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Last modified: November 18, 2021 ✓
Clebsch-Gordan Coefficients
---------------------------
The problem of angular momentum addition requires
the calculation of Clebsch-Gordan (CG) coefficients
Consider the addition of orbital and spin angular
momenta for two particles that reside in uncoupled
eigenstates. For example, consider spin orbit
coupling where J = L + S. The momentum can transfer
between L and S such that J is conserved. We know
how to find eigenalues of L and S by themselves but
how do we find the eigenvalues of the combined states
To accomplish this we construct a product state
composed of combinations of these individual states
Classically, angular momentum is:
L = r x p
Which is:
Lx = ypz - zpy
and so on.
Quantum mechanically this becomes:
Lx = -ih(y∂/∂z - z∂/∂y)
In Quantum Mechanics orbital angular momentum,
spin angular momentum and isospin share the same
mathematics. We will refer to this 'generic'
angular momentum as J.
The angular momentum operators satisfy the following
commutator:
[Ji,Jj] = ihεijkJk
Consider:
[Ji(1),Jj(1)] = ihεijkJk(1) in vector space V1
[Ji(2),Jj(2)] = ihεijkJk(2) in vector space V2
We can then form a combined operator as follows:
J = Ji(1) ⊗ 1 + 1 ⊗ Jj(2) in vector space V1 ⊗ V2
These also have the same commutator relationships.
Proof:
[Ji(1) ⊗ 1 + 1 ⊗ Ji(2),Jj(2) ⊗ 1 + 1 ⊗ Jj(2)]
= [Ji(1) ⊗ 1,Jj(1) ⊗ 1] + [1 ⊗ Ji(2),1 ⊗ Jj(2)]
= [Ji(1),Jj(1)] ⊗ 1 + 1 ⊗ [Ji(2),Jj(2)]
= ihεijk(Jk(1) ⊗ 1 + 1 ⊗ Jk(2))
= ihεijkJk
Like any vector, a magnitude can be defined for
J, the total momentum operator as:
J2 = Jx2 + Jy2 + Jz2
J2 is the Casimir invariant with the property:
[J2,Jx] = [J2,Jy] = [J2,Jz] = 0
Raising and Lowering Operators
------------------------------
The raising and lowering operators are:
J+ = Jx + iJy
J- = Jx - iJy
L+ = Lx + iLy
L- = Lx - iLy
S+ = Sx + iSy
S- = Sx - iSy
With the following commutators:
[Jz,J±] = ±hJ±
[J+,J-] = 2hJz
Since J2 and Jz commute they share common eigenstates
but will have different eingenvalues. Therefore, we
will need 2 indeces for each basis vector: j for J2
and m for Jz. j and m label the states so we can
write:
J2|j,m> = j|j,m>
and,
Jz|j,m> = m|j,m>
Therefore,
|j,m> ≡ |eigenvalue of J2|eigenvalue of Jz>
Recall [Jz,J+] = ihJ+
JzJ+ - J+Jz = ihJ+
JzJ+ = ihJ+ + J+Jz
JzJ+|j,m> = (ihJ+ + J+Jz)|j,m>
= ihJ+|j,m> + J+Jz|j,m>
But Jz|j,m> = m|j,m>. Therefore,
JzJ+|j,m> = ihJ+|j,m> + J+m|j,m>
Therefore,
Jz(J+|j,m>) = (m + ih)(J+|j,m>)
So (J+|j,m>) is an eigenvector of Jz with eigenvalue
(m + ih).
Similarly,
Jz(J-|j,m>) = (m - ih)(J-|j,m>)
Therefore, the effect of J± is to increase or decrease
the eigenvalue of Jz by the amount h. Let us now
consider J2.
[J2,J+] = 0
J2J+ - J+J2 = 0
J2J+ = J+J2
J2J+|j,m> = J+J2|j,m>
= J+j|j,m>
= jJ+|j,m>
Therefore,
J2(J+|j,m>) = j(J+|j,m>
So (J+|j,m>) is an eigenvector of J2 with eigenvalue,
j.
