8. spin and adding angular momentum we assumed that |r formed a basis and r( )|r = | r from...
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8. Spin and Adding Angular Momentum
• We assumed that |r formed a basis and R()|r = |r • From this we deduced R()(r) = (Tr) • Is this how other things work?• Consider electric field from a point particle• Can we rotate by R()E(r) = E(Tr)?
– Let’s try it• This is not how electric fields rotate• It is a vector field, we must also rotate the field components
– R()E(r) = E(Tr)• Maybe we have to do something
similar with ?
The Assumptions We Made 8A. Rotations Revisited
+
TR D r rR R R
• How do the D()’s behave?
• We want to find all matrices satisfying this relationship• Easy to show: when = 1, D() = 1• As before, Taylor expand D for small angles
• In a manner similar to before, then show
Spin Matrices
1 2 1 2R R R r rR R R R
2ˆ ˆ, 1D i O n n SR
TR D r rR R R
1 2 2 1 2 1 2
TTR D D r rR R R R R R R
1 2 2 1 1 2 2 1T T T TD D D r rR R R R R R R R 1 2 1 2D D DR R R R
ˆ ˆ, expD i n n SR
• We used to have identical expressions for the angular momentum L• From these we proved that L has the standard commutation relations• It follows that S has exactly the same commutation relations
• The three S’s are generalized angular momentum• But in this case, they really are finite dimensional matrices• Logically, our wave functions would now be labeled• But s is a constant, so just label them• There are 2s + 1 of them total:
We Already Know the Spin Matrices
,i j ijk kk
S S i S
1 2 1 2D D DR R R R ˆ ˆ, expD i n n SR
, , etc.x y zS S i S
, ss msm
1,
s
s
s
t
r
rr
r
• Recall, for angular momentum, we had to restrict l to integers, not half-integers
• Why? Because wave functions had to be continuous• Can we find a similar argument for spin? Consider s = ½
• Consider a rotation by 2:• This would imply if you rotate by 2, the state vector changes by | – | • But these states are indistinguishable, so this is okay!• Any value of s, integer or half-integer, is fine• The basic building blocks of matter are all s = ½• Other particles have other spins
Restrictions on s?
1 2 1 2D D DR R R R ˆ ˆ, expD i n n SR
~m imlY e
12S σ 1
2ˆ ˆ, expD i n n σR 1 1
2 2ˆcos sini n σ
ˆ ˆ, 2 cos sinD i n n σR 1
part. se- ½p+ ½ n0 ½
part. s 1, 0 0 ’s 3/2
• Basis states used to be labeled by |r• But now we must label them also by which
component we are talking about |r,ms• Comment: for spin ½, it is common to abbreviate the ms label:• The spin operators affect only the spin label:
• Operators that concern position, like R, P, and L, only affect the position label
• All these position operators must commute with spin operators
Basis States for Particles With Spin r r
,sm sm r r
2 2 2
2 2
, , , , , ,
, , 1
s s z s s s
s s s s
m s s m S m m m
S m s s m m m
S r r r r
r r
, ,s sm mR r r r
, , , 0i j i j i jS R S P S L
12, , r r
Sample ProblemDefine J = L + S. Find all commutators of J, J2, S2, and L2
• That’s 6 operators, so 65/2 = 15 possible commutators– I’ll just do five of them to give you the idea
,x yJ J ,x x y yL S L S , , , ,x y x y x y x yL L L S S L S S
0 0z zi L i S z z zi L S i J
2 , S J 2 , S L S 2 , S S
• Recall, for any angular momentum-like set of operators, [J2,J] = 0
0
2 2, S J 0
• Recall our Hamiltonian:• Note that S commutes with the Hamiltonian• We can diagonalize simultaneously H, L2, Lz, S2, and Sz:• It is silly to label them by s, because s = ½
• Degeneracy: ms takes on two values,doubling the degeneracy
Do all Hamiltonians commute with spin?• No! Magnetic interactions care about spin• Even hydrogen has small contributions (spin-orbit coupling) that depend on spin
Hydrogen Revisited2
21
2ek e
H
PR
, , , , sn l m s m
2 2 2
2 2 2
, , , , , , , ,
, , , , , , , ,
, , , , , , , ,
, , , , , , , ,
, , , , , , , ,
s n s
s s
z s s
s s
z s s s
H n l m s m E n l m s m
n l m s m l l n l m s m
L n l m s m m n l m s m
n l m s m s s n l m s m
S n l m s m m n l m s m
L
S
2 234 S , , , sn l m m
22nD E n
• Recall that:• Rewrite this in ket notation
• Define J:
• J is what actually generates rotations• If a problem is rotationally invariant, we would expect J to commute with H
– Not necessarily L or S
What Generates Rotations? TR D r rR R R
ˆ ˆ ˆ, , expR D i n n n LR R
ˆ ˆexp expi i n S n L
8B. Total Angular Momentum and Addition
ˆexp i n S L
J L S ˆ ˆ, expR i n n JR
Consider the rotation of the Earth around the Sun:•It has orbital angular momentumfrom its orbit around the Sun: L•It has spin angular momentumfrom its rotation around the axis: S•The total angular momentum is•It is another set of angular momentum-like operators•It will have eigenvectors |j,m with eigenvalues:
•Because L and S typically don’t commute with theHamiltonian, we might prefer to label our states byJ eigenvalues, which do•To keep things as general as possible, imagine any two angular momentum operators adding up to yield a third:
What are L, S and J?
J L S ,i j ijk kk
J J i J
2 2 2, ,
, ,z
j m j j j m
J j m m j m
J
1 2 J J J
• Commutation relations:• We could label states by their eigenvalues under the following four commuting
operators:
• Instead, we’d prefer to label them by the operators– These all commute with each other
• These have the same j1 and j2 values, so we’ll abbreviate them:
Two things we want to know:• Given j1 and j2, what will the states |j,m be?• How do we convert from one basis to another, i.e., what is:
– Clebsch-Gordan coefficients
Adding Angular Momentum,ai bj ab ijk ak
k
J J i J
2 21 2 1 2, , ,z zJ JJ J 1 2 1 2, ; ,j j m m
2 2 21 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2, ; , , ; , , , ; , , ; , .a a a az aj j m m j j j j m m J j j m m m j j m m J
1 2 J J J
2 2 21 2, , , zJJ J J
,j m
2 2 2, , , , , .zj m j j j m J j m m j m J
1 2 1 2, ; , ,j j m m j m
• It is easy to figure out what the eigenvalues of Jz are, because
• For each basis vector |j1,j2;m1,m2, there will beexactly one basis vector |j,m with m = m1 + m2
• The ranges of m1 and m2 are known• From this we can deduce exactly how many basis
vectors in the new basis have a given value of m• By looking at the distribution of m values, we
can deduce what j values must be around• Easier illustrated by doing it than describing it
The procedure
1 2z z zJ J J 1 2 J J J
1 2 1 2 1 2 1 2 1 2, ; , , ; ,z z zJ j j m m J J j j m m 1 2 1 2 1 2, ; ,m m j j m m
1 2 1 2,m m m m m
1 2 1 2, :N m N m m m m m
1 1 1 2 2 2, , , ,m j j m j j
, , for each value of .m j j j
Sample ProblemSuppose j1 = 2 and j2 = 1, and we change basis from |j1,j2;m1,m2 to |j,m.(a) What values of m will appear in |j,m, and how many times?(b) What values of j will appear in |j,m, and how many times?
• First, find a list of all the m1 and m2 values that occur– I will do it graphically
• Now, use the formula m = m1 + m2 to find the mvalue for each of these points
• From these, deduce the m values and how manythere are– I will do it graphically
• Note where the transitions are:
, ,a a am j j
1m
2m
m=3
m=2m=1
m=0m=-1
m=-2m=-3
m0
1 2j j1 2j j2 1j j1 2j j
Sample Problem (2)Suppose j1 = 2 and j2 = 1, and we change basis from |j1,j2;m1,m2 to |j,m.(a) What values of m will appear in |j,m, and how many times?(b) What values of j will appear in |j,m, and how many times?
• For any value of j, m will run from –j to j• Clearly, there is no j bigger than 3• But since m = 3 appears, there must be j= 3• This must correspond to m’s from –3 to 3 • Now, there are still states with m up to 2• It follows there must also be j = 2• This covers another set of m’s from –2 to 2• What remains has m up to 1• It follows there must be j = 1• And that’s it.Why did it run from j = 3 to j = 1?• Because it went from j1 + j2 down to j1 – j2
, ,m j j
m0
j0
• The set of all (m1,m2) pairs forms a rectangle• The largest value of m is m = j1 + j2, which can only happen one way• As m decreases from the max value, there is one more way of making each m
value for each decrease in m until you get to | j1 – j2 |• This implies that you get maximum jmax = j1 + j2 and minimum jmin = | j1 – j2 |• So, j runs from | j1 – j2 | to
j1 + j2 in steps of size 1
General Addition of Angular Momentum 1 2 J J J
1m
2m
1j1j
2j
2jm
1 2j j1 2j j 1 2j j1 2j j
1 2 1 2 1 2, 1, ,j j j j j j j
1 2 1 2 1 2 1 21j j j j j j j j
• For fixed j1 and j2, the number of basis vectors |j1,j2;m1,m2 is How many basis vectors |j,m are there?• For each value of j, there are 2j + 1.• Therefore the total is
Check Dimensions1 2 1 2 1 2, 1, ,j j j j j j j
1 22 1 2 1j j
1 2
1 2
2 1j j
j j j
D j
1 2
1 2
2 21j j
j j j
j j
2 22 2 2 2
1 2 1 2 1 2 1 2 1 2 1 21 1 1j j j j j j j j j j j j
2 2
1 2 1 21j j j j 1 2 1 2 1 2 1 21 1j j j j j j j j 1 22 1 2 1j j
• So dimensions work out
Sample ProblemSuppose we have three electrons. Define the total spin as S = S1 + S2 + S3. What are the possible values of the total spin s, the corresponding eigenvalues of S2, and how many ways can each of them be made?
