rot.qm.3d.part1

Angular momentum in classical physicsAngular momentum in classical physics

Consider a particle at the position r

i

k

j

rv

Where

r = ix + jy + kz

The velocity of this particle is given by

v = drdt = i

dxdt + j

dydt + k

dzdt

Classical Angular Momentum

Classical Angular MomentumThe linear momentum of the particle with mass m isgiven by

p = mv where e.i px = mvx = mdxdt

The angular momentum is defined as

L = rXp

L

L = r X p

Φ

| | | | r p sinΦ

r

p

The angular momentum is perpendicular to the plane .defined by r and p


We have in addition

L = rXp = (ix +jy + kz)X (ipx + jpy +kpz)

L = (rypz -rzpy)i + (rzpx -rxpz)j + (rxpy - rypx)k

or

i j k

rXp = r x ry rz

px py pz


Why are we interested in the angularmomentum ?

Consider the change of L with time

dLdt =

drdt Xp + rX

dpdt

dLdt = mvXv + rX

dpdt = rX

dpdt

dLdt = rX

ddt [m

drdt ] = mrX

d2rdt2

dLdt = mrXF F = force

r

F


r

F

dr L

dt=m

r r ×

r F


Examples :

movement of electron around nuclei movement of planets around sun

For such systems L is a constant of motion, e.g. doesnot change with time since

dLdt = 0

In quantum mechanics an operator O representing aconstant of motion will commute with the Hamiltonianwhich means that we can find eigenfunctions that areboth eigenfunctions to H and O

rF

Quantum mechanical representation of angular momentumQuantum mechanical representation of angular momentumoperatoroperatorWe have

L = rXp = iLx + jLy + kLz

where

Lx = rypz - rzpy ; Ly = rzpx - rxpy ; Lz = rxpy - rypx

In going to quantum mechanics we have

x --> x ; y --> y ; z --> z

px --> -ihδδx ; p y --> -ih

δδy ; p z --> -ih

δδz

:Thus

L x = -ih( yδδz - z

δδy ) ; L y = -ih( z

δδx - x

δδz )

L z = -ih( xδδy - y

δδx )

Rotation..Quantum Mechanics 3D

We have

L = iLx + jLy + kLz

thus

L.L = L2 =(iLx + jLy + kLz).(iLx + jLy + kLz)

L2 = Lx2 + Ly2 + Lz2


Can we find common eigenfunctions to

L2 , Lx , Ly , Lz ?

Only if all four operators commute

We must now look at the commutationrelations

The two operators L x and L y will

commute if

[L xL y - L yL x] f(x,y,z) = 0



We have

Lxf = -ih( yδfδz - z

δfδy ) = -ih ux

Ly = -f ih( zδfδx - x

δfδz ) = -ih uy

Next

LxLy = -f ih Lxuy

LxLy = -f ih [ -ih( yδuyδz - z

δuyδy ) ]

LxLy = -f h2 [ yδuyδz - z

δuyδy ]

We have

δuyδz =

δδz (z

δfδx - x

δfδz )

δuyδz =

δfδx + z

δ2fδzδx - x

δ2fδz2


Further

δ u

y

δ y

= δδy (z

δfδx - x

δfδz )

δuyδy = z

δ2fδyδx - x

δ2fδyδz

combining terms

Thus

LxLyf = -h2[ yδfδx + yz

δ2fδzδx - yx

δ2fδz2 - z2

δ2fδyδx +zx

δ2fδyδz ]

LxLy = -f h2[ yδfδx + yz

δ2fδzδx - yx

δ2fδz2 - z2

δ2fδyδx +zx

δ2fδyδz ]

It is clear that LxLy f can be evaluated by :interchanging x and y We get

LyLx = -f h2[ xδfδy + xz

δ2fδzδy - xy

δ2fδz2 - z2

δ2fδxδy +zy

δ2fδxδz ]


Rotation..Quantum Mechanics 3Dusing the relations

δ2fδzδy =

δ2fδyδz .etc

We have

[ LxL y - LyLx] = -f h2[ yδfδx - x

δfδy ] = -h2[ y

δδx - x

δδy ]

f

We have: Lz = -i h[ xδδy - y

δδx ]

Thus: [ LxLy - LyLx] f = ihLz f ; [Lx,Ly] = ihLz

We have shown [Lx,Ly] = ihLz

Rotation..Quantum Mechanics 3DBy a cyclic permutation

[ Ly,L z] = ihLx

[ Lz,Lx] = ihLy

We have shown that the three operators L x,L y,L zare non commuting

What about the commutation between Lx,Ly,Lz and L2

Y

Z X

X

Y

z

X

Yz

C3

Let us examine the commutation relation

between L2 and Lx

We have:

[ L2,Lx]=[Lx2 +Ly

2 +Lz2,Lx]

[ L2,Lx]=[Lx2,Lx]+[Ly

2,Lx]+[Lz2,Lx]

[Lx2,Lx]=Lx

2Lx −LxLx2 =Lx

3 −Lx3 =0

For the first term


For the second termRotation..Quantum Mechanics 3D

[Ly2,Lx]=Ly

2Lx −LxLy2

=Ly2Lx −LyLxLy +LyLxLy −LxLy

2

=Ly[LyLx −LxLy]+[LyLx −LxLy]Ly

=−ihLyLz −ihLzLy

Y

Z X

For the third termRotation..Quantum Mechanics 3D

[Lz2,Lx]=Lz

2Lx −LxLz2

=Lz2Lx −LzLxLz +LzLxLz −LxLz

2

=Lz[LzLx −LxLz]−[LzLx −LxLz]Lz

=ihLzLy +hLyLz

Y

Z X

[ L2,Lx]=[Lx2 +Ly

2 +Lz2,Lx]


In total

−ihLyLz −ihLzLy=0 +ihLzLy +hLyLz =0

Y

Z X

We have shown[L2,Lx] = [Lx2+Ly2+Lz2,Lx] = O

now by cyclic permutation

[Ly2+Lz2+Lx2,Ly] = [L2,Ly] = 0

[Lz2+Lx2+Ly2,Lz] = [L2,Lz] = 0

Thus Lx,Ly,Lz all commutes with L 2

and we can find common eigenfunctions for

L2 and Lx or L 2 and Ly or L 2 and Lz

the normal convention is to obtain eigenfunctions that areat the same time eigenfunctions to Lz and L2.

How do we find the eigenfunctions ?

Y

Z XRotation..Quantum Mechanics 3D

How do we find the eigenfunctions ?

The eigenfunctions f must satisfy

Lzf = af and L 2f = bf

The function f must in other wordssatisfy the differential equations

Lzf = af

as well as

L2f = bf


It is more convenient to solve the equations in

spherical coordinates

Θ

φ

r

(x,y,z)→

(r, Θ,φ )

We find after some tedious but straight forward

manipulations

Lz = -ih [ddφ

]

L2 = -h2[ d2

dΘ2 +cotΘ

d dΘ

+ 1

sin2Θ d2

dφ2 ]


We must now solve

L̂ zY(Θ,φ) = (b YΘ, φ )

and

L̂ 2 (Y Θ,φ) = (c YΘ, φ )

First let us solve

L̂ z (Y Θ , φ ) = -ihδδφ Y(Θ, φ ) = b Y(Θ, φ )

We shall try a solution of the form

Y(Θ, φ ) = S( Θ)T( φ )

as L̂ z only depends on φ


Rotation..Quantum Mechanics 3DWe have

-ihδδφ S(Θ)T(φ) = b S( Θ)T(φ)

or

- ihS(Θ)δ

δφ T(φ)= b S( Θ)T(φ)

multiplying with 1/ S( Θ) from left

δT(φ)

δφ =

ib

hT(φ)

The general solution is

T(φ) = AExp[

ib

hφ]

A general point in 3-D space is given by ( r,Θ, φ)

Θ

φ

r

( , , )x y z → ( ,r Θ,φ )

X

Y

ZrcosΘ

rsinΘ

We have the following relation = x rsinΘ cosφ = y rsinΘ sinφ = z rcosΘ

( ,The same point is represented by rΘ,φ+2π)

We must thus have

Exp[

ibh

φ] = Exp[

ibh

(φ+2π) = Exp[

ibh

φ] Exp[

ibh

2π]

A general point in 3-D space is given by ( r,Θ, φ)

Θ

φ

r

( , , )x y z → ( ,r Θ,φ )

X

Y

ZrcosΘ

rsinΘ

We have the following relation = x rsinΘ cosφ = y rsinΘ sinφ = z rcosΘ

( ,The same point is represented by rΘ,φ+2π)

We must thus have

Exp[

ibh

φ] = Exp[

ibh

(φ+2π) = Exp[

ibh

φ] Exp[

ibh

2π]

Thus

Exp[

ibh

2π] = cos

2πbh

⎛ ⎝

⎞ ⎠+ isin

2πbh

⎛ ⎝

⎞ ⎠=1

This equation is only satisfied if

bh= = 0,±1,±2,......m with m

Thus the eigenvalue b is quantized as

= b h m = 0,±1,±2,......m

The possible eigenfunctions are

(T φ) = [AExp imφ] , = 0,±1,±2,......m

rot.qm.3d.part1

Documents