Similarly,
J2(J-|j,m>) = j(J-|j,m>
Therefore, the net result is that J± has no effect
on the eigenvalue of J2, j.
Summarizing, J± change the eigenvalue of Jz but not
J2!!
Raising and Lowering the State, M
---------------------------------
We now need to figure out what J±|j,m> produces.
We know that these operators must raise or lower
the state, M. Therefore, we can write
J±|j,m> ∝ |j,m ± 1>
or,
J+|j,m> = C+|j,m + 1>
and,
J-|j,m> = C-|j,m - 1>
Consider:
J-J+|j,m> = (Jx - iJy)(Jx + iJy)|j,m>
= Jx2 + Jx2 + i[Jx,Jy]|j,m>
But Jx2 + Jx2 = J2 - Jz2 and [Jx,Jy] = ihJz
Therefore,
J-J+|j,m> = (J2 - Jz2 - hJz)|j,m>
There is an eigenstate |j,mmax> which cannot be
raised and a state |j,mmin> that cannot be lowered.
Because J+|j,mmax> = 0, J-J+|j,mmax> = 0. Therefore,
(J2 - Jz2 - hJz)|j,mmax> = 0
J2|j,mmax> - Jz2|j,mmax> - hJz|j,mmax> = 0
α|j,mmax> - mmax2|j,mmax> - hmmax|j,mmax> = 0
(α - mmax2 - hmmax)|j,mmax> = 0
(α - mmax2 - hmmax) = 0
α = mmax2 + hmmax
Because J-|j,mmin> = 0, J+J-|j,mmin> = 0 we can go
through the same procedure to get:
α = mmin2 - hmmin
If there are n steps between mmax and mmin then:
2mmax = nh ∴ mmax = nh/2
Now α = mmax2 + hmmax so,
α = n2h2/4 + nh2/2
= nh/2(nh/2 + h)
If we let j = n/2 and we get:
α = h2j(j + 1)
Therefore,
J2|j,m> = h2j(j + 1)|j,m>
<j,m|J+†J+|j,m> = <j,m|J-J+|j,m>
= <j,m|J2 - Jz2 - hJz|j,m>
= <j,m|J2 - Jz2 - hJz|j,m>
= <j,m|h2j(j + 1) - h2m2 - h2m|j,m>
= (j(j + 1) - h2m2 - h2m)<j,m|j,m>
= h2(j(j + 1) - m(m + 1))<j,m|j,m>
= C+2
Therefore,
J+|j,m> = h√[j(j + 1) - m(m + 1)]|j,m + 1>
Now, if we repeat the process for J+J- we get:
J+J- = (Jx + iJy)(Jx - iJy)
= J2 - Jz2 - i[JxJy]
= J2 - Jz2 + hJz
Which leads to:
C-2 = h2(j(j + 1) - m(m - 1))
and,
J-|j,m> = h√[j(j + 1) - m(m - 1)]|j,m - 1>
C± are the CLEBSCH-GORDAN coefficients. The C-G
coefficients represent the probability amplitude
for the corresponding product state.
Using the raising and lowering operators involves
starting at a particular state, applying the above
formulas and then equating
C±|j,m ± 1> = J±|state>
Where the RHS is the result of applying the
operator J2 = J ⊗ 1 + 1 ⊗ J for 2 momenta and
J3 = J2 ⊗ 1 + 1 ⊗ J for 3 momenta.
As an example consider combining the spins of
2 electrons:
|1,1> = |↑↑>
S-|1,1> = √[1(1 + 1) - 1(1 - 1)]|1,1 - 1>
= √2|1,0>
(S- ⊗ 1 + 1 ⊗ S-)|↑↑> = I-|↑>↑ + ↑I-|↑>
= |↓↑> + |↑↓>
Therefore,
√2|1,0> = |↓↑> + |↑↓>
or,
|1,0> = √(1/2)(|↓↑> + |↑↓>)
Now consider combining the isospins of 3 quarks
(same mathematics as spin):
|3/2,-3/2> = |uuu>
I-|uuu> = I-|u>uu + uI-|uu>
= duu + u[I-|u>u + uI-|u>]
= duu + u[du + ud]
= duu + udu + uud
and,
I-|3/2,3/2> = √[3/2(3/2 + 1) - (3/2)(3/2 - 1)]|3/2,1/2>
= √3|3/2,1/2>
Therefore,
√3|3/2,1/2> = duu + udu + uud
or,
|3/2,1/2> = √(1/3)(duu + udu + uud)
Angular Momentum Matrices
-------------------------
We can use J+ and J- to determine the matrix form
of the angular momentum operators. Consider the
possible states for spin 1/2.
- -
|1/2,1/2> = | 1 |
| 0 |
- -
- -
|1/2,-1/2> = | 0 |
| 1 |
- -
We start by raising from the lowest state:
S+|1/2,-1/2> = h√[1/2(1/2 + 1)
- (-1/2)(-1/2 + 1)]|1/2,1/2>
= h|1/2,1/2>
We now look for a matrix that carries out the
same thing:
- - - - - -
h/2| 0 1 || 0 | = h/2| 1 |
| 0 0 || 1 | | 0 |
- - - - - -
We continue by lowering the highest state.
S-|1/2,1/2> = h√[1/2(1/2 + 1)
- (1/2)(1/2 - 1)]|1/2,-1/2>
= h|1/2,-1/2>
We now look for a matrix that carries out the
same thing:
- - - - - -
h/2| 0 0 || 1 | = h/2| 0 |
| 1 0 || 0 | | 1 |
- - - - - -
Sx = (1/2)(S+ + S-) gives:
- -
Sx = h/2| 0 1 |
| 1 0 |
- -
Sy = (1/2i)(S+ - S-) gives:
- -
Sy = h/2| 0 -i |
| i 0 |
- -
For Sz we use:
Sz|s,m> = m|s,m> and find the following:
Sz|1/2,-1/2> = -h/2|1/2,-1/2>
- - - - - -
Sz = h/2| 1 0 || 1 | = h/2| 1 |
| 0 -1 || 0 | | 0 |
- - - - - -
- - - - - -
Sz = h/2| 1 0 || 0 | = h/2| 0 |
| 0 -1 || 1 | | 1 |
- - - - - -
These are the familiar Pauli matrices.
Finally,
- -
S2 = 3h2/4| 1 0 |
| 0 1 |
- -
Compare this with S2|1/2,1/2> = h2(1/2)(1/2 + 1)|1/2,1/2>
= 3h2/4
Now consider the 3 possible basis states for l = 1.
- -
| 1 |
|1,1> = | 0 |
| 0 |
- -
- -
| 0 |
|1,0> = | 1 |
| 0 |
- -
- -
| 0 |
|1,-1> = | 0 |
| 1 |
- -
We start by raising from the lowest state:
L+|1,-1> = h√[1(1 + 1) - (-1)(-1 + 1)]|1,0>
= h√2|1,0>
L+|1,0> = h2√[1(1 + 1) - 0]|1,9>
= h√2|1,1>
We now look for a matrix that carries out the
same thing:
- - - - - -
| 0 √2 0 || 0 | | 0 |
h| 0 0 √2 || 0 | = h√2| 1 |
| 0 0 0 || 1 | | 0 |
- - - - - -
- - - - - -
| 0 √2 0 || 0 | | 1 |
h| 0 0 √2 || 1 | = h√2| 0 |
| 0 0 0 || 0 | | 0 |
- - - - - -
We continue by lowering the highest state.
