Slide 1

Report
Angular momentum: definition and commutation
• classically, this is a central concept (no pun intended)!
• if the (conservative) force is central L is conserved
• the quantum mechanical implications are profound indeed
• if the situation is that one body ‘orbits’ around another one, do
the usual reduction to the equivalent ‘one-body’ problem, with m
the reduced mass replacing m in all formulae
ˆ : rˆ  pˆ  ir   in thepositionrepresentation,
 L : r  p  L
where  actson everything toits R
 
 



 
Lˆ x  i y  z  Lˆ y  i z  x  Lˆ z  i x  y 
y 
z 
x 
 x
 z
 y
• study of the commutation relations is most revealing
• all coordinates commute, and all momentum components commute
xˆ, pˆ x   i xˆ, pˆ y  xˆ, pˆ z   0 with analogousresultsfor p y and pz


xˆ, Lˆ   xˆ, pˆ zˆ xˆ, pˆ yˆ   0 because all theseoperatorscommute
x
y
z
 same thingfor y and z
Now for some non-zero commmutators


  
 
a useful theorem Aˆ , Bˆ Cˆ  Aˆ , Bˆ Cˆ  Bˆ Aˆ , Cˆ
xˆ, Lˆ   xˆ, zˆpˆ  xˆ, xˆpˆ   xˆ, zˆpˆ
y
x
z
x
 zˆxˆ, pˆ x 
 xˆ, xˆ  pˆ z  xˆxˆ, pˆ z   0  izˆ  0  0  izˆ
 by cyclicity xˆ, Lˆ y  izˆ yˆ , Lˆ z  ixˆ zˆ, Lˆ x  iyˆ
 
 
 
pˆ , Lˆ    pˆ , zˆpˆ   pˆ , xˆpˆ    pˆ , zˆpˆ
x
y
x
x
x
z
x
x
 zˆ pˆ x , pˆ x 
  pˆ x , xˆ  pˆ z  xˆ pˆ x , pˆ z   0  0   ipˆ z   0  ipˆ z
 by cyclicity pˆ x , Lˆ y  ipˆ z pˆ y , Lˆ z  ipˆ x pˆ z , Lˆ x  ipˆ y






The key non-zero commmutators
• most interesting of all is angular momentum with itself
• one can simply move around operators that commute, preserving
the order otherwise
 a 2nd

Lˆ , Lˆ   yˆpˆ
x
  
 
 
 
ˆ  Aˆ , Cˆ D
ˆ Bˆ  Cˆ Aˆ , D
ˆ Bˆ  Aˆ Bˆ , Cˆ D
ˆ  Aˆ Cˆ Bˆ , D
ˆ
useful theorem Aˆ Bˆ , CˆD
y
z


 zˆpˆ y , zˆpˆ x  xˆpˆ z 

 
  yˆ pˆ z , zˆpˆ x    yˆ pˆ z , xˆpˆ z   zˆpˆ y , zˆpˆ x  zˆpˆ y , xˆpˆ z
 {0  0  yˆ  pˆ z , zˆ  pˆ x  0}  {0  0  0  0}
 {0  0  0  0}  {0  xˆzˆ, pˆ z  pˆ y  0  0}



 iyˆ pˆ x  ixˆpˆ y  i xˆpˆ y  yˆ pˆ x  iLˆ z


 by cyclicity Lˆ x , Lˆ y  iLˆ z
Lˆ , Lˆ   iLˆ Lˆ , Lˆ   iLˆ
y
z
x
z
x
y
Expressing components of L in sphericals I
  r          
    
 
by thechain rule, where
x  x  r  x    x  
2
2 

x

y
 : arctan s 
r  x 2  y 2  z 2   arctan


z
z


   1 
 
 
2x
2
x
 



2
2
x  2 x 2  y 2  z 2  r  1  x 2y  2 z x 2  y 2  



z


 y
 x
  arctan 

 1

y2
1 2
x


  y  
 x 2  



2
 x    x 2   y  
 x   z
 2 
     2 2  
  2
2 
 r  r  z  s  zs    x  y  x  
 x    xz     y  
    2 
 2 
 r  r  r s    s  
• this is in a succinct and ‘hybrid’ notation
  y    yz     x  
    2 
 2 
y  r  r  r s    s  
just swapping x  y and a sign flip
Expressing components of L in sphericals II
• for z, things proceed a bit differently and it is a lot simpler!
 
2z

z  2 x 2  y 2  z 2

   1
 
 r  1  x 2  y 2

z2

  x 2  y 2


z2

 

0
 

2
  s    z    s  
z   z
     2 2  2 
    2 
 r  r  z  s  z    r  r  r  
• we can now construct the angular momentum operators
   z    s      y    yz    x    
 
 
Lˆ x  i y  z   i y     2    z      2 
  2   
y 
 z
   r  r  r      r  r  r s    s    
  yz zy    ys yz 2      xz   
 y  
xz  

 i       2  2       2    i  
 2


 r

r  r  r
r s      s   
  s   s  

 r sin sin  r sin cos  r cos    

 Lˆ x  i

2
2
 
r sin 
 r sin 


 

 i sin
 cot  cos 

 

Expressing components of L in sphericals III, and an
astounding eigenresult
   x    xz     y      z    s    
ˆ
Ly  i z     2 
  2    x     2   
   r  r  r s    s      r  r  r    


 

 Lˆ y  i  cos
 sin  cot

 

   y    yz    x      x    xz     y    
ˆ
Lz  i x     2 
  2    y     2 
  2   
   r  r  r s    s      r  r  r s    s    
  xy xy    xyz xyz    x 2 y 2   
 i      2  2 
  2  2  
r  r  r s r s    s
s   
 r

1 im
 Lˆ z  i
 Lˆ z F( )  mF( ) since F( ) 
e

2
• Sweet!! The F functions are eigenfunctions of the Lz operator,
with eigenvalue mħ!!
• therefore, the spherical harmonics are eigenfunctions as well with
that same eigenvalue
Working out the square of L: another astounding
eigenresult
2


1


1



ˆ
ˆ
ˆ
ˆ
ˆ
ˆ


L  L  L  L  L  L   
 sin 
 2
2 
  sin   
 sin   
2
2
x
2
y
2
z
2
• compare this with Legendre’s equation: exactly the same thing
2 
 1  

1



 Pl ( )  l (l  1) Pl ( )
sin
 2


 sin  
  sin   2 

 Lˆ2 Pl ( )  l (l  1) 2 Pl ( )  Lˆ2 Pl m ( )  l (l  1) 2 Pl m ( )
 Lˆ2Y m ( ,  )  l (l  1) 2 Y m ( ,  ) and Lˆ Y m ( ,  )  m Y m ( ,  )
l
l
z l
l
• evidently, one can simultaneously determine both the magnitude
of the angular momentum, and its z component
• claim: [L2, Lz] = 0 proof: you do it on paper!
• claim: [L2, L] = 0
L2 commutes with any component of L
• we’ve already established that the components don’t commute
Two new angular momentum operators, built from Lx
and Ly, and a bizarre identity
Lˆ : Lˆ x  iLˆ y and these will be raising and lowering operators

• interesting commutation relations
Lˆ x , Lˆ  Lˆ x , Lˆ x  i Lˆ x , Lˆ y  0  i iLˆ z  Lˆ z
       
Lˆ , Lˆ  Lˆ , Lˆ  iLˆ , Lˆ  iLˆ  i0  iLˆ
Lˆ , Lˆ  Lˆ , Lˆ  iLˆ , Lˆ   iLˆ  i iLˆx  iLˆ
y
y
z
x
z
y
x
y
z
z
y
y
• since [L2, L] = 0  [L2, L ± ]
• a fantastic and bizarre operator identity:



z

 

Lˆ Lˆ  Lˆx  iLˆ y Lˆx  iLˆ y  Lˆ2x  Lˆ2y  i Lˆx Lˆ y  Lˆ y Lˆx
 
 Lˆ2x  Lˆ2y  i Lˆx , Lˆ y  Lˆ2x  Lˆ2y  i iLˆz  Lˆ2x  Lˆ2y  Lˆz


but since Lˆ2 : Lˆ2x  Lˆ2y  Lˆ2z  Lˆ2x  Lˆ2y  Lˆ2  Lˆ2z
Lˆ Lˆ  Lˆ2  Lˆ2z  Lˆz  Lˆ2  Lˆ Lˆ  Lˆ2z  Lˆz
 
 
y

 Lˆ x  Lˆ

Learning about raising and lowering
• now to ask: suppose some function Y is an eigenfunction of both
L2 (with eigenvalue l) & Lz (with eigenvalue m)
• what is the effect on Y of L ± ? first, test it with L2:
 
 
 
Lˆ2 LˆY  Lˆ Lˆ2Y  Lˆ lY   l LˆY
 LˆY is also an eigenstate of Lˆ2 with eigenvalue l
• second, test it with Lz :
Lˆ z Lˆ Y  Lˆ z Lˆ Y  Lˆ Lˆ z  Lˆ Y  mLˆ Y  Lˆ Y  m    Lˆ Y







 Lˆ Y is also an eigenstateof Lˆ z with eigenvaluem  

• we see how the raising and lowering takes place: l is unchanged,
while m is raised or lowered by one unit of angular momentum
• there is therefore a ‘ladder’ of states for a given l
• but once the z component of Lz gets as big as (or nearly as big as)
L itself the process must stop: there must be a’top’ state Ytop
 


Lˆ Ytop  0 where Lˆ zYtop : m topYtop

 
Trying to raise the top, or lower the floor
• we therefore have Lˆ Ytop  0 where Lˆ zYtop : m topYtop



 Lˆ2Ytop  Lˆ Lˆ  Lˆ2z  Lˆ z Ytop  Lˆ Lˆ Ytop  Lˆ2zYtop  Lˆ zYtop
 
 
2
 lYtop  0  m top
Ytop  m topYtop  l  m top m top   
• now lower the states one by one with the lowering operator
• same l each time, but m is knocked down by ħ
• not yet clear what the multiplicative factor might be…
• finally we arrive at the ‘unlowerable’ bottom state Ybot
Lˆ Ybot  0 where Lˆ zYbot : m bot Ybot



 Lˆ2Ybot  Lˆ Lˆ  Lˆ2z  Lˆ z Ybot  Lˆ Lˆ Ybot  Lˆ2zYbot  Lˆ zYbot
 
 
2
 lYbot  0  mbot
Ybot  mbotYbot  l  mbot mbot   
 m top m top     mbot mbot   
what does thismean,since we know mbot  m top ?
Sorting our the relationship between top and bot
• two m must differ by some integer nћ
mtop  mbot  n where n  0
 m bot  n m bot  n  1   m bot m bot   
2
2
mbot
 m bot 2n  1  n(n  1) 2  m bot
 m bot 
mbot 2n  2  n(n  1)  2m bot  n  mbot   n2
and so we have that m top   n2  n 
n
2
• in principle n may be odd or even
• it may be shown (Griffiths problem 4.18)
Lˆ Yl m   l l  1  mm  1 Yl m1   l  m l  m  1 Yl m1

• the spherical harmonics are not eigenfunctions of the
raising and lowering operators

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