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Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 3 for Electrical Engineering
Lecturers: Daniel Rohrlich, Ron Folman
Teaching Assistants: Daniel Ariad, Barukh Dolgin
Week 9. Quantum mechanics – angular momentum operators •
ˆ2
commutation relations • eigenvalues and eigenvectors of Lˆ z , L
Sources: Merzbacher (2nd edition) Chap. 9;
Merzbacher (3rd edition) Chap. 11.
The time-independent Schrödinger equation for a particle in three
dimensions is
 2 2

Eψ(r)  
  V (r) ψ(r)
 2m

where r = (x,y,z), and  2 
2
x
2

2
y
2

2
z
2
.
A special, but important, class of potentials is the class of central
potentials, which depend only on r = (x2 + y2 + z2)1/2 , i.e.
 2 2

Eψ(r)  
  V (r ) ψ(r) .
 2m

The time-independent Schrödinger equation for a particle in three
dimensions is
 2 2

Eψ(r)  
  V (r) ψ(r)
 2m

where r = (x,y,z), and  2 
2
x
2

2
y
2

2
z
2
.
A special, but important, class of potentials is the class of central
potentials, which depend only on r = (x2 + y2 + z2)1/2 , i.e.
 2 2

Eψ(r)  
  V (r ) ψ(r) .
 2m

The time-independent Schrödinger equation for a particle in three
dimensions is
 2 2

Eψ(r)  
  V (r) ψ(r)
 2m

where r = (x,y,z), and  2 
2
x
2

2
y
2

2
z
2
.
A special, but important, class of potentials is the class of central
potentials, which depend only on r = (x2 + y2 + z2)1/2 , i.e.
 2 2

Eψ(r)  
  V (r ) ψ(r) .
 2m

If the only force acting is a central force, we know from classical
physics about an important conserved quantity:
angular momentum
.
Angular momentum L = r × p is conserved because F(r) and r
are always parallel:
dL
 dp  d


0  r  F(r )  r    
rp 
dt
 dt  dt
.
If the only force acting is a central force, we know from classical
physics about an important conserved quantity:
angular momentum
.
Angular momentum L = r × p is conserved because F(r) and r
are always parallel:
dL
 dp  ? d


0  r  F(r )  r    
rp 
dt
 dt  dt
© 2005-2009 George Coghill
.
If the only force acting is a central force, we know from classical
physics about an important conserved quantity:
angular momentum
.
Angular momentum L = r × p is conserved because F(r) and r
are always parallel:
dL
 dp  d


0  r  F(r )  r    
rp 
dt
 dt  dt
.
If the only force acting is a central force, we know from classical
physics about an important conserved quantity:
angular momentum
.
Angular momentum L = r × p is conserved because F(r) and r
are always parallel:
dL
 dp  d


0  r  F(r )  r    
rp 
dt
 dt  dt
Is there a quantum
operator for angular
momentum? Is it
conserved?
.
Angular momentum operators
In classical mechanics, angular momentum is defined as
L=r×p .
So in quantum mechanics, is angular momentum defined as
ˆ  rˆ  pˆ ?
L
Angular momentum operators
In classical mechanics, angular momentum is defined as
L=r×p .
So in quantum mechanics, is angular momentum defined as
ˆ  rˆ  pˆ ?
L
(Isn’t there a problem with the ordering of r̂ and p̂?)
?
© 2005-2009 George Coghill
Angular momentum operators
In classical mechanics, angular momentum is defined as
L=r×p .
So in quantum mechanics, is angular momentum defined as
ˆ  rˆ  pˆ ?
L
Angular momentum operators
In classical mechanics, angular momentum is defined as
L=r×p .
So in quantum mechanics, angular momentum is defined as
 
 
ˆ
 ,
Lx  yˆ pˆ z  zˆ pˆ y  i y
z
y 
 z
 
 
ˆ
L y  zˆ pˆ x  xˆ pˆ z  i z
x
 ,
z 
 x
 
 
ˆ

Lz  xˆ pˆ y  yˆ pˆ x  i x
y
x 
 y
.
Angular momentum operators
In classical mechanics, angular momentum is defined as
L=r×p .
So in quantum mechanics, angular momentum is defined as
 