• Electrons have spin s = ½, so • Combine the first two electrons:• Now add in the third:
• If s1+2 = 0, this says:
• If s1+2 = 1, this says
• Final answer for s:– The repetition means there are two ways to combine to make s = ½
• For S2:
11 2 3 2s s s
1 2 1 2 1 2, ,s s s s s 0, ,1 0,11 2 3 1 2 3, ,s s s s s
1 12 20 , ,0s 1 1
2 2, , 12
1 12 21 , ,1s 31
2 2, , 312 2,
31 12 2 2, ,s
2 2 2, ,s ss m s s s m S 2 2 2 23 3 154 4 4, ,S
• Recall hydrogen states labeled by• Because of relativistic corrections, these aren’t eigenstates• Closer to eigenstates are basis states
– j = l ½ • States with different mj are related by rotation
– Indeed, the value of mj will depend on choice of x, y, z axis– And they are guaranteed to have the same energy
• Therefore, when labelling a state we need to specify n, l, j• We label l values by letters, in a not obvious way
– Good to know the first four: s, p, d, f• We then denote j by a subscript, so example state could be 4d3/2
• Remember restrictions: l < n and j = l ½ • Often, we don’t care about j, so just label it 4d• Remember, number of states for given n,l is
Hydrogen Re-Revisited, , , sn l m m
, , , jn l j m
l let0 s1 p2 d3 f4 g5 h6 i7 k8 l9 m10 n11 o12 q
, 2 2 1d n l l
• We wish to interchange bases |j1,j2;m1,m2 |j,m• These are complete orthonormal basis states in the same vector space• We can therefore use completeness either way
• The coefficients are called Clebsch-Gordancoefficients, or CG coefficients for short
• Our goal: Show that we can find them (almost) uniquely• Note that the states |j1,j2;m1,m2 are all related by J1
and J2
– There are no arbitrary phases concerning how they are related• The |j,m states with the same j’s and different m’s are related by J• But there is no simple relation between |j,m’s different j’s – convention choice
How do we change bases? 8C. Clebsch-Gordan Coefficients
1 2 1 2 1 2 1 2, ; , , , , ; ,j m
j j m m j m j m j j m m1 2
1 2 1 2 1 2 1 2, , ; , , ; , ,m m
j m j j m m j j m m j m
1 2 1 2, ; , ,j j m m j m
• If you ever have to look them up, be warned, different sources use different notations
• Recall that the other states are alsoeigenstates of J1
2 and J22
• People also get lazy and drop some commas
• In addition, the Clebsch-Gordan coefficients are defined only up to a phase– Everyone agrees on phase up to sign
• As long as you use them consistently, itdoesn’t matter which convention you use.
• They will turn out to be real, and therefore• Because of this ambiguity, people get lazy and often use what is logically the
wrong one
Convention Confusion
1 2 1 2 1 1 2 2 1 1 2 2, ; , , ; , , ,j j m m j m j m j m j m
1 2, , ; ,j m j j j m
1 2 1 2j j m m jm
1 2 1 2 1 2 1 2, , ; , , ; , ,j m j j m m j j m m j m
When are the coefficients meaningful and (probably) non-zero?(1) j range:
(2) m range:
(3) conservation of Jz:•Let’s prove the last one using
•Act on the left with Jz and on the right with J1z and J2z:
•Must be zero unless
Nonzero Clebsch-Gordan (C-G) Coefficients1 2 1 2, ; , ,j j m m j m 1 2 J J J 1 2 1 2 1 2, 1, ,j j j j j j j
1 2 1 2j j j j j
1 1 1 2 2 2, ,j m j j m j j m j
1 2m m m
1 2z z zJ J J 1 2 1 2 1 2 1 2 1 2, , ; , , , ; ,z z zj m J j j m m j m J J j j m m
1 2 1 2 1 2 1 2 1 2, , ; , , , ; ,m j m j j m m j m j j m m m m
1 2 1 2 1 20 , , ; ,m m m j m j j m m
1 2m m m
j1 + j2 – j is an integer
j – m is an integer, etc.
• Largest value for m is j, therefore
• Recall in general
• We therefore have
• Recall: only if m1 + m2 = m (= j) are non-zero• This relates all the non-zero terms for m = j, all relative sizes determined• To get overall scale, use normalization
• This determines everything up to a phase– We arbitrarily pick
Finding C-G Coefficients for m = j1 2 J J J 1 2 1 2 1 2, 1, ,j j j j j j j
, 0J j j 0 ,j j J
2 2, , 1J j m j j m m j m
1 2 1 20 , , ; ,j j J j j m m 1 2 1 2 1 2, , ; ,j j J J j j m m
2 2 2 21 1 1 1 1 2 1 2 2 2 2 2 1 2 1 20 , , ; 1, , , ; , 1j j m m j j j j m m j j m m j j j j m m
1 , ,j j j j1 2
1 2 1 2 1 2 1 2, , ; , , ; , ,m m
j j j j m m j j m m j j1
2
1 2 1 1, , ; ,m
j j j j m j m
1 2 1 1, , ; , 0j j j j j j j
1 2 1 2, ; , ,j j m m j m
• We now have CG coefficients when m = j• I will now demonstrate that if we have them for m, we can get them for m – 1 • First note
• Dagger this• So
• So if we know them for m, we know them for m – 1• Since we know them for m = j, we know them for m = j – 1, j – 2, etc. • Hence we have a (painful) procedure for finding all CG coefficients• Sane people don’t do it this way, they look them up or use computers
Finding C-G Coefficients for m – 1 from m1 2 J J J
2 2, , 1J j m j j m m j m 2 2 , 1 ,j j m m j m j m J
2 21 2 1 2 1 2 1 2, 1 , ; , , , ; ,j j m m j m j j m m j m J j j m m
1 2 1 2 1 2, , ; ,j m J J j j m m
2 2 2 21 1 1 1 1 2 1 2 2 2 2 2 1 2 1 2, , ; 1, , , ; , 1j j m m j m j j m m j j m m j m j j m m
1 2 1 2, ; , ,j j m m j m
• Adding j1 and j2 is thesame as adding j2 and j1
– Corollary: if j1 = j2, then the combinations of spins is symmetric if j1 + j2 – j is even, anti-symmetric if it is odd
• You can work your way up from m = –j in the same way we worked our way down from m = j:
• Adding j1 = 0 or j2 = 0 is pretty trivial,because these imply J1 = 0 or J2 = 0
• If you ever look things up in tables, they will assume j1 j2 > 0, and assume you will use the first or third rule to get other CG coefficients
• Or you can use computer programs to get them
Properties of CG-coefficients
1 2
1 2 1 2 2 1 2 1, ; , , 1 , ; , ,j j j
j j m m j m j j m m j m
1 2
1 2 1 2 1 2 1 2, ; , , 1 , ; , ,j j j
j j m m j m j j m m j m
,,0; ,0 , 0, ;0, , m mj m j m j m j m
> clebsch(1,1/2,1,-1/2,3/2,1/2); 31 1 1 12 2 2 2 31, ;1, ,
1 2 1 2, ; , ,j j m m j m
• For j2 small, we can find simple formulas for the CG coefficients• If j2 = ½, then j = j1 ½
CG coefficients when j2 = ½
• Example:
• For one electron, J = L + S. Let j1 l, m mj, drop j2 = s = ½, m2 = ½
• For adding two electron spins, drop s1 and s2, abbreviate mi = ½
1 11 12 21 1 1 1 1 1 1
1 1 12 2 2 2 2 2 21 1
, , ; , , ; ,2 1 2 1
j m j mj m j m j m
j j
1 12 21 1 1
2 2 2, , , , ,2 1 2 1
j jj j j
l m l ml m l m l m
l l
12
12
1, 1 , 1,0 , 1, 1
0,0
11 21 1 1 1 1 1 1 1
1 1 1 12 2 2 2 2 2 2 21
, ; , , , ; , ,2 1
j mj m j m j m j m
j
Sample ProblemHydrogen has a single electron in one of the states |n,l,m,ms = |2,1,1,– or |2,1,0,+ , or in one of the states |n,l,j,mj = |2,1,3/2,1/2 or |2,1,1/2,1/2 . In all four cases, write explicitly the wave function
• For s = ½, wave function looks like • Spin state ms tells us which component exists• This lets us immediately write the wave function
for the first two:
• For the |j,mjstates we have:
1 02,1,1, 21 1 2,1,0, 21 1
0 1, , ,
1 0R r Y R r Y
1 1 1 12 2 2 23 1 1 1 1 1
2 2 2 2 2 2
1 12,1, , 2,1, , 2,1, ,
2 1 1 2 1 1
1 23 32,1,1, 2,1,0,
1 1 1 12 2 2 21 1 1 1 1 1
2 2 2 2 2 2
1 12,1, , 2,1, , 2,1, ,
2 1 1 2 1 1
2 13 32,1,1, 2,1,0,
1 12 21 1 1
2 2 2, , , , ,2 1 2 1
j jj j j
l m l ml m l m l m
l l
Sample Problem (2)… or in one of the states |n,l,j,mj = |2,1,3/2,1/2 or |2,1,3/2,1/2 . In all four cases, write explicitly the wave function
1 02,1,1, 21 1 2,1,0, 21 1
0 1, , ,
1 0R r Y R r Y
