Download 3rd Set of Note for Chem 655

3rd Set of Notes for Chem 655
Angular Momentum
Magnitude of the ancular momentum is related to its components by l2 = lx2 + ly2 + lz2
Previously derived the angular momentum operators
We said that lz, lx, and ly cannot all be specified simultaneously.
Lets explore their commutation relations:
[lx, ly] = [ypz - zpy, zpx-xpz]
=
similarly one finds that [ly, lz] = iћ lx and [lz, lx] = iћ ly
Lets see if each component commutes with l2
Start with lz
Find [lz2 , lz], [lx2 , lz], [ly2 , lz] and then sum them all together.
[lz2 , lz] =
A)
[lx2 , lz] =
[ly2 , lz] =
[l2 , lz] =
by symmetry
[l2 , lq] =
B)
The ;
The commutation relations in A) and B) therefore define the angular momentum
observable properties.
THUS an observable is an angular momentum if its operators satisfy these commutation
relations.
Have seen lz does not commute with either lx or ly so one can be specified if l is known
but the others cannot.
Note that l2 does commute with all of the components so the magnitude of the angular
momentum may be specified simultaneously with any of its components.
THUS we have previously mentioned the con modet to represent a state of angular
momentum with specified l and lz component but lx and ly cannot be specified.
These commutation relations can be written as:
l x l = iħ l
Ladder Operators or Shift Operators for Angular Momentum
As with the Harmonic oscillator, one can find a set of Shift Operators that will define the
properties of the angular momentum and the matrix elements of the operators.
raising operator l+ = lx + ily
lowering operator l- = lx - ily
The inverse relations are
lx = (1/2) (l+ + l-)
ly = (1/2i) (l+ - l-)
The commutators of the shift operators can be determined:
[lz, l+] =
[lx, l+] =
[lz, l-] =
Similarly
And
[l2 ,l+] = [l2 , l-] = 0
Now see how the commutation relations govern the values of the permitted eigen values
of l2 and any one of the components. We will focus on lz.
We'll find the wavefunctions and
as well.
Shift operations
ħ has the same units as and angular momentum l or lz so it make sense that lz is some
multiple of ħ
lz | ml> = ml ħ | ,ml>
Because l2 commutes with lz then |,ml> is also an eigenstate of l2
l2 still may depend on both quantum numbers.
l2 |ml> = f(, ml) ħ2 |,ml>
We will determine f (we know it is equal to _____________________).
1. Know that f is real to l2 is Hermitian
2. l2 has same units as ħ2 so f is unitless
3. l2 is sum of squares so its eigenvalues are non-negative
Also the eigenvalues of l2 - lz2 are non-negative
(l2 -lz2 )|,ml> = (lx2 +ly2 )|, ml> ≥ 0
So f(,ml) ≥ ml2
AND (l2 -lz2 )|,ml> = f(,ml)-ml2 |ħ2 |ml>
since l2 and l+ commute
l2 l+|, ml> = l+ l2 |, ml> = l+ f(, ml)ħ2 |,ml> = f(,ml)ħ2 l+|ml>
Since the eigenvalue of l2 for ,ml> is the same as for l2 the state l+|,ml>
Then l+ has no effect on the magnitude of the angular momentum when it acts.
Does it change the component on the z axis?
lzl+|ml> = (l+lz + [lz,l+]|,ml> =
So we find that l+|,ml> = c+(,ml)ħ|,ml+1>
c+ is a dimensionless numerical coeff.
Similarly find that l-|,ml> = c-(,ml)ħ|,ml-1)
Thus shift operators change ml by + /- 1 each time they operate
Also know that ml2 ≤ f, based on arguments above, so ml must have a maximum value
we will call it l.
Thus operating on state in which ml = l, must generate 0.
This gives a way to find out what l is.
Since l+|,l> = 0
Then l-l+|,l> = 0
Also
Expanding l-l+
l-l+|,l> = (l2 -lz-ħlz)|,l> = 0
Rearranging
Thus f(l) = l(l+1)
Already shown that l- acting on a state leaves the eigenvalue of l2 unaffected, sot all the
states |,l> , |,l-1>, etc all have the same value of l2 . So
f(ml) = l(l+1) for ml = l, l-1, …
so eigenvalue of l2 does not depend on the value of ml.
Know that there is a lower bound on ml because the value of lz2 cannot exceed the
eigenvalue of l2 . Suppose the lower bound is k
One can show that k = -l
Thus f(l,ml) = l(l+1) for ml = l, l-1, …, -l
So know that l2 |l,ml> = l(l+1)ħ2 |l, ml>
and from before lz|l,ml> = ml ħ|l,ml>
Just need to decide on the allowed values of l and ml. Since the shift operators step the
states |l,ml> between |l,+l > and |l, -l> in unit steps.
So it can only be integral or half integral.
So in summary on the basis of the commutation relations and Hermicity
1. Magnitude of the angular momentum is confined to the values {l(l+1)}1 / 2 ħ, with l = 0,
½, 1, … (integral or half-integral)
2. The component on an arbitrary z-axis is limited to the 2l+1 values mlħ with ml = l, l-1,
… -l
Note that the half integral quantum numbers do not necessarily apply to a particular
physical situation. For cyclical boundary conditions only integral values are admissible.
Now we consider that angular momentum can be either orbital angular momentum,
spin angular momentum, or applicable to either.
orbital angular momentum states denoted as |l,ml>
internal spin angular momentum states denotes as |s,ms>
general angular momentum will be denoted as |j,mj>
Finding the values of c+ and cthese were introduced when we explored the effects of the shift operators
j+/-|j,mj> = c+/-(j,mj)ħ|j,mj+/-1>
also <j',mj' |j,mj> =
since the states form and orthonormal set
One can show using the definition of Hermicity the
c+(j,mj) = {j(j+l) - mj(mj+1)}1 / 2
c-(j,mj) = {j(j+1)} - mj(mj-1)}1 / 2
Can use these to calculate other quantities of interest such as
<j,mj+1|jx|j,mj> or <j,mj+2|jx |j,mj> or
<j,mj+2|jx2 |j,mj>
Examples:
Let's look at the orbital angular momentum eigenfunctions in more detail
Applies to particle on a sphere, rigid rotor and is limited to integral values of the quantum
numbers l, and ml.
Show now how to get the spherical harmonics solutions using the raising and lowering
operators.
Begin by finding the wavefunction for the state with the maximum value of ml, the |l,l>
state. Then once this is found, one can use the lowering operator to get the othes.
use eqn. l+|l,l> = 0
(remember the raising and lowering operators only change ml)
in spherical polar coordinates
lx
ly
lz
and one can show that the raising and lowering operators become:
l+ = ħ ei φ (d/d + i cot d/d)
l- = -ħ ei φ (d/d - i cot d/d)
So l+|l,l> = 0
becomes
ħ ei φ (d/d + i cot d/d) ll() = 0
Perform separation of variables by:
and find that:
Both sides must equal a constant
This leads to:
ll = Nsinl e- i l φ
N is ???
Now can use l- to find the rest of the function with a given value of l.
Example:
Lets look at the spin angular momentum
Uhlenbeck and Goudsmit realized that a great simplification of the description of atomic
spectra could be obtained if electron had an intrinsic angular momentum with quantum
number
s=½
ms = + ½
the state
sz = ½ ħ
sz =
|½, ½>
and
s2  =
ms = - ½
the  state | ½ , - ½ >
s2 
and the effect of the shift operators are:
s+  = 0
s+ =
It follows that
<|s+|> =
Matrix elements
s-
s- =
Angular momentum of a system where orbital angular momenta are combined. or
spin angular momenta are combined or spin and orbital angular momenta are
combined.
What do the commutation rules imply for the total angular momentum of the combined
(coupled) system.
What is required to specify the coupled state?
The first angular momentum is specified by reporting the quantum numbers j1 and mj1
and the j2 and mj2 specify the second angular momentum that is coupled to the 1st.
The overall combined state is then |j1,mj1;j2,mj2>
Do each of these operators commute? For independent sources of angular momentum,
angular momentum components of one particle (q) coupled with angular momentum
components of a 2nd particle (q'), the operators do commute:
[j1q, j2q'] = 0
for all the components q = x, y, z and q' = x, y, z.
Because j12 and j22 can be expressed in terms of their components, they also commute
and Remember, they also commute with j1z, and j2z. That is j12 ,j1z, j22 , and j2z all
commute with one another.
So it is permissible to express the state as |j12 , mj1 ; j22 , mj2>
What about the total angular momentum of the entire system j = j1 + j2? Can it be
specified?
1st need to know if it acts like an angular momentum?
2nd need to discover the values of j that are allowed in the system
1st Task is determined by evaluating the commutators of its components
[jx,jy] = [j1x+j2x, j1y+j2y]
if this gives iħj1z and the other two are cyclic permutations
of the coordinate labels, then it acts like angular momentum
Note that j = j1 - j2 is not
Since j appears to be an angular momentum, then its magnitude is [j(j+1)]½ ħ with j
having integral or half integral values and the zth component having the magnitude mjħ
where mj = -j, -j+1, …., +j also j2 commutes with its components jz etc.
Now 2nd task - determine what values of j can exist in the system.
One question is if j1 and j2 have been specified, can j be specified??
Does j or j2 commute with j1 and j2 or j12 and j22 ?
[j2 , j12 ] = [j2 , j22 ] = 0
since both j12 and j22 both commute with all of their components and j2 can be expressed
in terms of all of those components.
Since they commute then:
Also since j2 commutes with its components including jz, jz = j1z + j2z then ml can be
specified as well as j.
So it appears that the state of the coupled angular momenta could be specified as
|j1, j2, j, ml> but can it be specified as |j1,mj1,j2,mj2;j,mj> ??
For this to be allowed then j2 must commute with the zth component of each of the
angular momenta being coupled, OR
[j1z, j2 ] (must be equal to zero) must commute.
[j1z,j2 ] = [j1z,jx2 ] + [j1z,jy2 ] + [j1z, jz2 ]
So these ________ ______________ _____________________.
Since they do not commute then if j is specified, mj1 and mj2 cannot be specified.
So need to make a choice when specifying the system, either use:
the uncoupled picture |j1m1j;j2mj2> which leaves the total angular momentum unspecified
saying nothing about the relative orientation of the
two momenta
OR the coupled picture |j1,j2;j,mj> which leaves the individual components unspecified
Either is valid, one may be more useful for a particular situation.
So now what are the permitted values of j?
If one uses the coupled picture, then what are the allowed values of j and mj
j is allowed to be integral or half integral but for a given j1 and j2 what are the allowed
values of j (answer after looking at mj
The allowed values of mj follow from jz= jz1 + jz2
and are mj = mj1 +mj2
so the total component of angular momentum about an axis is the sum of the components
of the two contributing momenta
To get the allowed values of j 1st look at the total number of states in the uncoupled
picture based on the degeneracy
(2j1+1)(2j2+1) = 4j1j2 + 2j1 + 2j2 + 1
One state where max values of mj1 and mj2 are had
mj1 = j1
and
mj2 = j2
so mj = j1+j2
Max value of mj is equal to j by definition, so j = j1 + j2
There are 2j + 1 = 2j1 + 2j2 + 1 states corresponding to this value of j, so there are 4j1j2
more states to find.
Although state mj = j1 + j2 can only occur in one way, states such as mj = j1 + j2 - 1 or mj
= j1 + j2 - 2 … down to j = |j1 - j2| (j is a positive number) can occur in multiple ways
For example the m state with mj = j1 + j2 - 1 can occur in two ways
The m state with m = mj = j1 + j2 -2 can occur in three ways
and etc.
One can reason that the permitted states of the angular momentum arising from a system
combined from two sources of angular momentum are given by the Clebsch Gordon
series
j = j1 + j2 , j1+ j2 -1, …, |j1 - j2|
Example:
Note that when the Pauli exclusion principle is operative, then there are some restrictions
on the allowed values of ms1 and ms2 if they are in the same l state. Here one must use
microstates to determine the allowed values.
Vector model
Attempt to represent the features of coupled angular momenta deduced from the
commutation relations.
Features of te vector diatrams of coupled momenta.
1. The length of the vector representing the total angular momentum is {j(j+1)}½ , with j
one of the values permitted by the Clebsch-Gordon series.
2. The vector must lie at an indeterminate angle on a cone about the z axis, (jx and jy
cannot be specified if jz is specified.
3. The lengths of the contributing angular momenta vectors ar {j1(j1+1)}1 / 2 and
{j2(j2+1)}1 / 2 . These have definite values even when j is specified.
4. The projection of the total angular momentum on the z-axis is mj; in the coupled
picture (in which j is specified), the values of mj1 and mj2 are indefinite bust their sum is
equal to mj
In the uncoupled picture (j not specified), the individual components mj1 and mj2 may be
specified, and their sum is equal to mj.
See example figures
Important example that comes up is for two elecxtrons each with spin ½ .
In the coupled picture the electrons may be in any of the four states
12,12,12,12,
see figure
Now in the coupled picture the Clebsch Gordan series says the the toal spin S can have
values of 1 and 0. When S = 0 only one possible value of the z-component, only 0
Ms=0, a Single. When S= 1 Ms = 1, 0, -1, three possible states so a triplet
See figures


Relation between coupled and uncoupled schemes
The state |j1j2;jmj> is built from all values of mj1 and mj2 where mj1 + mj2 = mj
So seems reasonable to expect that the coupled state is a sum over all the uncoupled
states |j1j2:j,mj> which have mj1 + mj2 = mj
OR
|j1j2;jmj> =  Cmj1mj2 |j1mj1;j2mj2>
The Cmj1mj2 are called the Clebsch-Gordan coeffeicients , Wigner coefficients or 3jsymbols
Consider Singlet and Triplet coupled states
of two spin ½ particles
See table
Coupling of several angular momenta
What happens for three or more momenta coupled together.
Three momenta could
1. couple j1 and j2 to form j12 and then couple to j3 to give the overall j.
OR
2. couple j1 to j3 to form j13 and then couple to j2 to give the overall j.
The states obtained in these ways are different, but the states obtained by the first
coupling procedure can be expressed as linear combinations of the states obtained by the
second procedure
The coefficients of the linear combination are known as the Rach coefficients or the 6jsymbols.