 
ˆ
 ,
Lx  yˆ pˆ z  zˆ pˆ y  i y
z
y 
 z
 
 
ˆ
L y  zˆ pˆ x  xˆ pˆ z  i z
x
 ,
z 
 x
 
 
ˆ

Lz  xˆ pˆ y  yˆ pˆ x  i x
y
x 
 y
.
Is L̂ conserved?
Theorem:


ˆ , Hˆ  0,
If a Hermitian operator  commutes with Ĥ, i.e. A
then we can find eigenstates of  that are also eigenstates of Ĥ.
Proof: If Ĥ commutes with an observable Â, then for any
state ψ , Hˆ Aˆ ψ  Aˆ Hˆ ψ . So if ψ n is an eigenvector of Ĥ
with eigenvalue En, i.e. Hˆ ψ n  En ψ n , then so is Aˆ ψ n :



Hˆ Aˆ ψ n  Hˆ Aˆ ψ n  Aˆ Hˆ ψ n  Aˆ En ψ n  En Aˆ ψ n

If En is nondegenerate, then Aˆ ψ n  a ψ n for some number
a, so ψ n is an eigenstate of Â. If En is degenerate, then
eigenstates of  form a basis for the subspace of eigenvectors
of Ĥ with eigenvalue En.
.
Corollary:


ˆ , Hˆ  0,
If a Hermitian operator  commutes with Ĥ, i.e. A
then the expectation value of  in any state (t) is constant
in time, i.e.
d ˆ
A 0 .
dt
Proof: Define the state ψni such that Hˆ ψ ni  En ψ ni
and Aˆ ψ ni  ani ψ ni . Write  (t) 
and then
Aˆ 

n,i
does not depend on time.

n,i
2
cni ani
cnie iEnt /  ψ ni
So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:


 


ˆ
Lx , V (r )  i  y
z
, V (r )
y
 z

 
 
 V (r )
 i  y
z
y 
 z
 
 

 iV (r ) y
z
y 
 z
So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:


 


ˆ
Lx , V (r )  i  y
z
, V (r ) ψ
y
 z

 
 
 
 
 [V (r ) ψ]  iV (r ) y
ψ
 i  y
z
z
y 
y 
 z
 z
Remember: the derivatives act also on a wave function ψ.
So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:


 


ˆ
Lx , V (r )  i  y
z
, V (r )
y
 z

 
 

 V (r )
  i  y
z
y 
 z
dV (r )  r
r 
 y

 i
z
dr  z
y 
y
 dV (r )   z
  i 
  y  z  V (r )  0 .
r
 dr   r
So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:


 


ˆ
Lx , V (r )  i  y
z
, V (r )
y
 z

 
 

 V (r )
  i  y
z
y 
 z
dV (r )  r
r 
 y

 i
z
dr  z
y 
y
 dV (r )   z
  i 
 y  z  0 .
r
 dr   r
So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:





2
2
ˆ
Lx ,   i  y
z
, 
y
 z

 
  2
   i 2
 i  y
z
y 
 z
 
 
 y

z
y 
 z
 2 


 
2
 i   y
 z
 2y    2z   
z
y
z
y 

So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:





2
2
ˆ
Lx ,   i  y
z
, ψ
y
 z

 
  2

 
2[ 
ψ
   i  y
 ψ]
 i  y
z
z
y 
y 
 z
 z
 2 



2
 i  y
 z
 2y    2z   
z
y
z
y 

Remember: the derivatives act also on a wave function ψ.
So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:


 

2
2
ˆ
Lx ,   i  y
z
, ψ
y
 z

 
  2ψ

  ψ]
2[ 
   i  y

 i  y
z
z
y 
y 
 z
 z
 2 



2
 i  y
 z
 2y    2z   
z
y
z
y 

Remember: the derivatives act also on a wave function ψ.
So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:





2
2
ˆ
, 
z
Lx ,   i  y
y

 z
 2 
 


2
 2y    2z  
 z
 i  y

y

z

y

z



    

 i 
 0 .
 y z z y 
So is L̂ conserved?
That is, do Ĥ and L̂ commute? Let’s see:





2
2
ˆ
, 
z
Lx ,   i  y
y

 z
 2 
 


2
 2y    2z  
 z
 i  y

y

z

y

z



  
  