3 1 1 22 2 3 32,1, , 2,1,1, 2,1,0, 1 1 2 1
2 2 3 32,1, , 2,1,1, 2,1,0,
3 1
2 2
01
212,1, , 11
2 ,1
,3
YR r
Y
1 12 2
01
212,1, , 11
,1
3 2 ,
YR r
Y
• You can also get the CG coefficients from Maple:
> clebsch(1,1/2,1,-1/2,3/2,1/2);> clebsch(1,1/2,0,1/2,3/2,1/2);> clebsch(1,1/2,1,-1/2,1/2,1/2);> clebsch(1,1/2,0,1/2,1/2,1/2);
Sample ProblemHydrogen has a single electron in the state |n,l,j,mj = |2,1,3/2,1/2. If one of the following is measured, what would the outcomes and corresponding probabilities be, and what would the state afterwards look like: E, J2, Jz, L2, S2, Lz,Sz
• For the first five choices, our state is an eigenstate of the operator
• The eigenstate will be unchanged by this measurement• For the last two, we write it in terms of
eigenstates of Lz or Sz
• Then we have• State afterwards is
• Or we have• State afterwards is
3 1 1 22 2 3 32,1, , 2,1,1, 2,1,0,
2
13.6 eV3.40 eVE
n 2 2 2 215
4j j J 12z jJ m 2 2 2 22l l L
2 2 2 234s s S
231 1
2 2 22,1,1, 2,1, ,z zP L P S 13
2,1,1, 2
31 12 2 20 2,1,0, 2,1, ,z zP L P S 2
32,1,0,
• A scalar operator S is anything that is unchanged under rotation• Examples:• Scalar operators commute with the generator of rotations J:• Vector operators V are operators that rotate like a vector:• Examples: • They have commutation relations with J given by• A rank 2 tensor Tij under rotation rotates as:• Can show that
• Rank k tensor has k indicesand commutation relations:
• Scalar = rank 0 tensor, Vector = rank 1 tensor• Rank 2 tensor is sometimes just called a “tensor”
Definition and Commutation with J 8D. Scalar, Vector, Tensor
†R SR SR R 2 2 2 2 2, , , , , ,V R P R L S J L S
, 0S J
†R R V VR R R, , , ,R P L S J
,i j ijk kkJ V i V
†ij ik jl kl
k l
R T R TR R R R
,i jk ijl lk ikl jllJ T i T T
1 2 1 2 3 2 1 1 2
,k k k ki j j j ij l lj j j ij l j l j ij l j j ll
J T i T T T
• If V and W are any two vector operators,then we can define a rank-2 tensor operator:– One can similarly define higher rank tensor operators
• This tensor has nine independent components
• But it has pieces that aren’t very rank-2 tensor-like:– Dot product VW is a scalar operator– Cross product VW is a vector operator– The remaining five pieces are the truly rank-2 part
• We want figure out how to extract the various pieces
How to Make a Tensor From Vectors
ij i jT VW
,i jk ijl lk ikl jllJ T i T T
• We start with a vector operator V• Define the three operators Vq by:• You can then show the following:
– Proof by homework problem• Compare this with:• Another way to write it:
Generalize this formula• Define a spherical tensor of rank k as 2k + 1 operators:• It must have commutation relations:
• Trivial example: A scalar is a spherical tensor of rank 0
Spherical Tensors
10 1 2
,z x yV V V V iV
21, , , 2 .z q q q qJ V qV J V q qV
21, 1, , 1, 2 1, 1zJ q q q J q q q q 1
1
, 1, 1,q qq
V V q q
J J
, 1, ,kqT q k k k
, , ,k kq q
q
T T k q k q
J J , k kz q qJ T qT
2 21, k k
q qJ T k k q qT
00, 0S J
• Those matrix elements are CG coefficients• Proof:
Combining Spherical Tensors (1)
, , ,k kq q
q
T T k q k q
J J
Theorem: Let V and W be spherical tensors of rank k1 and k2 respectively. Then we can build a new spherical tensor T of rank k defined by: 1 2
1 2
1 2
1 2 1 2,
, ; , ,k k kq q q
q q
T V W k k q q k q
1 2
1 2
1 2
1 2 1 2,
, , , ; , ,k k kq q q
q q
T V W k k q q k q J J
1 2 1 2
1 2 1 2
1 2
1 2 1 2,
, , , ; , ,k k k kq q q q
q q
V W V W k k q q k q J J
1 2 1 2
1 2 1 2
1 2 1 2
1 1 1 1 2 2 2 2 1 2 1 2,
, , , , , ; , ,k k k kq q q q
q q q q
V W k q k q V W k q k q k k q q k q
J J
1 2
1 2 2 2 1 1
1 2 1 2
1 1 1 1 , 2 2 2 2 , 1 2 1 2, ,
, , , , , , ; , ,k k kq q q q q q q
q q q q
T V W k q k q k q k q k k q q k q
J J J
Combining Spherical Tensors (2)
1 2
1 2 2 2 1 1
1 2 1 2
1 1 1 1 , 2 2 2 2 , 1 2 1 2, ,
, , , , , , ; , ,k k kq q q q q q q
q q q q
T V W k q k q k q k q k k q q k q
J J J
1 2
1 2
1 2 1 2
1 2 1 2 1 1 2 1 2
1 2 1 2, , 1 2 1 2 2 1 2 1 2
, ; , , ; ,, ; , ,
, ; , , ; ,k k
q qq q q q
k k q q k k q qV W k k q q k q
k k q q k k q q
J
J
• We have complete set of states |k1,k2;q1,q2
• Now insert complete set of states |k’,q’:
• J doesn’t change the k value, so k’ = k• So we have proven it
1 2
1 2
1 2
1 2 1 2 1 2,
, , ; , ,k k kq q q
q q
T V W k k q q k q
J J J
, , ,k kq q
q
T T k q k q
J J
1 2
1 2
1 2
1 2 1 2,
, ; , ,k kq q
q q
V W k k q q k q
J
1 2
1 2
1 2
1 2 1 2, ,
, , ; , , , ,k k kq q q
k q q q
T V W k k q q k q k q k q
J J
1 2
1 2
1 2
1 2 1 2,
, ; , ,k k kq q q
q q
T V W k k q q k q
,
, ,kq
k q
T k q k q
J
, ,kq
q
T k q k q
J
How it Comes Out
• This sum only makes sense if CG coefficients are non-zero
• Only non-zero terms are when q1 + q2 = q– So it’s really just a single sum
• By combining two vectors, we can get k = 0, 1, 2– k = 0: Scalar (dot product)– k = 1: Vector (cross product)– k = 2: Truly rank 2 tensor part
• We can then combine rank 2 tensors with more vectors to make rank 3 spherical tensors
1 2 1 2k k k k k
1 2
1 2
1 2
1 2 1 2,
, ; , ,k k kq q q
q q
T V W k k q q k q
Sample ProblemIf we combine two copies of the position operator R, what are the resulting components of the rank-2 spherical tensor Tq
(2)?
1 2
1 2
1 2
1 2 1 2,
, ; , ,k k kq q q
q q
T V W k k q q k q 10 1 2
,z x yV V V V iV
1(1) 10 1 2
,R Z R X iY
2 1 12 1 1 1,1;1,1 2,2T R R 21
2 1X iY 2 21 12 2X iXY Y
2 1 1 1 11 1 0 0 11,1;1,0 2,1 1,1;0,1 2,1T R R R R 1 1 1
2 2 2X iY Z XZ iYZ
2 1 1 1 1 1 10 1 1 0 0 1 11,1;1, 1 2,0 1,1;0,0 2,0 1,1; 1,1 2,0T R R R R R R
21 1 1 22 36 6
X iY X iY Z 2 2 216
2Z X Y 2 1 1 1 11 0 1 1 01,1;0, 1 2, 1 1,1; 1,0 2, 1T R R R R 1 2
2 2X iY Z XZ iYZ
2 1 12 1 1 1,1; 1, 1 2, 2T R R 21
2 1X iY 2 21 12 2X iXY Y
• Suppose we have an atom or other rotationally invariant system• Eigenstates should be eigenstates of J2, Jz, probably other stuff• It is common to need matrix elements
of operators between these states:• We know how the ket and bra rotate• If we also know how the operator in the middle rotates, we should be able to find
relations between these various quantities• Suppose the operator is a spherical tensor,
or combinations thereof• Then we know how T rotates, and we should be able to find relations• This helps us because:
– If the calculation is hard, we do it a few times and deduce the rest– If the calculation is impossible, we measure it a few times and deduce the rest
Why it should work 8E. The Wigner-Eckart Theorem
, ,j m
, , , ,j m O j m
, , , ,kqj m T j m
Similarities With CG coefficients (1)
• I want to compare the matrix element above to the CG coefficient above
• Recall relations for Jz:
• Use commutationrelation:
• Let Jz act on the braor the ket on the left:
• Hence matrix elements are zero unless
• Compare to the CG coefficient above:– This vanishes unless
, , , ,kqj m T j m
, k kz q qJ T qT , , , ,zJ j m m j m
, , , , , , , ,k k kz q q z qj m J T T J j m j m qT j m
, , , , , , , ,k kq qm m j m T j m q j m T j m
m q m
, , ; ,j m j k m q
m q m
Similarities With CG coefficients (2)
• Recall relations for J:• For m = j,
• Implies:• Our commutation relations tell us:
• Compare to the CG coefficients:
, , , ,kqj m T j m
2 21, k k
q qJ T k k q qT
2 2, , , , 1J j m j j m m j m
, , ; ,j m j k m q
2 2, , , , , , , ,k k kq q qj j J T T J j m k k q q j j T j m
, , 0J j j
, , 0j j J
2 2 2 210 , , , , 1 , , , ,k k
q qj j m m j j T j m k k q q j j T j m
2 2 2 20 , , ; 1, , , ; , 1j j m m j j j k m q k k q q j j j k m q
Similarities With CG coefficients (3)
• We have:• Equivalent to
• Our commutation relations tell us:
, , , ,kqj m T j m
2 21, k k
q qJ T k k q qT 2 2, , , , 1J j m j j m m j m
, , ; ,j m j k m q
2 2, , , , 1J j m j j m m j m
2 2, , , , 1j m J j j m m j m
1 2 21, , , , , , , ,k k k
q q qj m J T T J j m k k q q j m T j m
2 2
2 2 2 21
, , 1 , ,
, , , , 1 , , , ,
kq
k kq q
j j m m j m T j m
j j m m j m T j m k k q q j m T j m
2 2
2 2 2 2
, 1 , ; ,
, , ; 1, , , ; , 1
j j m m j m j k m q
j j m m j m j k m q k k q q j m j k m q
These Matrix Elements are CG Coefficients
• We have three relations that are identical for these two expressions:
• These expressions were all that were used to find the CG coefficients– Plus, we had a normalization condition
• Hence, these two expressions are identical– Up to normalization
, , , ,kqj m T j m , , ; ,j m j k m q
m q m 2 2 2 2
10 , , , , 1 , , , ,k kq qj j m m j j T j m k k q q j j T j m
2 2
2 2 2 21
, , 1 , ,
, , , , 1 , , , ,
kq
k kq q
j j m m j m T j m
j j m m j m T j m k k q q j m T j m
, , , , , , ; ,kqj m T j m j m j k m q
The Wigner-Eckart Theorem
What can the proportionality constant depend on?•Not m, m’, nor q•It can depend on , ’, j, j’, and of course T •The Wigner-Eckart Theorem:
•The square root in the denominator is a choice, neither right nor wrong•That other thing is called a “reduced matrix element”•You don’t calculate it (directly)
– You may be able to calculate left side for one value of m, m’, q– Or you may be able to measure left side for one value of m, m’, q
•Then you deduce the reduced matrix element from this equation•Then you can use it for all the other values of m, m’, q
, , , , , , ; ,kqj m T j m j m j k m q
1, , , , , , , ; , ,
2 1k k
qj m T j m j T j j k m q j mj
Why Is the Wigner-Eckart Theorem Useful?
• The number of matrix elements is – For example, if j = 3, j’ =2, k = 1, this is 105 different matrix elements
• Calculating them computationally may be difficult or impossible• Measuring them may be a great deal of work• By doing one (difficult) computation or one (difficult) measurement you can
deduce a lot of others
Comment: why is the factor of 2j + 1 there?• If T0
(k) is Hermitian, then you can show
1, , , , , , , ; , ,
2 1k k
qj m T j m j T j j k m q j mj
2 1 2 1 2 1j k j
*
, , , ,k kj T j j T j
Sample ProblemThe magnetic dipole transition of hydrogen causing the 21 cm line is governed by the matrix element , where F is the total angular momentum quantum number and mF is the corresponding z-component, and S and L are spin and orbital angular quantum operators for the electron. Deduce as much as you can about these matrix elements for mF = +1, 0, or –1.
121 , 0, 0 1 , 1,F Fs F m s F m S L
We have no idea what most of this means, but it’s clear:•F and mF are angular quantum number, effectively, j F and m mF
•S and L are vector operators
•Call reduced matrix element A:•Non-vanishing only if q + mF = 0•Get the CG coefficients from program
•All other matrix elements vanish
12 V S L 1 1 1
0 1 2,z x yV V V V iV
1 112 0 1
1 ,0,0 1 ,1, 1 ,0 1 ,1 1,1; , 0,0q F Fs V s m s V s m q
1,1; , 0,0FA m q
1 11 3
1 10 3
1 11 3
1 ,0,0 1 ,1, 1
1 ,0,0 1 ,1,0
1 ,0,0 1 ,1, 1
s V s A
s V s A
s V s A
Sample ProblemDeduce as much as you can about these matrix elements for mF = +1, 0, or –1.
121 , 0, 0 1 , 1,F Fs F m s F m S L
• For mF = + 1, we also have
• Vx and Vy: two equations, two unknowns• We therefore have:
• Similarly:
1 12 3
13
1 12 3
1 ,0,0 1 ,1, 1
1 ,0,0 1 ,1,0
1 ,0,0 1 ,1, 1
x y
z
x y
s V iV s A
s V s A
s V iV s A
1
21 ,0,0 1 ,1, 1 0
1 ,0,0 1 ,1, 1 0
x y
z
s V iV s
s V s
16
16
1 ,0,0 1 ,1, 1
1 ,0,0 1 ,1, 1
x
y
s V s A
s V s iA
1 1
2 6ˆ ˆ1 ,0,0 1 ,1, 1s s A i S L x y
1 12 6
ˆ ˆ1 ,0,0 1 ,1, 1s s A i S L x y
1 12 3
ˆ1 ,0,0 1 ,1,0s s A S L z
• Consider the product of any two spherical harmonics:• By completeness, this can be written as a sum of spherical harmonics:
• The coefficients clm can be found usingorthogonality:• Think of the expression as an operator acting on a wave function:• It is not hard to see that this operator is a spherical tensor operator
• Think of clm then as a matrix element• By the Wigner-Eckart theorem:• All that remains is to find
the reduced matrix elements
Products of Spherical Harmonics 8F. Integrals of Spherical Harmonics
1 2
1 2,
, , ,m m ml l lm l
l m
Y Y c Y 1 2
1 2
*, , ,m mm
lm l l lc d Y Y Y
1 2
1 2, ,m m
l lY Y 2 1
2 1,m m
l lY Y
2 2 2 2
2 2 2 2
12 22 2 2 2 2, , , , , , ,m m m m
z l l l lL Y m Y L Y l l m m Y
2
11 1 2 1 22 1
, , ; ,lm llc l Y l l m l l m m
1 2
1 2, ,m m
l lY Y
2
2 1 1, ,mlm lc l m Y l m
Working on the Reduced Matrix Element
• Substitute the top equation in the bottom
• Multiply this expression by and sum over m1, m2:
• Rename l’ as l
1 2
1 2 2
* 11 1 2 1 22 1
, , , , , ; ,m mmlm l l l ll
c d Y Y Y l Y l l m l l m m
1 2
1 2 2
11 1 2 1 22 1
,
, , , , ; , ,m m ml l l ll
l m
Y Y l Y l l m l l m m Y
1 2 1 2, ; , ,0l l m m l
1 2
1 2,
, , ,m m ml l lm l
l m
Y Y c Y
1 2
1 2
1 2
2
1 2
1 2 1 2,
11 1 2 1 2 1 2 1 22 1
, ,
, , , ; , ,0
, , ; , , ; , ,0 ,
m ml l
m m
ml ll
m m l m
Y Y l l m m l
l Y l l m l l m m l l m m l Y
sum over complete states
1 2
1 2 2
1 2
11 2 1 2 12 1
, ,
, , , ; , ,0 , ,0 ,m m ml l l ll
m m l m
Y Y l l m m l l Y l l m l Y
2
01
1,
2 1l ll Y l Y
l
Finishing the Computation
• Must be true at all angles• Evaluate at = 0• Formula for the Y’s at = 0 is simple
• Now we solve for thereduced matrix element
• We therefore have
1 2
1 2 2
* 11 1 2 1 22 1
, , , , , ; ,m mml l l ll
d Y Y Y l Y l l m l l m m
1 2
1` 2 2
1 2
01 2 1 2 1
,
1, , , ; , ,0 ,
2 1m m
l l l lm m
Y Y l l m m l l Y l Yl
,0
2 10,
4m
l m
lY
2
1 21 2 1
2 1 2 1 1 2 1, ;0,0 ,0
4 4 42 1l
l l ll l l l Y l
l
2 1 1 2 1 2
12 1 2 1 , ;0,0 ,0
4ll Y l l l l l l
1 2
1 2
* 1 21 2 1 2 1 2
2 1 2 1, , , , ;0,0 ,0 , ; , ,
4 2 1m mm
l l l
l ld Y Y Y l l l l l m m l m
l
When doesn’t it vanish?
We want to know when this is non-zero, or likely to be non-zero:•We need:
•We need:
•Under parity, each of the spherical harmonics transforms to•So the whole integral satisfies
•We need
1 2
1 2
* 1 21 2 1 2 1 2
2 1 2 1, , , , ;0,0 ,0 , ; , ,
4 2 1m mm
l l l
l ld Y Y Y l l l l l m m l m
l
1 2m m m
1 2 1 2l l l l l
1 lm ml lY Y
1 21 2 1 2
1 2 1 2
* *, , , 1 , , ,
l l lm m m mm ml l l l l ld Y Y Y d Y Y Y
1 2 evenl l l 1 2 1 2 1 2 1 2, 2, 4, ,l l l l l l l l l
Sample ProblemAtoms usually decay spontaneously by the electric dipole process, in which case the rate is determined by the matrix element , whereI and F are the initial and final states. For hydrogen in each of the following states, which states might be the final n and l quantum states if the initial state is: 4s, 4p, 4d, 4f?
F I R
• The initial state has n’ = 4 and l’ = 0, 1, 2, or 3• Final state has unknown n and l, but
– Must have n < 4 because energy goes down– Must have l < n
• The position operators can be written in terms of l = 1 spherical harmonics • So we have
• To not vanish, we need• For 4s, l = 0:• Must have l < n < 4
4, , 4 ,ml m l lR r Y
, , ,mn l m nl lR r Y
*2, 4, 10
, , ,m q mF I n l l l lR r rR r r dr d Y Y Y
R
1 1 and 1 evenl l l l l 1 1 and 0 1 evenl l 1l
4 2 ,3s p p
4l n
Sample Problem (2)
• For 4p: l’ = 1, so • Must have l < n < 4
• For 4d: l’ = 2, so • Must have l < n < 4
• For 4f:• Must have l < n < 4
1 1 and 1 evenl l l l l 4l n
0 2 and 1 1 evenl l 0,2l 4 1 ,2s,3s,3dp s
1 3 and 2 1 evenl l 1,3l
4 2 ,3d p p
2 4 and 3 1 evenl l 2,4l 4 3f d
Atoms usually decay spontaneously by the electric dipole process, in which case the rate is determined by the matrix element , whereI and F are the initial and final states. For hydrogen in each of the following states, which states might be the final n and l quantum states if the initial state is: 4s, 4p, 4d, 4f?
F I R