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Lecture 13: Angular Momentum-I.
The material in this lecture covers the following in Atkins.
Rotational Motion
Section 12.6 Rotation in two dimensions
Lecture on-line
Angular Momentum in 2D (PowerPoint)
Angular Momentum in 2D (PDF format)
Handout for this lecture
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Tutorials on-line
Vector conceptsBasic Vectors
More Vectors (PowerPoint)
More Vectors (PDF)
Basic conceptsObservables are Operators - Postulates of Quantum
Mechanics
Expectation Values - More Postulates
Forming OperatorsHermitian Operators
Dirac Notation
Use of Matricies
Basic math background
Differential Equations
Operator Algebra
Eigenvalue EquationsExtensive account of Operators
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Audio-visuals on-line
Rigid Rotor (PowerPoint)
(Good account from the Wilson Group,****)
Rigid Rotor (PDF)
(Good account from the Wilson Group,****)
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Motion in 1D
Free translation
Confined translation1D
Harmonic Oscillation1D
Vibrating diatomic molecule
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Motion in 2D
Confined translation2D Rotation in 2D The rigid rotor
Rotating Diatomics Vibrating Diatomics
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Quantum Mechanical Rotation in 2D
let us consider a particle of mass mmoving in the xy plane in a circle
of fixed radius a
k
j
i
x
y
z
a
The position of the particleis given by
r = i x + j y
where
r = r r
vv r
r
= + =x y a2 2
With the velocity given by
dr
dt= i
dx
dt+ j
dy
dt
v = iv + j vx y
vv r
r v r
Position and velocityof circular motion
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Quantum Mechanical Rotation in 2D
k
j
i
x
y
z
r
It is more informative towork in the sphericalcoordinates (r, )
We have
r = i x + j y
r = i rcos + j rsin
The velocity is given by
v = i drcos + j rsin
r r r
r r v
r v
dt
ddt
or
dtddt
since r is constant (r = a)
v = - i rsin d + j rcosr v
Position and velocityof circular motion in
spherical coordinates
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Quantum Mechanical Rotation in 2D
We note thatv | = v v
r v =
+=
a
d
dt
ad
dt
2 2 2 2
2
( ) [sin cos ]
( )
k
j
i
x
y
z
r
v
r r v
r v
r = i rcos + j rsin
v = -i rsin
d
+ j rcos
dt
d
dt
also v r v r v r
ad
dta
d
dt
Velocity
x x y y:
sin cos
sin cos
(v) perpendicular to
position vector (r).
= +
= +
=
2
2 0
s
r
Position and velocityof circular motion in
spherical coordinates
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Quantum Mechanical Rotation in 2D
k
j
i
x
y
z
r
v
Let us now evaluateAngular momentum:
L = r p = m r vv r r r r
We have
L = r p = ( ix + jy) ( ipxr r r r r r r
+ jpy )
= i i (x i j (x
j i (y
r r r r
r r r r
v v
v r
v r
+
+ + =
=
=
p p
p j j ypkxp kyp
L xp yp k
L m xv yv k
x y
x y
y x
y x
y x
) )
) ( )
( )
( )
Angular momentumof circular motion
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Quantum Mechanical Rotation in 2D
k
j
i
x
y
z
r
v
L
r rL m xv yv k
ory x= ( )
r r v
r v
r = i rcos + j rsin
v = - i rsin dr + j rcos
dt
ddt
r
r
L mrd
dt
mr ddt
k
= +[ cos
sin ]
2 2
2 2
r r
r r r
L m rd
dtk
L m r v k r p k
=
= =
2
| | | | | | | |
Angular momentumof circular motion in
spherical coordinates
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Quantum Mechanical Rotation in 2D
k
j
i
x
y
z
r
v
L
r rL mr d
dtk= 2
The total energy is E = Ekin +=
E
Epot
pot
0
|p mr ddt= 2
Ep
m
m rd
dt
m
mrd
dt
mrkin= = =
2
2 2 2 22
22 2 2
( ) ( )
EL
mr
EL
I
kinz
kinz
=
=
2
2
2
2
2
or since I = mr is moment of inertia2
Total energy ofcircular motion
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Quantum Mechanical Rotation in 2D...