2
 i 2
 0 .
z y 
 y z
And what holds for L̂ x holds for L̂ y and L̂ z .
So L̂ is conserved!
What are the possible values of L̂ ?
Let’s calculate commutation relations for L̂.
Commutation relations
 Lˆ x , Lˆ y   yˆ pˆ z  zˆ pˆ y , zˆ pˆ x  xˆ pˆ z 

 
  yˆ pˆ z , zˆ pˆ x    yˆ pˆ z , xˆ pˆ z   zˆ pˆ y , zˆ pˆ x  zˆ pˆ y , xˆ pˆ z

  yˆ pˆ z , zˆ pˆ x   zˆ pˆ y , xˆ pˆ z




 yˆ  pˆ z , zˆ  pˆ x  xˆ zˆ , pˆ z  pˆ y  i xˆ pˆ y  yˆ pˆ x  iLˆ z ,

   
using Aˆ Bˆ , Cˆ  Aˆ Bˆ , Cˆ  Aˆ , Cˆ Bˆ . Similarly,
 Lˆ y , Lˆ z   iLˆ x
 Lˆ z , Lˆ x   iLˆ y
,
.
Commutation relations
 Lˆ x , Lˆ y   iLˆ z
,
 Lˆ y , Lˆ z   iLˆ x
,
 Lˆ z , Lˆ x   iLˆ y
.
We can choose a basis of eigenstates of L̂ x, or of L̂ y, or of L̂ z,
but only one of these bases at a time!
Also, from our generalized uncertainty principle,
  
1
ˆ
ˆ
A B 
2
 Aˆ , Bˆ 
,
we conclude
 ˆ
 ˆ
 ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
Lx L y 
Lz , L y Lz 
Lx , Lz Lx 
Ly
2
2
2
.
ˆ 2.
Eigenvalues and eigenvectors of Lˆ z , L
Let’s prove that Lˆ   Lˆ x  iLˆ y is a raising operator:
 Lˆ z , Lˆ    Lˆ z , Lˆ x  iLˆ y   iLˆ y  Lˆ x   Lˆ
.
Similarly, Lˆ   Lˆ x  iLˆ y is a lowering operator:
 Lˆ z , Lˆ    Lˆ z , Lˆ x  iLˆ y   iLˆ y  Lˆ x   Lˆ
Suppose Lˆ z l  Lz l . Then
Lˆ z Lˆ l  Lˆ Lˆ z l  Lˆ l
.
ˆ 2.
Eigenvalues and eigenvectors of Lˆ z , L
Let’s prove that Lˆ   Lˆ x  iLˆ y is a raising operator:
 Lˆ z , Lˆ    Lˆ z , Lˆ x  iLˆ y   iLˆ y  Lˆ x   Lˆ
.
Similarly, Lˆ   Lˆ x  iLˆ y is a lowering operator:
 Lˆ z , Lˆ    Lˆ z , Lˆ x  iLˆ y   iLˆ y  Lˆ x   Lˆ
Suppose Lˆ z l  Lz l . Then


Lˆ z Lˆ l  Lˆ Lˆ z  Lˆ l  Lz    Lˆ l
,
.
ˆ 2.
Eigenvalues and eigenvectors of Lˆ z , L
Let’s prove that Lˆ   Lˆ x  iLˆ y is a raising operator:
 Lˆ z , Lˆ    Lˆ z , Lˆ x  iLˆ y   iLˆ y  Lˆ x   Lˆ
.
Similarly, Lˆ   Lˆ x  iLˆ y is a lowering operator:
 Lˆ z , Lˆ    Lˆ z , Lˆ x  iLˆ y   iLˆ y  Lˆ x   Lˆ
.
Suppose Lˆ z l  Lz l . Then


Lˆ z Lˆ l  Lˆ Lˆ z  Lˆ l  Lz    Lˆ l
,
so Lˆ l is an eigenvector of L̂ zwith eigenvalue Lz  .
Similarly, Lˆ l is an eigenvector of L̂ zwith eigenvalue Lz  .
ˆ2.
Eigenvalues and eigenvectors of Lˆ z , L
Eigenvalues of L̂ z: ..., Lz  2, Lz  , Lz , Lz  , Lz  2,...;
   L̂ we must have
but since Lz is bounded by Lˆ z
Lˆ  mmax
2
2
 
