & = y / x
r
dl = d ' uˆ ' + ' d& uˆ& + dzuˆ z
ds 2 = dr 2 + r 2 d& 2 + dz 2
! Volume element: dV = ' d ' d& dz
'ˆ ,&ˆ, ẑ
r
"
1 "
"
% = uˆ '
+ uˆ&
+ uˆ z
"'
' "&
"z
2
2
1
1
"
"
"
"
'
%2 =
+ 2
+ 2
2
' "' "' ' "&
"z
Notes 8
Quantum Physics F2005
4
Spherical
x = r sin & cos ( ; y = r sin & sin ( ; z = r cos&
2
2
2
r = x + y + z ; tan ( = y / x; tan& =
r
dl = drrˆ + rd&&ˆ + r sin& d((ˆ
x2 + y 2
z
dV = r 2 sin& drd& d(
r
"
1 "
1
"
% = uˆr
+ uˆ&
+ uˆ(
"r
r "&
r sin& "(
1 "  2 " 
1
" 
" 
1
"2
% = 2
+ 2
sin&
+ 2
r




"&  r sin2 & "( 2
r "r  "r  r sin& "& 
2
Notes 8
Quantum Physics F2005
5
The Rectangular 3D Box –
Quantum well
Notes 8
Quantum Physics F2005
6
5-11 The 3-D Box
a a


x in $ , 
0
 2 2

b b


V ( x, y , z ) = 
y in $ , 
 2 2


c c

) otherwise x in $ 2 , 2 



The 3D Box problem is a straightforward extension of the
1D infinite well or 1D box problem where

$ a , a 
0
x
in

 2 2 
V ( x) = 
) x > a

2
Notes 8
Quantum Physics F2005
7
Px
1- D
V (x)
Px * h"
i"x
E * ih
"
"t
2
Px
+ V ( x) = E
2m





Notes 8
2
h" #
i "x
+V ( x)# = ih " #
2m
"t





2 "2
h
$
# +V ( x)# = ih" #
2m "x2
"t
Quantum Physics F2005
8
3 $D
Px, Py , Pz
V ( x, y , z )
Px * h " , Py * h " , Pz * h "
i "z
i "x
i "y
E * ih
2
2
Px + Py + Pz
2m
"
"t
2
+ V ( x, y , z ) = E
"
h 2  " 2
"2
" 2 
$
#+
#+
# + V ( x , y , z ) # = ih #
"t
2m  "x 2
"y 2
"z 2 
Notes 8
h2 2
"
)# =F2005
$
% # + V (Quantum
x, y, zPhysics
ih #
2m
"t
9
1- D
• Volume element d+ = dx
• Probability of finding a particle in d+ at time t
2
)
P( x, t ) = ∫$) # ( x, t ) dx
• Normalization condition
2
)
∫$) # ( x, t ) dx = 1
• Stationary State wave function
$ iEt
#( x,t) =! ( x)e h
&&dinger equation
• Time independen t Schro
2 2
$ h " ! +V ( x)! = E!
2m "x2
• Eigenfunction
Notes 8
! (x )
Quantum
n Physics F2005
10
• Volume d+ = dx dy dz
3 -D
• Probabilit y of finding a particle in d+ at time t
2
P( x, y, z,t) = #( x, y, z,t) dx dy dz
• Normalization conditon
)
$)
)
$)
2
)
$)
∫ dx ∫ dy ∫ dz # ( x, y, z , t ) = 1
• Stationary state wave function
# ( x, y , z , t ) = ! ( x , y , z ) e
$
iEt
h
• Eigenfunction ! n( x)
&&
• Time independent Schrodinger
equation
2
$ h %2! ( x, y, z) +V ( x, y, z)! = E!
2m
• Eigenfunction ! n ( x), ! n ( y), ! n ( z)
Notes 8
1 Quantum2 Physics F20053
11
1D
! n ( x)
a
x,
2
! n ( x) =
2  cos  n- x
a sin  a
x>
a
2
! n ( x) = 0
Subsitute ! back into time independent Schr&o&dinger
eq. to obtain allowed energies for the particle
2
2 h2-F2005
Notes 8
Quantum
En = nPhysics
2ma2
12
3D
! n n n ( x, y, z) =! n ( x) ! n ( x) ! n ( z)
1 2 3
1
a
x, ,
2
! n n n ( x, y, z) =
1 2 3
2
b
c
y , ,z ,
2
2
2  cos  n1- x
a sin  a
3
n

2
cos
•   2 y
b sin  b
n

2
• c  cos  3c z
 sin 
Notes 8
Quantum Physics F2005
13
x > a, y > b, z > c
2
2
2
! n n n ( x, y, z) = 0
1 2 3

2
2
2
n   n   n   h 2- 2
En n n = 1  +  2  +  3  
1 2 3 a   b   c   2m








Notes 8

Quantum Physics F2005
14
Degeneracy & symmetry
A degeneracy always reflects the existence of some
symmetry in a given problem.
E.g. 1 - 3D box with a = b . c

2
2
2
+
n1 n2 n3  h2- 2
+ 
En n n =
a2
c2  2m
1 2 3









Exchange n1 & n2, En1n2n3 stays the SAME.
But
! n n n &! n n n are DISTINCT.
1 2 3
because
!n n n
213
( x, y, z,t ) &
!n n n
( x, y, z,t )
1 2 3
213
Notes 8
Quantum Physics F2005
differ in their dependence
on x & y.
15
• Symmetry here is interchang e coordinate s x & y.
•Degeneracy here is two DISTINCT eigenfunctions
having the SAME energy value.
See Fig. 5-27
112
a = b . c box case $
$211,121
E.g. 2
Fig. 5-27
a=b=c
box case $ 211,121,112
See Figure 5 - 27 a . b . c box
Notes 8
$ 112
$ 211
$ 121Quantum Physics F2005
16
Notes 8
Quantum Physics F2005
17
Solutions for Central Potentials
The 3D time-independent Schrodinger Eq:
h2 2
ˆ
% # + V # = E#
H# = $
2m
" 
"# 
" 2# 
1
1
h 2  1 "  2 "# 
$
+ (V $ E ) # = 0
r
+
 sin &
+
"&  r 2 sin 2 & "( 2 
2m  r 2 "r  "r  r 2 sin & "& 
2
2
 h2 " 2 ( r# )
 " ( r# ) 
1 " ( r# )  
h2  " ( r# )
$
+ cot & 
$
  + (V $ E ) ( r # ) = 0
+ 2
2 
2
2
 2m "r 2
"
"
"
&
&
&
(
2
mr
sin

 



Multiply both sides by r 2 , letting V = V (r ) and separating variables:
2
2
2
 " ( r# ) 
1 " ( r# ) 
h2 " ( r# ) 2
h2  " ( r# )
$r
+ r (V $ E ) ( r # ) =
+ cot & 


+ 2
2
2m "r 2
2m  "& 2
sin
"
"
&
&
(



r # = u (r ) F (& , ( )
2
2
2
2
 "(F ) 
h2 1 " ( u (r ) ) 2
1
1 " (F )
h2  " ( F )
$r
+ r (V $ E ) =
+ cot & 

+ 2
2 
2m u (r ) "r 2
&
&
(
"
"
F (& , ( ) 2m  "& 2
sin



2
=Notes
= l (l + 1)
a constant
8
Quantum Physics F2005
18
The ( equation
Notes 8
Quantum Physics F2005
19
( solution for central potential
1 2
1 1 "  2 "#  1
1
" 
"#  1
1
" 2#
2mE
&
% #=
r
+
sin
+
=
$


h2
#
# r 2 "r  "r  # r 2 sin & "& 
"&  # r 2 sin 2 & "( 2
# = R/0
1 2 "  2 "R  1
" 
"/  2 2 2m( E $ V (r )) 1 " 20
2
&
&
&
$ sin &  r
$
$
=
=
$
sin
sin
r
sin
m
l

h2
"r  "r  #
"& 
"& 
0 "( 2
R
0 = e± iml(
Because the wavefunction must be single valued ml must be an integer.
Notes 8
Quantum Physics F2005
20
The & equation
Notes 8
Quantum Physics F2005
21
& solution for central potential 1
1 "  2 "R  2 2m( E $ V (r ))
ml2
1 1 " 
"/ 
&
r
$
r
=
+
$
sin
= l (l + 1)


2
2


R "r  "r 
sin & / sin & "& 
"& 
h
The angular part of Laplace's equation is called the Legendre Equation.
ml2
1 1 " 
"/ 
&
sin
+ 2 $
= l (l + 1)


sin & / sin & "& 
"& 
Making the substitution x=cos&
d
dx d
d
d
=
= $ sin &
= $ 1 $ x2
, so
d& d& dx
dx
dx
d 
m2
2 d /lm 
(1 $ x )
)/lm = 0
+ (l (l + 1) $
2


1$ x
dx 
dx 
Notes 8
Quantum Physics F2005
22
& solution for central potential 2
We will start by solving the special case when m=0.
d 
2 dPl 
(1 $ x )
+ (l (l + 1)) Pl = 0


dx 
dx 
To find the solution we use the power series method,
)
assume a solution of the form, y ( x) = ∑ an x n
n =0
and substitution into gives
)
∑ an n( n $ 1) x
n=2
n$2
)
)
n
)
$ ∑ an n(n $ 1) x $ 2 ∑ an nx + l (l + 1) ∑ an x n = 0
n=2
n
n =1
n =0
from which we get the recursion relation,
(n $ l )(n + l + 1)
an + 2 =
an
(n + 1)(n + 2)
Notes 8
Quantum Physics F2005
23
& solution for central potential 3
We can deduce the solution for n=0 and use
with the recursion relation.
(2n $ 2 j )! x n $ 2 j
Pn ( x) = ∑ ( $1) n
j =0
2 l !(n $ 2 j )!(n $ 1)!
where N=n/2 for n even and N=(n-1)/2 for n odd.
N
j
A second set of possible solutions yields unphysical results.
The first few Legendre polynomials (Pn ) are:
P0 ( x) = 1, P1 ( x) = x, P2 ( x) = ( 3 x 2 $ 1) / 2
or P0 = 1, P1 (& ) = cos & , P2 (& ) = ( 3cos 2 & $ 1) / 2
Notes 8
Quantum Physics F2005
24
& solution for central potential 4
• Example plots of the first few Legendre
functions (Legendre_plots.mws)
1
2
3
Notes 8
Quantum Physics F2005
25
& solution for central potential 5
Orthogonality and normalization
of Legendre Polynomials
1
2
∫ [ Pn ( x) Pm ( x) ] dx = 1 nm
2n + 1
$1
Orthogonality means that we can express
the angular part of any wavefunction using a
sum of Legendre polynomials.
Notes 8
Quantum Physics F2005
26
& solution for central potential 6
We will not solve the m . 0 case now, but
we will state the relation between the Legendre
functions (m=0) and the full solutions (the
associated Legendre functions).
m
/lm ( x) = (1 $ x 2 )
m /2
d Pl
m
dx
from which we can find
/00 = 1,
/10 = x, /1±1 = (1 $ x 2 )1/ 2 ,
/02 = 1 $ 3 x 2 , / ±2 1 = (1 $ x 2 )1/ 2 x
Notes 8
Quantum Physics F2005
27
& and ( together
Sperical harmonics
Notes 8
Quantum Physics F2005
28
Associated Legendre functions
from Gasiorowicz
Notes 8
Quantum Physics F2005
29
Angular solutions put together
The & and ( solutions can now be combined
Flm (& , ( ) = /lm (& )0 m (( ) and when normalized, yields the
Spherical Harmonics:
 2l + 1  (l $ m)! m
im(
/
Yl = 
(cos
)
e
&
l

+
4
(
l
m
)!
 - 
The orthonormality relation is:
m
m m'
Y
∫∫ l Yl ' d& d( = 1 ll '1 mm '
Notes 8
Quantum Physics F2005
30
Spherical Harmonic Functions
Notes 8
l
m
0
0
1
0
1
1
2
0
Ylm (( ,& )
( 4- )
$1/ 2
1/ 2
( 3 / 4- )
1/ 2
$ ( 3 / 8- )
1/ 2
( 5 /16- )
cos &
sin & ei(
( 3cos & $ 1)
Quantum Physics F2005
2
31
The radial equation
Notes 8
Quantum Physics F2005
32
The r part of the 3D TISE
" 2 ( u (r ) )
l (l + 1)
2m
$
+
u (r ) + 2 (V $ E )u ( r ) = 0
2
2
"r
r
h
# is finite, continuous, and smooth,
so u(r) must go to zero at r=0.
First, let's look at u(r) solutions for l = 0.
2
h 2 " ( u (r ) )
$
+ (V $ E )u ( r ) = 0
2
2m "r
Notes 8
Quantum Physics F2005
33
u(r) example for l=0: particle in a box
0 r < a
assume V = 
) r 2 a
2
h 2 " ( u (r ) )
$
$ Eu (r ) = 0 inside
2
2m "r
u (r ) = A sin kr + B cos kr
= A sin kr because u must = 0 at r=0
Boundary condition at a gives: kn =
na
n- r
2 2
2 2 2
2 2 2
k
n
nh
h
h
a ; E =
n
=
=
n
r
2m
2ma 2
8md 2
A sin
# (r ) =
Notes 8
Quantum Physics F2005
34
The radial equation for the
Coulomb potential
$
" 2 ( u (r ) )
"r 2
l (l + 1)
2m
e2
+
u (r ) + 2 ($
)u (r ) = Eu (r )
2
h
r
4-3 0 r
h2
r
Let ' =
where a0 = 4-3 0
a0
me 2
E
me 4
Let 3 =
where ER =
2
ER
( 4-3 0 ) 2h 2
and so: $
Notes 8
" 2 (u( ' ) )
"'
2
+
l (l + 1)
'
2
u(' ) +
Quantum Physics F2005
2
'
u( ' ) = 3 u( ' )
35
l=0 + Coulomb potential
Let's look at forms of the radial equation
for l = 0 in the large r limit:
d 2u
$ 3'
' *)
3
+
=
0
⇒
4
u
u
e
d'2
Remember that 3 is negative because these are bound
states and the potential is negative.
We now search for solutions to the full
equation using the polynomial expansion approach.
The lowest order polynomial with the right behavior
as ' * 0 is u ( ' ) = re $
3'
.
(Think about the number of roots for a polynomial.)
Notes 8
Quantum Physics F2005
36
l=0 Coulomb solutions
Substituting back into
the original diff eq
and solving for 3 .
(
)
$ $ 31 $ 31 ' $ 1 = 31 '
⇒ 31 = $1
E1 =
me 4
( 4-3 0 )
2
2h 2
= $13.6eV (Rydberg)
Notes 8
Quantum Physics F2005
37
l=0 Coulomb solutions
We look for solution of
increasing polynomial order in
similar manner and can find:
u2 4  2 ' $ ' 2  e $ ' / 2
u3 4  27 ' $ 18' 2 + 2 ' 3  e $ ' / 3
and we find that the energies
are:
E1
En = 2
n
Notes 8
Quantum Physics F2005
38
Energy levels for solutions so far (1)
$1
En = 2
n
n=4
n=3
n=2
n =1
3 r-nodes
2 r-nodes
1 r-node
0 r-nodes
$
1
Ry
4
$1 Ry
l =0
Notes 8
Quantum Physics F2005
39
The l=0 wavefunctions
!1 := e
( $' )
!2 := ( 2 $ ' ) e
2
!3 := ( 27 $ 18 ' + 2 ' ) e
Notes 8
Quantum Physics F2005
 1 
 $ ' 
 2 
 1 
 $ ' 
 3 
40
Probability of finding electron at r:
radial probability
# *# is the probability of finding the electron in
a particular position. If we want the probability
of finding the electron at a distance between r and r+dr
from the nucleus, then we have to integrate # *# around the
sphere.
2- -
w( ' ) = ∫ ∫ r 2 sin & R 2 ( ' )Ylm* (& , ( ) Ylm (& , ( ) d& d(
0 0
For l=0, w( ' ) = 4- r 2 R 2 = 4- u 2
Notes 8
Quantum Physics F2005
41
Radial probability for l=0 solutions
w1
w2
w3
Notes 8
Quantum Physics F2005
42
Radial probability
At what distance is the electron most likely to be found?
For the n = 1, l = 0 state:
d ( ' 2 R2 )
2
du
du
=
=2
=2
d'
d'
d'
' = 1!!!
r = a0 = 0.5 Angstroms
Notes 8
d ( ' e$ ' )
d'
= 2 ( e$ ' $ ' e$ ' ) = 0
Quantum Physics F2005
43
Solutions to Coulomb potential with
l>0
" 2 ( u ( ' ) )  l (l + 1) 2 
$
+
$  u(' ) = 3 u(' )
2
2
"'
'
 '
This is equivalent to an effective potential of
 l (l + 1) 2 
$ 
Veff = 
2
'
 '
On the next page we plot effective potential for various l.
Notes 8
Quantum Physics F2005
44
Effective potential with l>0
So higher l will push the wavefunctions out
and push the energies up.
Notes 8
Quantum Physics F2005
45
Radial solutions for l>0 using r limits
d 2u
$
5
$
3
⇒
4
u
u
e
For large ':
d'2
d 2u
For small ' :
d'2
l (l + 1)
'2
3'
u ⇒ u 4 ' l +1 or ' $ l
6Look for solutions of the form: u 4 ' l +1e $
3'
If we plug this form as it stands back into the
1
full equation, we get 3 = $
.
2
(l + 1)
The solution we have found corresponds to the
l =0 solutions for l = n -1.
Notes 8
Quantum Physics F2005
46
Radial solutions with l>0
1
.
2
(l + 1)
Note that this for l > 0 has the same energy
as the l =0 solutions for l = n -1.
Remember these are solutions for u (r ) with no nodes.
(Radially, they actually look like the function for l = 0, n = 1.
In & they have increasing number of nodes.)
For a given l we have 3 = $
Notes 8
Quantum Physics F2005
47
Energy levels for solutions so far
$1
En = 2
n
n=4
n=3
n=2
l =3
l=2
3 r-nodes
2 r-nodes
1 r-node
$
1
Ry
4
l =1
0 r-nodes
Because the energies line up,
n =1
0 r-nodes
l =0
Notes 8
$1 Ry
these states are labeled
by the corresponding n number.
Quantum Physics F2005
48
What about solutions with more rnodes?
Let's look at a solution with one node:
Guess: u ( ' ) = ( c0 $ c1 ' ) ' l +1e $ '
3
We find that this can be a solution if: 3 = $
1
(l + 2)
2
.
This is the same energy as our nodeless
solution with l = n - 2.
Notes 8
Quantum Physics F2005
49
Energy levels for solutions so far
$1
En = 2
n
n=4
n=3
n=2
l =3
l=2
3 r-nodes
2 r-nodes
l =1
1 r-node
0 r-nodes
n =1
l=2
l =1
1 r-node
0 r-nodes
l =0
Notes 8
Quantum Physics F2005
50
A table of the radial wavefunctions
n
1
2
l=0
e$ '
(1 $
'
2
)e $ ' / 2
3 ( 27 $ 18 ' + 2 ' ) e
2
Notes 8
$' /3
l=1
no solution
l=2
no solution
' e$ ' / 2
no solution
' (6 $ ' )e$ ' / 3
Quantum Physics F2005
' 2e$ ' / 3
51
Atomic energy structure
•
•
•
•
•
•
This should begin to look familiar.
The energy levels correspond to the energies of the
Bohr model.
For n=1 energy level, only l=0 is found.
For the n=2 energy level, only l=0 and l=1 solutions
are found.
For the n=3 level, only l=0,1,2 are found.
In chemistry, we designate the l=0 case as s, l=1 as
p, l=2 as d, and l=3 as f.
Note the ml does not affect the energy of a state
because it does not appear in the radial equation.
Notes 8
Quantum Physics F2005
52
A big summary of energy levels and quantum numbers
(from Semat 1972)
Notes 8
Quantum Physics F2005
53
A big summary of hydrogen wavefunctions
(from Semat 1972)
Notes 8
Quantum Physics F2005
54
A big summary of pictures of wavefunctions
(from Semat 1972)
Notes 8
Quantum Physics F2005
55
What we observe:
emission from transitions between states
E photon
Notes 8
 1
1 
= 1Ry ×  2 $ 2 
n

n
f
i


Quantum Physics F2005
56
Calculating transition probabilities
The probability for an transition between two
atomic levels with the emission of a photon
polarized along the z axis is proportional to
the matrix element:
M fi7 ! f z! i = ! f r cos &! i .
It can be shown that M fi is non-zero only when
l f $ li = ±1 and
8ml = 0 or ± 1
Notes 8
Quantum Physics F2005
57
Angular momentum and magnetic
quantum numbers:
l and ml
Notes 8
Quantum Physics F2005
58
Another look at the 3D Schrodinger
equation
Another way to construct the 3D Schrodinger equation
is to compare it to the classical equation:
p2
+V = E
2m
Breaking p up into components:
p 2 = px2 + p y2 + pz2 = pr2 + pt2
L
and in terms of angular momentum: pt = .
r
L2
pr2
+
+V = E
2
2m 2mr
Notes 8
Quantum Physics F2005
59
Angular momentum
If we now compare this expression to the Laplacian version
we see that L2 should be an operator and we can test if it
has eigenvalues and functions.
We already know from our prior math that the angular part
of the Laplacian has the same eigenfunctions as the Hamiltonian.
Let's look more closely at the L operator.
We think: Lˆ2! = L2! .
r r r
Classically: L = r × p, so let's do the math.
Notes 8
Quantum Physics F2005
60
Angular momentum operators
In Cartesian coordinates:
i
r r
r×p= x
px
j
y
py
k
z
pz
" 
h "
Lx = ypz $ zp y =  y $ z 
i  "z
"y 
" 
h "
Ly = zpx $ xpz =  z $ x 
i  "x
"z 
" 
h "
Lz = xp y $ ypx =  x $ y 
i  "y
"x 
Notes 8
Quantum Physics F2005
61
Angular momentum in spherical
coordinates
h
"
" 
Lx =  $sin(
$ cot& cos(
i
"&
"( 
h
"
" 
Ly =  $cos(
$ cot& sin(
i
"&
"( 
h "
Lz =
i "(
Notes 8
Quantum Physics F2005
62
What can we tell about L?
We can learn quite a bit about angular momentum without doing
integrals.
We can show that Lz , and Lx cannot have a complete
set of common eigenstates by considering the commutator [ Lz , Lx ] .
Lz = xp y $ ypx ; Lx = ypz $ zp y ; Ly = zpx $ xpz
L2 = L2x + L2y + L2z
[ Lx , Lz ] =  ypz $ zp y , xp y $ ypx  =  ypz , xp y  $ [ ypz , ypx ] $  zp y , xp y  +  zp y , ypx 
=  ypz , xp y  + 0 + 0 +  zp y , ypx 
= x  ypz , p y  + [ ypz , x ] p y +  zp y , ypx  = $ x  p y , ypz  $ [ x, ypz ] p y +  zp y , ypx 
= $ xy  p y , pz  $ x  p y , y  pz $ y [ x, pz ] p y $ [ x, y ] pz p y +  zp y , ypx  =
= 0 $ x  p y , y  pz + 0 $ 0 +  zp y , ypx  = x  y, p y  pz +  zp y , ypx 
= x  y, p y  pz + y  zp y , px  +  zp y , y  px = x  y, p y  pz $ y  px , zp y  $  y, zp y  px
= x  y, p y  pz $ yz  px , p y  $ y [ px , z ] p y $ z  y, p y  px $ [ y, z ] p y px
= x  y, p y  pz $ 0 $ 0 $ z  y, p y  px $ 0 = ih { xpz $ zpx } = $ihLy
Similarly:  Lx , Ly  = ihLz and  Ly , Lz  = ihLx
Notes 8
Quantum Physics F2005
63
What can we tell about L?
• We can also show (homework!) that [L2,Lz]=0.
• This tells us that there exists a complete set
of eigenfunctions for both operators, or that
we can measure both quantities
simultaneously with infinite accruracy.
• (We already knew that because we did the
work to find the eigenfunctions and values,
but we could have avoided that work if all we
wanted was to know whether it could be
done.)
Notes 8
Quantum Physics F2005
64
Useful commutator/operator relations
[ A, B ] = AB $ BA
[ A, B ] = $ [ B, A]
[ A, B + C ] = [ A, B ] + [ A, C ]
[ A, BC ] = [ A, B ] C + B [ A, C ]
 A, [ B, C ] +  B, [C , A] + C , [ A, B ] = 0
Notes 8
Quantum Physics F2005
65
Angular Momentum in Spherical
coordinates 2
2
2