quantitative derivation
k
j
i
x
y
z
r
x r= cos ; y = rsin
The Hamiltonian is againgiven by
H = E + Ekin pot
or since E = 0pot
Constructing quantummechanical Hamiltonian
of circular motion inCartesian coordinates
Hm x y
= + = +
p
2m
p
2mx2
y2
h2 2
2
2
22
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Quantum Mechanical Rotation in 2D...
quantitative derivation
k
j
i
x
y
z
r
2
2
2
2
2
2 2
2
2
2 22 2
22
1 1
2
1 1
x y r r
Hm r r
+ = + +
= + +
r r
r r ( )
( ) ( ) ( )h
Since does not depend on r,
the first twoderivatives with respect to r can beneglected and we have
H m r=
h2
2
2
22
1
H = - 2m
2h d
dx
d
dy
H E
2
2
2
2+
= ( ) ( )
Next making use ofspherical coordinates
Constructing quantummechanical Hamiltonian
of circular motion inspherical coordinates
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k
j
i
x
y
z
r
Quantum Mechanical Rotation in 2D...
quantitative derivation
or since I = mr2 : HI
=
h2 2
22
Constructing Schrdingerequation of circular motion
in spherical coordinates
Hm r
=
h2
2
2
22
1
The Schrdingerequation is
= = h
h
2 2
2
2
2 22
2
I E orIE
2
22
22
2=
=
m
mIE
l
l
h
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The general solution is m
l
m
l ml lN im
where
IE
( ) exp= [ ]
= 2
hhere m is just
at the moment a real number
l
We must have that
represent the same point inspace
m m
(r, ) ( ,
l l
ce and r
( ) ( )
sin )
= +
+
2
2
Quantum Mechanical Rotation in 2D...
quantitative derivationSolutions to Schrdingerequation of circular motion
in spherical coordinates
k
j
i
x
y
z
r
22
2
22
2
=
=
m
mIE
l
l
h
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ml
=
( )
exp ( )
exp exp
+ =
+[ ]
[ ] [ ]
2
1
22
12
2
im
im im
l
l l
Quantum Mechanical
Rotation
in 2D...
quantitative derivation
= m
m
1
21
21
2
2
exp ( )
( )
i m
m
l
l
ll
[ ]( )
= ( )
Thus for to beequal to
m
m
m
l
l
lml
or
( )
, ; ; ;....;
( ):+
( ) =
=
2
1 1
0 1 2 3
2
k
j
i
x
y
z
r
The general solution is
ml ml lN im( ) exp= [ ]
Solutions to Schrdingerequation of circular motion
in spherical coordinates
Nml=
1
2
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We have that
mm
;
m
ll
l
= =
=
2
20 1 2
2 2IEThus E
Ih
h
; ; , ..
Quantum Mechanical
Rotation
in 2D...
quantitative derivation
Negative m values corrspondto rotation in one directionpositive m values to rotations
in the other direction. Form values of the same absolutevalue but opposite signs theenergies are the same. The two
states are degenerate
l
l
l
Properties of solutions toSchrdinger equation of circularmotion in spherical coordinates
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Quantum Mechanical
Rotation
in 2D...
quantitative derivation
The real parts of the wavefunctionsrepresents a particle on a ring.
EI
=
=
h2 2
20 1 2
m;
m
l
l ; ; , ..
Properties of solutions toSchrdinger equation of circularmotion in spherical coordinates
ml l
l l
im
m i m
( ) exp
{cos sin }
= [ ]
[ ] + [ ]
1
21
2
Note that the number of nodes in the realpart of increases with m .lm l ( )
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Quantum Mechanical
Rotation
in 2D...
quantitative derivation
L xp ypz y x= ( )
Angular momentum of solutions toSchrdinger equation of circularmotion in spherical coordinates
or quantum
xp yp
ix
yy
x
y x
mechanically
L
L
z
z
=
=
s
h s
or in spherical coordinates
Lz =h
i
We
iml l
note that
is an eigenfunction to L
m
z
( ) exp
= [ ]12
with eigenvalues;L m mz l l= = h 0 1 2; ; ;....