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
L  L L  Lz   Lz  L L  Lz   Lˆ z
since
 0  Lˆ mmin for some mmax and mmin. Note


    
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
L L  Lx  iL y Lx  iL y   Lx   L y   Lˆ z
,
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
L L  Lx  iL y Lx  iL y  Lx  L y  Lˆ z ,


ˆ 2 , Lˆ  0 .
Note also L
z
.
ˆ2.
Eigenvalues and eigenvectors of Lˆ z , L
To make the rest of the calculations easier, we should change to
spherical coordinates:
x  r sin  cos  , y  r sin  sin  , z  r cos  .


x  y 


ˆ
.


 y  x
Now
so Lz  i

  x  y
x
y
The eigenfunctions of L̂ z are
1
2
e im with eigenvalues m so
2
2
ˆ2 m


L


m


mmax
max
max
2
2
ˆ2 m


L


m


mmin .
min
min
Therefore mmax = –mmin and, by convention, mmax = l.
ˆ2:
Eigenvalues and eigenvectors of Lˆ z , L
Summary: For a given value of l, the eigenstates of L̂ z are
 l ,  l  1 ,..., m ,..., l 1 , l , with respective eigenvalues
 l,  l  1,..., m,..., l  1, l . These 2l+1 eigenvectors
of L̂ z are also eigenvectors of L̂2 with degenerate eigenvalue
l (l  1) 2 .
Schrödinger’s equation for a central potential is
 2 2

Eψ(r)  
  V (r ) ψ(r) .
 2m

A vector identity:
ˆ2
L

2
 r     r      r 2 2  r    r   2
.
x
y
z
 x, r
 y, r
 z, it is
What is r  ? Since r
r
r
r
hence
ˆ2
L
 x  y  z  

  r
r    r 


,
r
 r x r y r z 

  
 r   r  r  r
r
r  r
2
2 2

2 2   2 
  r    r

r  r 

.
Schrödinger’s equation for a central potential is
 2 2

Eψ(r)  
  V (r ) ψ(r) .
 2m

A vector identity:
ˆ2
L

2
 r     r      r 2 2  r    r   2
x
y
z
 x, r
 y, r
 z, it is
What is r  ? Since r
r
r
r
hence
 x  y  z  

  r
r    r 


,
r
 r x r y r z 
ˆ2
L
  2 
 r    r

2
r 
r 

2 2
.
.
ˆ2
L
  2 
2

Solving
for
we obtain
 r    r

2
r 
r 

2 2
2
ˆ

1


L


2 
r2

 2  ,

2 r
r   r   
and by comparing this with the expression for  2 in spherical
coordinates,
2

1


ψ
1

ψ
1  
ψ 


2
2
 ψ


 sin 
 ,
 r
2 r
2
2
sin   
 
r   r  sin  
ˆ2
L
  2 
2

Solving
for
we obtain
 r    r

2
r 
r 

2 2
2
ˆ

1


L


2 
r2

 2  ,

2 r
r   r   
and by comparing this with the expression for  2 in spherical
coordinates,
2

1


ψ
1

ψ
1  
ψ 


2
2
 ψ


 sin 
 ,
 r
2 r
2
2
sin   
 
r   r  sin  
ˆ2
L
  2 
2

Solving
for
we obtain
 r    r

2
r 
r 

2 2
2
ˆ

1


L


2 
r2

 2  ,

2 r
r   r   
and by comparing this with the expression for  2 in spherical
coordinates we conclude
2

1

1 
 
2
2
ˆ
L   

sin 

2
2 sin  
 
 sin  
.
Now back to Schrödinger’s equation in spherical coordinates:
ˆ2 
2    2   L
Eψ(r , ,  )  
r
  ψ  V (r )ψ ;



r   2 
2mr 2  r 
2
since the eigenvalues of L̂ are l (l  1) 2, we can solve this
equation by expressing ψ(r,θ,φ) as a product of two functions:
ψ(r,θ,φ) = R(r) Ylm(θ,φ) ,
where
ˆ 2Y m  l (l  1) 2Y m ,
L
l
l
Lˆ zYlm  m Ylm .
ˆ 2 and Lˆ :
Here are the lowest eigenfunctions Ylm(θ,φ) of L
z
Here’s how they look: