Spherical Harmonics and Selection Rule

Chemistry 180-345A

Rules for all problem sets
B. C. Sanctuary

1. All problems sets must be hand-written. Computer printed material is not
acceptable.
2. Do all problems sequentially and draw a line across the page between problems.
3. If the answer is numerical, place it in a box.
4. Any written exercises must use proper, grammatically correct, English.
5. Identify yourself, with student number on each page.
6. Staple all pages together.
7. The problems sets must be handed in on time. Overdue problem sets will have
10% deducted from the grade.

PLEASE:
Discuss with your peers but do not copy
Do not wait until the last day. It is a good idea to let these ideas mature, so start early.
Do not miss class because you have not got the problems done on time.


Chemistry 180-345A

Problem set 5
(Maximum 50)

Due: November 17, 2004
at 9:30 (no later)

1. (5 points) Explain why the one electron atoms have energy that is degenerate
with respect to l and ml, but not with respect to n in the absence of a magnetic
field.

(Motivation: When we get to the multielectron atoms, the energy of the various
states depends on both n and l. This is because when there is more than one
electron, there are electron-electron repulsions. These disrupt the orbital angular
momentum and lift the degeneracy with respect to l.)

2. (5 points) The range of the polar coordinates are 0 r , 0 , and
0 2 and the range of the Cartesian coordinates are - < x, y, z < . In the
integrals we must relate the two. This is done by the Jacobian, and the result is,

dxdydz = r 2 dr sin d d

Look up the definition of the Jacobian and prove the above result.
(motivation: Jacobians are the general way of making sure that the volume element
in one coordinate system is the same in another. In the case of polar coordinates, the
volume element is often called the "solid angle".




3. (5 points) Use the generating functions for Legendre polynomials and work out
the first 5 (n=0 to 4)
1 dn 2
n (
z - 1)
n
Pn ( z ) = n
2 n ! dz

and
( n + 1) Pn+1 ( z ) - ( 2n + 1) zPn ( z ) + nPn-1 ( z ) = 0
(Motivation: Many functions have a certain structure that is maintained for higher
orders. The generating functions relate these functions together. In the case of
Legendre polynomials, the basic symmetry is rotational invariance. You have studied
point groups and found character tables and studied symmetry. Rotations also form a
group, but rotations are continuous and so the group is not a point group
(discontinuous jumps from one symmetry to another). The different orders of
Legendre polynomials have unique rotational symmetry: l=0, scalar; l=1, vector; l=2,
quadrupole; l=3, octapole, etc.)

4. (5 points) Check orthogonality and the expression for l=0 to 3 (10 easy integrals,
but it might be easier to change variables to x=cos.)

2
d sin P ( cos ) P (cos ) = 2l + 1
0
l m lm




(Motivation: it is useful to prove to yourself that these really are orthogonal for the
first few terms. Then you feel comfortable with this property.)

5. (5 points) Work out the associated Legendre functions for l=0,1,2 using

d ml
Pl ml ( ) = ( -1) l (1 - 2 ) ml / 2 Pl ( )
m

d ml
for positive m only.

(Motivation: another example of the many generating functions that start from either
low orders or known functions from which others or higher orders can be obtained).
6. (5 points) Prove the relationship:

cos ab = cos a cos b + sin a sin b cos ( a - b )
where ab is the angle between two unit vectors a and b and each of these is
oriented in an arbitrary coordinate system with anlges a a and b b
respectively.
(Motivation: Recall the angle between two vectors is independent of any
coordinate system. However, we usually do want to specify a coordinate system
that orients the two vectors with the polar angles. There is a simple way of
finding the relation between the two, and that is what you have to try to find out.)

7. (5 points) Use the spherical harmonics for l=0,1,2 and verify the addition
theorem

4 l
Pl (cos ) = l Ylm ( R ,R )Ylm ( r , r )
(2l + 1) m =-
*



Compare the result for l=1 to the result in question 6 and show that they give the
same result.
(Motivation: I want you to remember the importance of the addition theorem. It is
very useful. In question 6, you found the relationship between the angle between two
vectors and their separate orientations in a coordinate system. Here that result is
generalized from vectors to all types of rotational symmetry. Note that a product of
two spherical arise and these separate the angle dependences as seen in problem 6.)

8. (5 points) Using the definition for the moments, and write down the actual
expressions for l=0,1,2.(9 cases)

4
M lm = ( r ) (r)Ylm ( r , r )dr
l

(2l + 1)
(Since the electron density, or charge distribution, (r ) , is not specified, just leave the
integrals for each case.)

Motivation: First I am not sure if the above is the accepted definition of moments only
because of the numerical prefactor. I simply used the addition theorem and separated the
R and r dependences and combined the two separately. Still, it is, up to a constant, the
usually definition. By working out the first few, note that the spherical harmonics
determine the symmetry: dipole (vector), quadrupole, etc. Suppose, for example, that the
charge distribution is a dipole, and this is a vector, then the orthogonality of the spherical
harmonics, (problem 9), means that only the l=1 term will survive. That is, the symmetry
of the charge distribution, (r ) , determines the symmetry of the problem.)

9. (10 points) Consider the orthonormality of the spherical harmonics
2

Y ( , )Yl m ( , ) sin d d = ll mm
*
lm
0 0

This can be expressed as a matrix element that can be written in Dirac notation as
2
lm l m = Ylm ( , )Yl m ( , ) sin d d = ll mm
*

0 0



In spectroscopy, we treat atoms and molecules quantum mechanically and
calculate transition probabilities which we use to understand spectra. If a
molecule has a dipole moment, it is given by µe = qz = q cos where q is a
constant charge. Calculate the matrix element for this by using this property,

cos Ylm ( , )
1/ 2 1/ 2
1 l - m -1 (l + m + 1)(l - m + 1) 1 l - m (l + m)(l - m)
= ( -1) Yl +1m ( , ) + ( -1) Yl -1m ( , )
3 (2l + 3)(2l + 1) 3 (2l + 1)(2l - 1)


That is, evaluate the transition probability amplitude

lm µe l m = q lm cos l m
Find the selection rules, (that is the relationship between lm and l m which are
called the electric dipole selection rules).

2
Also find the probability of the transition, lm µe l m .

Finally work out the probability for a transition: 2s 2p and for 2p 3p .

(Motivation: after they have built their apparatus, this is what the majority of
spectroscopists spend a lot of time doing. Computer programs and general
methods are used for these calculations. The matrix elements determine all the
selection rules (which transitions can occur and which cannot), and give the
strength (intensity) of the transition. This is not the only condition (the Bohr
frequency condition must also be satisfied that simply says that the energy of the
incoming light (frequency) must match the difference between the two energy
levels in the molecule. Although other types of states exist, in addition to the
angular momentum states described in this problem (e.g. rotation, vibration,
electronic, Raman, electron spin (EPR), nuclear spin (NMR), and many others),
the technique is the same. Exceptions occur with intense lasers when the response
is non-linear, but this is rare.)

Chemistry 180-345A

Rules for all problem sets
B. C. Sanctuary

1. All problems sets must be hand-written. Computer printed material is not
acceptable.
2. Do all problems sequentially and draw a line across the page between problems.
3. If the answer is numerical, place it in a box.
4. Any written exercises must use proper, grammatically correct, English.
5. Identify yourself, with student number on each page.
6. Staple all pages together.
7. The problems sets must be handed in on time. Overdue problem sets will have
10% deducted from the grade.

PLEASE:
Discuss with your peers but do not copy
Do not wait until the last day. It is a good idea to let these ideas mature, so start early.
Do not miss class because you have not got the problems done on time.


Chemistry 180-345A

Problem set 5
(Maximum 50)

Due: November 17, 2004
at 9:30 (no later)

1. (5 points) Explain why the one electron atoms have energy that is degenerate
with respect to l and ml, but not with respect to n in the absence of a magnetic
field.

(Motivation: When we get to the multielectron atoms, the energy of the various
states depends on both n and l. This is because when there is more than one
electron, there are electron-electron repulsions. These disrupt the orbital angular
momentum and lift the degeneracy with respect to l.)

2. (5 points) The range of the polar coordinates are 0 r , 0 , and
0 2 and the range of the Cartesian coordinates are - < x, y, z < . In the
integrals we must relate the two. This is done by the Jacobian, and the result is,

dxdydz = r 2 dr sin d d

Look up the definition of the Jacobian and prove the above result.
(motivation: Jacobians are the general way of making sure that the volume element
in one coordinate system is the same in another. In the case of polar coordinates, the
volume element is often called the "solid angle".




3. (5 points) Use the generating functions for Legendre polynomials and work out
the first 5 (n=0 to 4)
1 dn 2
n (
z - 1)
n
Pn ( z ) = n
2 n ! dz

and
( n + 1) Pn+1 ( z ) - ( 2n + 1) zPn ( z ) + nPn-1 ( z ) = 0
(Motivation: Many functions have a certain structure that is maintained for higher
orders. The generating functions relate these functions together. In the case of
Legendre polynomials, the basic symmetry is rotational invariance. You have studied
point groups and found character tables and studied symmetry. Rotations also form a
group, but rotations are continuous and so the group is not a point group
(discontinuous jumps from one symmetry to another). The different orders of
Legendre polynomials have unique rotational symmetry: l=0, scalar; l=1, vector; l=2,
quadrupole; l=3, octapole, etc.)

4. (5 points) Check orthogonality and the expression for l=0 to 3 (10 easy integrals,
but it might be easier to change variables to x=cos.)

2
d sin P ( cos ) P (cos ) = 2l + 1
0
l m lm




(Motivation: it is useful to prove to yourself that these really are orthogonal for the
first few terms. Then you feel comfortable with this property.)

5. (5 points) Work out the associated Legendre functions for l=0,1,2 using

d ml
Pl ml ( ) = ( -1) l (1 - 2 ) ml / 2 Pl ( )
m

d ml
for positive m only.

(Motivation: another example of the many generating functions that start from either
low orders or known functions from which others or higher orders can be obtained).
6. (5 points) Prove the relationship:

cos ab = cos a cos b + sin a sin b cos ( a - b )
where ab is the angle between two unit vectors a and b and each of these is
oriented in an arbitrary coordinate system with anlges a a and b b
respectively.
(Motivation: Recall the angle between two vectors is independent of any
coordinate system. However, we usually do want to specify a coordinate system
that orients the two vectors with the polar angles. There is a simple way of
finding the relation between the two, and that is what you have to try to find out.)

7. (5 points) Use the spherical harmonics for l=0,1,2 and verify the addition
theorem

4 l
Pl (cos ) = l Ylm ( R ,R )Ylm ( r , r )
(2l + 1) m =-
*



Compare the result for l=1 to the result in question 6 and show that they give the
same result.
(Motivation: I want you to remember the importance of the addition theorem. It is
very useful. In question 6, you found the relationship between the angle between two
vectors and their separate orientations in a coordinate system. Here that result is
generalized from vectors to all types of rotational symmetry. Note that a product of
two spherical arise and these separate the angle dependences as seen in problem 6.)

8. (5 points) Using the definition for the moments, and write down the actual
expressions for l=0,1,2.(9 cases)

4
M lm = ( r ) (r)Ylm ( r , r )dr
l

(2l + 1)
(Since the electron density, or charge distribution, (r ) , is not specified, just leave the
integrals for each case.)

Motivation: First I am not sure if the above is the accepted definition of moments only
because of the numerical prefactor. I simply used the addition theorem and separated the
R and r dependences and combined the two separately. Still, it is, up to a constant, the
usually definition. By working out the first few, note that the spherical harmonics
determine the symmetry: dipole (vector), quadrupole, etc. Suppose, for example, that the
charge distribution is a dipole, and this is a vector, then the orthogonality of the spherical
harmonics, (problem 9), means that only the l=1 term will survive. That is, the symmetry
of the charge distribution, (r ) , determines the symmetry of the problem.)

9. (10 points) Consider the orthonormality of the spherical harmonics
2

Y ( , )Yl m ( , ) sin d d = ll mm
*
lm
0 0

This can be expressed as a matrix element that can be written in Dirac notation as
2
lm l m = Ylm ( , )Yl m ( , ) sin d d = ll mm
*

0 0



In spectroscopy, we treat atoms and molecules quantum mechanically and
calculate transition probabilities which we use to understand spectra. If a
molecule has a dipole moment, it is given by µe = qz = q cos where q is a
constant charge. Calculate the matrix element for this by using this property,

cos Ylm ( , )
1/ 2 1/ 2
1 l - m -1 (l + m + 1)(l - m + 1) 1 l - m (l + m)(l - m)
= ( -1) Yl +1m ( , ) + ( -1) Yl -1m ( , )
3 (2l + 3)(2l + 1) 3 (2l + 1)(2l - 1)


That is, evaluate the transition probability amplitude

lm µe l m = q lm cos l m
Find the selection rules, (that is the relationship between lm and l m which are
called the electric dipole selection rules).

2
Also find the probability of the transition, lm µe l m .

Finally work out the probability for a transition: 2s 2p and for 2p 3p .

(Motivation: after they have built their apparatus, this is what the majority of
spectroscopists spend a lot of time doing. Computer programs and general
methods are used for these calculations. The matrix elements determine all the
selection rules (which transitions can occur and which cannot), and give the
strength (intensity) of the transition. This is not the only condition (the Bohr
frequency condition must also be satisfied that simply says that the energy of the
incoming light (frequency) must match the difference between the two energy
levels in the molecule. Although other types of states exist, in addition to the
angular momentum states described in this problem (e.g. rotation, vibration,
electronic, Raman, electron spin (EPR), nuclear spin (NMR), and many others),
the technique is the same. Exceptions occur with intense lasers when the response
is non-linear, but this is rare.)