L-|1,1> = h√[1(1 + 1) - (1)(1 - 1)]|1,0>
= h√2|1,0>
L-|1,0> = h2√[1(1 + 1) - 0]|1,-1>
= h√2|1,-1>
We now look for a matrix that carries out the
same thing:
- - - - - -
| 0 0 0 || 1 | | 0 |
h| √2 0 0 || 0 | = h√2| 1 |
| 0 √2 0 || 0 | | 0 |
- - - - - -
- - - - - -
| 0 0 0 || 0 | | 0 |
h| √2 0 0 || 1 | = h√2| 0 |
| 0 √2 0 || 0 | | 1 |
- - - - - -
Lx = (1/2)(L+ + L-) gives:
- -
| 0 1 0 |
Lx = h/√2| 1 0 1 |
| 0 1 0 |
- -
Ly = (1/2i)(L+ - L-) gives:
- -
| 0 1 0 |
Ly = h/i√2| -1 0 1 |
| 0 -1 0 |
- -
For Lz we use:
Lz|l,m> = m|l,m> and find the following:
- - - - - -
| 1 0 0 || 1 | | 1 |
h| 0 0 0 || 0 | = +h| 0 |
| 0 0 -1 || 0 | | 0 |
- - - - - -
- - - - - -
| 1 0 0 || 0 | | 0 |
h| 0 0 0 || 1 | = 0| 1 |
| 0 0 -1 || 0 | | 0 |
- - - - - -
- - - - - -
| 1 0 0 || 0 | | 0 |
h| 0 0 0 || 0 | = -h| 0 |
| 0 0 -1 || 1 | | 1 |
- - - - - -
- -
| 1 0 0 |
Lz = h| 0 0 0 |
| 0 0 -1 |
- -
Finally,
- -
| 1 0 0 |
L2 = 2h2| 0 1 0 |
| 0 0 1 |
- -
Compare this with L2|1,1> = h21(1 + 1)|1,1>
= 2h2
NOTE: THESE Lx, Ly AND Lz NATRICES SHOULD
NOT BE CONFUSED WITH THE MATRICES FOR SO(3).
Irreducible Representations
---------------------------
The Clebsch–Gordan coefficients define a unitary
transformation that allow us to decompose the
tensor product space into a direct sum of irreducible
representations of the angular momentum. To show
this we use the example of combining orbital and spin
angular momentum.
We start with the highest and lowest states:
|3/2,3/2> = |1,1> ⊗ |1/2,1/2>
|3/2,-3/2> = |1,-1> ⊗ |1/2,-1/2>
Using the formula for lowering the state |3/2,3/2> we
get:
J-|3/2,3/2> = √3|3/2,1/2> ... 1.
and,
J-|1,1> ⊗ |1/2,1/2>
= L-|1,1> ⊗|1/2,1/2> + |1,1> ⊗ S-|1/2,1/2>
= √2|1,0> ⊗ |1/2,1/2> + |1,1> ⊗ |1/2,-1/2> ... 2.
Equating 1 and 2 gives:
|3/2,1/2> = √(2/3){|1,0> ⊗ |1/2,1/2>} + √(1/3){|1,1> ⊗ |1/2,-1/2>}
Likewise raising the state |3/2,-3/2> gives:
|3/2,-1/2> = √(2/3){|1,0> ⊗ |1/2,-1/2>} + √(1/3){|1,-1> ⊗ |1/2,1/2>}
We use orthogonality with |3/2,1/2> to get:
|1/2,1/2> = √(1/3){|1,0> ⊗ |1/2,1/2>} - √(2/3){|1,1> ⊗ |1/2,-1/2>}
and orthogonality with |3/2,-1/2> to get:
|1/2,-1/2> = -√(1/3){|1,0> ⊗ |1/2,-1/2>} + √(2/3){|1,-1> ⊗ |1/2,1/2>}
We can write the raising operator in matrix form as:
- -
⊗ | 0 0 0 1 0 0 |
⊗ | 0 0 0 0 1 0 |
S+ ⊗ I = | 0 0 0 0 0 1 |
⊗ | 0 0 0 0 0 0 |
⊗ | 0 0 0 0 0 0 |
⊗ | 0 0 0 0 0 0 |
- -
and,
- -
⊗ | 0 √2 0 0 0 0 |
⊗ | 0 0 √2 0 0 0 |
I ⊗ L+ = | 0 0 0 0 0 0 |
⊗ | 0 0 0 0 √2 0 |
⊗ | 0 0 0 0 0 √2 |
⊗ | 0 0 0 0 0 0 |
- -
J+ = S+ ⊗ I + I ⊗ L+ produces:
- -
| 0 √2 0 1 0 0 |
| 0 0 √2 0 1 0 |
J+ = | 0 0 0 0 0 1 |
| 0 0 0 0 √2 0 |
| 0 0 0 0 0 √2 |
| 0 0 0 0 0 0 |
- -
- - - - - -
| 0 √2 0 1 0 0 || 0 | | 0 |
| 0 0 √2 0 1 0 || 0 | | 0 |
| 0 0 0 0 0 1 || 0 | = | 1 |
| 0 0 0 0 √2 0 || 0 | | 0 |
| 0 0 0 0 0 √2 || 0 | | √2 |
| 0 0 0 0 0 0 || 1 | | 0 |
- - - - - -
This corresponds to:
√3|3/2,-1/2> = √2{|1,0> ⊗ |1/2,-1/2>} + 1{|1,-1> ⊗ |1/2,1/2>}
Likewise, we can write the lowering operator in
matrix form as:
- -
| 0 0 0 0 0 0 |
| √2 0 0 0 0 0 |
J- = | 0 √2 0 0 0 0 |
| 1 0 0 0 0 0 |
| 0 1 0 √2 0 0 |
| 0 0 1 0 √2 0 |
- -
- - - - - -
| 0 0 0 0 0 0 || 1 | | 0 |
| √2 0 0 0 0 0 || 0 | | √2 |
| 0 √2 0 0 0 0 || 0 | = | 0 |
| 1 0 0 0 0 0 || 0 | | 1 |
| 0 1 0 √2 0 0 || 0 | | 0 |
| 0 0 1 0 √2 0 || 0 | | 0 |
- - - - - -
This corresponds to:
√3|3/2,1/2> = √2{|1,0> ⊗ |1/2,1/2>} + 1{|1,1> ⊗ |1/2,-1/2>}
It is easy to see that the elements of the resulting
column vector has the following values.
- -
| - | <- |1,1> ⊗ |1/2,1/2> = |3/2,3/2>
| - | <- |1,0> ⊗ |1/2,1/2> = |3/2,1/2>
| - | <- |1,-1> ⊗ |1/2,1/2> = |3/2,-1/2>
| - | <- |1,1> ⊗ |1/2,-1/2> = |3/2,1/2>
| - | <- |1,0> ⊗ |1/2,-1/2> = |3/2,-1/2>
| - | <- |1,-1> ⊗ |1/2,-1/2> = |3/2,-3/2
- -
For completeness we also show the matrix for JZ:
- -
| 3/2 0 0 0 0 0 |
| 0 1/2 0 0 0 0 |
JZ = | 0 0 -1/2 0 0 0 |
| 0 0 0 1/2 0 0 |
| 0 0 0 0 -1/2 0 |
| 0 0 0 0 0 -3/2 |
- -
We can now go to a new basis by applying unitary
transformation, U. This matrix puts the transpose
of the column vectors into successive rows.
- -
| 1 0 0 0 0 0 |
| 0 √(2/3) 0 √(1/3) 0 0 |
U = | 0 0 √(1/3) 0 √(2/3) 0 |
| 0 0 0 0 0 1 |
| 0 √(1/3) 0 -√(2/3) 0 0 |
| 0 0 √(2/3) 0 -√(1/3) 0 |
- -
- - - - - -
| 1 0 0 0 0 0 || a | | a |
| 0 √(2/3) 0 √(1/3) 0 0 || b | | √(2/3)b + √(1/3)d |
| 0 0 √(1/3) 0 √(2/3) 0 || c | = | √(1/3)c + √(2/3)e |
| 0 0 0 0 0 1 || d | | f |
| 0 √(1/3) 0 -√(2/3) 0 0 || e | | √(1/3)b - √(2/3)d |
| 0 0 √(2/3) 0 -√(1/3) 0 || f | | √(2/3)c - √(1/3)e |
- - - - - -
Where:
a = |3/2,3/2>
b = |1,1> ⊗ |1/2,-1/2>
c = |1,-1> ⊗ |1/2,1/2>
d = |1,0> ⊗ |1/2,1/2>
e = |1,0> ⊗ |1/2,-1/2>
f = |3/2,-3/2>
- -
| 1 0 0 0 0 0 |
| 0 √(2/3) 0 0 √(1/3) 0 |
U† = | 0 0 √(1/3) 0 0 √(2/3) |
| 0 √(1/3) 0 0 -√(2/3) 0 |
| 0 0 √(2/3) 0 0 -√(1/3) |
| 0 0 0 1 0 0 |
- -
For the operators we get:
- -
| 0 √3 0 0 0 0 |
| 0 0 2 0 0 0 |
UJ+U† = | 0 0 0 √3 0 0 |
| 0 0 0 0 0 0 |
| 0 0 0 0 0 1 |
| 0 0 0 0 0 0 |
- -
- -
| 0 0 0 0 0 0 |
| √3 0 0 0 0 0 |
UJ-U† = | 0 2 0 0 0 0 |
| 0 0 √3 0 0 0 |
| 0 0 0 0 0 0 |
| 0 0 0 0 1 0 |
- -
- -
| 3/2 0 0 0 0 0 |
| 0 1/2 0 0 0 0 |
UJzU† = | 0 0 -1/2 0 0 0 |
| 0 0 0 -3/2 0 0 |
| 0 0 0 0 1/2 0 |
| 0 0 0 0 0 -1/2 |
- -
This is a block diagonal matrix that inicates the
representation is irreducible. These matrices can
all be written as the direct sum. For example, in
the last case this decomposes as:
- -
| 3/2 0 0 0 | - -
| 0 1/2 0 0 | ⊕ | 1/2 0 |
| 0 0 -1/2 0 | ⊕ | 0 -1/2 |
| 0 0 0 -3/2 | - -
- -
Therefore, we can write:
3 ⊗ 2 = 4 ⊕ 2
In summary, the transformation matrix that encodes
the change of basis between the tensor product and
the irreducible direct sum is composed of the
Clebsch-Gordan coefficients.
Use of Tables
-------------
Combining the angular momenta of complex systems
can be extremely tedious. Fortunately, a table of
Clebsch-Gordan coefficients exists which greatly
simplifies the task. Here we illustrate their basic
use by using sections of the tables for some simple
cases.
For compactness the numbers in the blocks are the
coefficients squared times their sign. Thus -1/2
stands for -√(1/2).
Use of the tables take a little bit of explanation.
The 1/2 x 1/2 in the upper left corner tells us that
we are combining 2 particles each with angular
momentum = 1/2 for . The 1 x 1/2 tells us that we are
combining 2 particles - 1 with angular momentum = 1
and one with angular momentum = 1/2.
Along the top row are possible values of j = j1 ± j2.
Along the second row is the value of m = m1 ± m1.
The columns give the eigenstates of the overall
momenta in terms of the eigenstates of the individual
momenta. Therefore,
|j,m> = ΣC(|j1,m1> ⊗ |j2,m2>)
= A(|j1,m1> ⊗ |j2,m2>) + B(|j1,m1> ⊗ |j2,m2>)
For example, we can construct j = 3/2 as 1 + 1/2
and m = 1/2 as 1 - 1/2 or 0 + 1/2. The state |3/2,1/2>
is constructed using the bottom tables as:
|3/2,1/2>> = √(1/3)(|1,1> ⊗ |1/2,-1/2>)
+ √(2/3)(|1,0> ⊗ |1/2,1/2>)
The rows give the eigenstates of the individual
momenta in terms of the overall momenta. Therefore,
|j1,m1> ⊗ |j2,m2> = √(2/3)|j,m> - √(1/3)|j,m>
For example,
|1,0> ⊗ |1/2,1/2> = √(2/3)|3/2,1/2> - √(1/3)|1/2,1/2>
Note that the rows and columns of the table are
mutually orthogonal. That is, the dot product of
a row with any other row must be zero. Likewise,
for the dot product of a column with any other
column. Note also that the dot product of a row
or column with itself is unity thereby satisfying
the normalization condition.
(√(1/3),√(2/3)).((√(2/3),-√(1/3)) = 0
(√(1/3),√(2/3)).(√(1/3),√(2/3)) = 1
This follows because the entries in the table give
the expansion coefficients of the mutually orthogonal
eigenstates with unit norms that form the eigenbasis
of the tensor product space.
We use the previous 2 examples to illustrate the
process. For combining the spins of 2 electrons
we use the top table to get:
|1,0> = √(1/2)(|1/2,1/2> ⊗ |1/2,-1/2>)
+ √(1/2)(|1/2,-1/2> ⊗ |1/2,1/2>)
= √(1/2)(↑↓ + ↓↑)
|1,-1> = 1|1/2,-1/2> ⊗ 1|1/2,-1/2>
= |↓↓>
|0,0> = √(1/2)(|1/2,1/2> ⊗ |1/2,-1/2>)
- √(1/2)(|1/2,-1/2> ⊗ |1/2,1/2>)
= √(1/2)(↑↓ - ↓↑)
And, for the case of combining the isospins of 3
quarks we get:
|3/2,3/2> = 1(|1,1> ⊗ |1/2,1/2>)
= |uuu>
|3/2,1/2> = √(1/3)(|1,1> ⊗ |1/2,-1/2>)
+ √(2/3)(|1,0> ⊗ |1/2,1/2>)
Now,
|1,0> = √(1/2)(|1/2,1/2> ⊗ |1/2,-1/2>)
+ √(1/2)(|1/2,-1/2> ⊗ |1/2,1/2>)
= √(1/2)(|1/2,1/2> ⊗ |1/2,-1/2>
+ |1/2,-1/2> ⊗ |1/2,1/2>)
Therefore,
|3/2,1/2> = √(1/3)(|1,1> ⊗ |1/2,-1/2>)
+ √(2/3)(√(1/2)(|1/2,1/2> ⊗ |1/2,-1/2>
+ |1/2,-1/2> ⊗ |1/2,1/2>) ⊗ |1/2,1/2>)
= √(1/3)(uud + udu + duu)
|3/2,-1/2> = √(2/3)(|1,0> ⊗ |1/2,-1/2>)
+ √(1/3)(|1,-1> ⊗ |1/2,1/2>)
= √(1/3)(udd + dud + ddu)
|3/2,-3/2> = 1(|1,-1> ⊗ |1/2,-1/2>)
= ddd
|1/2,1/2> = √(2/3)(|1,1> ⊗ |1/2,-1/2>)
- √(1/3)(|1,0> ⊗ |1/2,1/2>)
= √(2/3)uud - √(1/3)√(1/2)(ud + du)u
= 2√(1/6)uud - √(1/6)(udu + duu)
= √(1/6)(2uud - udu - duu)
|1/2,-1/2> = √(1/3)(|1,0> ⊗ |1/2,-1/2>)
- √(2/3)(|1,-1> ⊗ |1/2,1/2>)
= √(1/3)√(1/2)(ud + du)d - √(2/3)ddu
= √(1/6)(udd + dud) - 2√(1/6)ddu
= √(1/6)(udd + dud - 2ddu)
We can also construct the states |1/2,1/2> and
|1/2,- 1/2> as follows:
|1/2,1/2> = |0,0> ⊗ |1/2,1/2>
and,
|1/2,-1/2> = |0,0> ⊗ |1/2,-1/2>
Where,
|0,0> = √(1/2)(|1/2,1/2> ⊗ |1/2,-1/2>)
- √(1/2)(|1/2,-1/2> ⊗ |1/2,1/2>)
= √(1/2)(ud - du)
Therefore,
|1/2,1/2> = √(1/2)((ud - du) ⊗ u)
= √(1/2)(udu - duu)
and,
|1/2,-1/2> = √(1/2)((ud - du) ⊗ d)
= √(1/2)(udd - dud)