"
"
"
1
2
2
ˆ
+ 2
L = $ h  2 + cot &
2 
&
&
&
(
"
"
"
sin


• We have already found the eigenfunctions of this
operator. They are the spherical harmonics.
• We have previously found that the eigenvalues of
L2 are l(l+1) with l=integers 0, 1, 2, 3... if the
potential is central.
• Angular momentum manifests itself as a magnetic
dipole moment when the particle with L has charge.
• It is most useful to know the projection of the dipole
onto an applied magnetic field. (Let’s say in the z
direction.)
Notes 8
Quantum Physics F2005
66
Angular momentum projection
We want to know now whether the projection
of the angular momentum onto the z-axis
can be an eigenvalue.
h "
ˆ
Lz! =
! = Lz!
i "(
In the previous section we found: ! = R/0
with 0 = eiml( , thus Lˆ ! = hm ! .
z
l
! is simultaneously an eigenstate of Lˆ2 and Lˆ z .
For L2 = hl (l + 1), ml can take on integer values from
-l to + l.
Notes 8
Quantum Physics F2005
67
Pictorial representation of L and ml
Notes 8
Quantum Physics F2005
68
Angular momentum and magnetic moment
An electron moving in a plane orbit of area A
e
is equivalent to a current given by I =
cT
where T is the period of the orbit.
Such a plane current has a magnetic moment
given by: µ = IA.
So the magnetic moment of a classical orbiting
eA
electron is: µ =
.
cT
We can relate the area and period to the
angular momentum.
Notes 8
Quantum Physics F2005
69
Angular momentum and magnetic moment
1 2- 2
Area can be expressed as: A = ∫ r d(
2 0
2 d(
Angular momentum is: L = mr
dt
r
r LT
So A =
2m
e r
eh
r
L=$
l (l + 1) Lˆ.
Therefore: µ = $
2mc
2mc
Notes 8
Quantum Physics F2005
70
Angular momentum and magnetic moment
r
If an atom has a magnetic moment µ, then it will
r
experience a force when it is placed in a magnetic field B.
If the magnetic field is uniform, then the atom will
r r r
experience a torque M = µ × B and precess about the
direction of the field. The shift in energy due to
eh
Bml .
precession is: 8E = µ B cos & =
2mc
Notes 8
Quantum Physics F2005
71
Electron spin
• It is observed that a free electron exhibits a
magnetic moment which is quantized.
• We have associated this moment with a new
quantum number which we call spin s so that
µs = 2s
eh
1
with s =
2mc
2
• When we compute the total magnetic moment
of an atom, we must include both electron
spin and orbital angular momentum as a
vector sum.
Notes 8
Quantum Physics F2005
72
Appendix – general radial wavefunctions for V~1/r
using series solutions (from Eisberg and Resnick)
d 2 F  2  dF  9 $ 1 l (l + 1) 
F = 0.
+  $ 1
+
$
2
2

d'  '  d'  '
' 
a0 . 0 and s 2 0.
Assume F ( ' ) = ' s ∑ ak ' k
k =0
This form is used because it assures that F will
be finite at ' =0.
)
∑ {[ ( s + k )( s + k + 1) $ l (l + 1) ] ak '
k =0
s+k $2
$ ( s + k + 1 $ 9 ) ak ' s + k $1} = 0
It can be shown that:
[( s)( s + 1) $ l (l + 1)] a0 ' s $2
)
+ ∑ {[ ( s + j + 1)( s + j + 2) $ l (l + 1) ] a j +1 $ ( s + j + 1 $ 9 )a j } ' s + j $1 = 0
j =0
Notes 8
Quantum Physics F2005
73
General radial solutions
Collecting terms in powers of ' , we see that
the sum above will only be zero for all ' if:
[( s)( s + 1) $ l (l + 1)] = 0
(This relation requires that s = l or s = -(l + 1).
Only the s = l solution can be physically allowed.)
and
(l + j + 1 $ 9 )
a j +1 =
aj
(l + j + 1)(l + j + 2) $ l (l + 1)
Notes 8
Quantum Physics F2005
74