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What you should learn from this lecture
1. The angular momentum a particle with poition r
and momentum p is given as L = r p = m r v .
ofv
rv
r r r r 2. For a circular motion in the xy - plane the angular momentumis L (xp yp ) and pointing along the z - axis.z y x=
3
2
2. For a circular motion the total energy is all kinetic and given by
where I = mr is the moment of inertia2EL
Ikin
z=
4. The possible energies for the circular motion of
a particle are withEI
= = h
2 2
20 1 2
mm
ll ; ; , ..
51
2
.
( ) exp
The corresponding wavefunctions
are
the real part has m nodesl
ml lim
where
= [ ]
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What you should learn from this lecture
6
1
2. ( ) expThe wavefunctions
also eigenfunctions to L with eigenvalue mz l
ml limare
= [ ]h
71
2
2
. ( ) exp
.
Thus represents a system
with energy E =m
2Iand angular momentum m
m values represent counter clockwise motion
whereas negative values represent clockwise motion
l2
l
l
ml lim
Positive
= [ ]
hh
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Appendix : Quantum Mechanical Rotation in 2D...ualitative origin of quantization
The angular momentum of a particle of mass m on a circular pathof radius r in the xy-plane is represented by a vector of magnitude
prperpendicular to the plane.
We
EL
Ikin
zhave
=2
and
L prI mr
z == 2
and
L pr
I mr
z == 2
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+ =
[ ]
=
+( )
( , )
cos sin
x t AExpiEt Exp ikx
AExp
i
Et kx kx
h
h
p k= h
Consider a free pariclemoving to the right Its wavefunction is given by
Its momentum is given by
The wavelength isdetermined from the condition
+ += +( , ) ( , )x t x t
AExpiEt kx kx
AExpiEt k x k x
+( ) =
+( ) + +( )( )
h
h
cos sin
cos sin
k = 2 k = 2 /
p h= =h( / ) /2
de Broglie
Appendix : Quantum Mechanical Rotation in 2D...qualitative origin of quantization
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Consider a particle moving at a ring
Its wavefunction ( ) must satisfy
acceptable Not acceptable
( ) ( )= + 2
Same physical situation
Appendix : Quantum Mechanical Rotation in 2D...qualitative origin of quantization
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Accordingh
to QM a particle of momentum p
moves as a wave with wave length =p
Standing wave must repete itself
thus wavelength must be a integer
fraction of the circumperence
=2 r
n
de Broglie
Appendix : Quantum Mechanical Rotation in 2D...qualitative origin of quantization
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Thusr
m
h
p
2=
=2 r
m
=h
p
We have
Boundary cond.
de Broglie
m = 0 1 2 3, , , ,...
or
phm
r
m = 1,2,3,4,...= 2
0
,
Appendix : Quantum Mechanical Rotation in 2D...qualitative origin of quantization
A di Q M h i l R i i 2D
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The angular momentum of a particle confined to aplane can be represented by a vector of length |ml|
units along the z-axis and with an orientation thatindicates the direction of motion of the particle. Thedirection is given by the right-hand screw rule.
phm
r
pr m
=
= =
20
m = 1, 2, 3, 4,...
Lz
,
h
E prI
h mI
Em
I
= =
=
( )( )
,
2 2 2
2
2 2
2
0
hm = 1, 2, 3, 4,...
Appendix : Quantum Mechanical Rotation in 2D...qualitative origin of quantization
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Vibrational spectroscopy
E
1
2
h
3
2h
5
2h
7
2h
x
9
2h
11
2h
v = 0
v = 1
v = 2
v = 3
v = 4
v = 5
v = 6 E v= +h( )
1
2
A vibrating diatomicmolecule AB
Has the vibrational energy levels
=k
=
+
m m
m m
A B
A B
A photon of energy h
will be absorbed if
h = h = / 2
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Vibrational spectroscopy
Molecules in general absorbs photonsof specific frequencies (fingerprints)