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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:
L_{x} = yp_{z} - zp_{y}
and so on.
Quantum mechanically this becomes:
L_{x} = -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:
[J_{i},J_{j}] = ihε_{ijk}J_{k}
Consider:
[J_{i}^{(1)},J_{j}^{(1)}] = ihε_{ijk}J_{k}^{(1)} in vector space V_{1}
[J_{i}^{(2)},J_{j}^{(2)}] = ihε_{ijk}J_{k}^{(2)} in vector space V_{2}
We can then form a combined operator as follows:
J = J_{i}^{(1)} ⊗ 1 + 1 ⊗ J_{j}^{(2)} in vector space V_{1} ⊗ V_{2}
These also have the same commutator relationships.
Proof:
[J_{i}^{(1)} ⊗ 1 + 1 ⊗ J_{i}^{(2)},J_{j}^{(2)} ⊗ 1 + 1 ⊗ J_{j}^{(2)}]
= [J_{i}^{(1)} ⊗ 1,J_{j}^{(1)} ⊗ 1] + [1 ⊗ J_{i}^{(2)},1 ⊗ J_{j}^{(2)}]
= [J_{i}^{(1)},J_{j}^{(1)}] ⊗ 1 + 1 ⊗ [J_{i}^{(2)},J_{j}^{(2)}]
= ihε_{ijk}(J_{k}^{(1)} ⊗ 1 + 1 ⊗ J_{k}^{(2)})
= ihε_{ijk}J_{k}
Like any vector, a magnitude can be defined for
J, the total momentum operator as:
J^{2} = J_{x}^{2} + J_{y}^{2} + J_{z}^{2}
J^{2} is the Casimir invariant with the property:
[J^{2},J_{x}] = [J^{2},J_{y}] = [J^{2},J_{z}] = 0
Raising and Lowering Operators
------------------------------
The raising and lowering operators are:
J_{+} = J_{x} + iJ_{y}
J_{-} = J_{x} - iJ_{y}
L_{+} = L_{x} + iL_{y}
L_{-} = L_{x} - iL_{y}
S_{+} = S_{x} + iS_{y}
S_{-} = S_{x} - iS_{y}
With the following commutators:
[J_{z},J_{±}] = ±hJ_{±}
[J_{+},J_{-}] = 2hJ_{z}
Since J^{2} and J_{z} commute they share common eigenstates
but will have different eingenvalues. Therefore, we
will need 2 indeces for each basis vector: j for J^{2}
and m for J_{z}. j and m label the states so we can
write:
J^{2}|j,m> = j|j,m>
and,
J_{z}|j,m> = m|j,m>
Therefore,
|j,m> ≡ |eigenvalue of J^{2}|eigenvalue of J_{z}>
Recall [J_{z},J_{+}] = ihJ_{+}
J_{z}J_{+} - J_{+}J_{z} = ihJ_{+}
J_{z}J_{+} = ihJ_{+} + J_{+}J_{z}
J_{z}J_{+}|j,m> = (ihJ_{+} + J_{+}J_{z})|j,m>
= ihJ_{+}|j,m> + J_{+}J_{z}|j,m>
But J_{z}|j,m> = m|j,m>. Therefore,
J_{z}J_{+}|j,m> = ihJ_{+}|j,m> + J_{+}m|j,m>
Therefore,
J_{z}(J_{+}|j,m>) = (m + ih)(J_{+}|j,m>)
So (J_{+}|j,m>) is an eigenvector of J_{z} with eigenvalue
(m + ih).
Similarly,
J_{z}(J_{-}|j,m>) = (m - ih)(J_{-}|j,m>)
Therefore, the effect of J_{±} is to increase or decrease
the eigenvalue of J_{z} by the amount h. Let us now
consider J^{2}.
[J^{2},J_{+}] = 0
J^{2}J_{+} - J_{+}J^{2} = 0
J^{2}J_{+} = J_{+}J^{2}
J^{2}J_{+}|j,m> = J_{+}J^{2}|j,m>
= J_{+}j|j,m>
= jJ_{+}|j,m>
Therefore,
J^{2}(J_{+}|j,m>) = j(J_{+}|j,m>
So (J_{+}|j,m>) is an eigenvector of J^{2} with eigenvalue,
j.
Similarly,
J^{2}(J_{-}|j,m>) = j(J_{-}|j,m>
Therefore, the net result is that J_{±} has no effect
on the eigenvalue of J^{2}, j.
Summarizing, J_{±} change the eigenvalue of J_{z} but not
J^{2}!!
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> = (J_{x} - iJ_{y})(J_{x} + iJ_{y})|j,m>
= J_{x}^{2} + J_{x}^{2} + i[J_{x},J_{y}]|j,m>
But J_{x}^{2} + J_{x}^{2} = J^{2} - J_{z}^{2} and [J_{x},J_{y}] = ihJ_{z}
Therefore,
J_{-}J_{+}|j,m> = (J^{2} - J_{z}^{2} - hJ_{z})|j,m>
There is an eigenstate |j,m_{max}> which cannot be
raised and a state |j,m_{min}> that cannot be lowered.
Because J_{+}|j,m_{max}> = 0, J_{-}J_{+}|j,m_{max}> = 0. Therefore,
(J^{2} - J_{z}^{2} - hJ_{z})|j,m_{max}> = 0
J^{2}|j,m_{max}> - J_{z}^{2}|j,m_{max}> - hJ_{z}|j,m_{max}> = 0
α|j,m_{max}> - m_{max}^{2}|j,m_{max}> - hm_{max}|j,m_{max}> = 0
(α - m_{max}^{2} - hm_{max})|j,m_{max}> = 0
(α - m_{max}^{2} - hm_{max}) = 0
α = m_{max}^{2} + hm_{max}
Because J_{-}|j,m_{min}> = 0, J_{+}J_{-}|j,m_{min}> = 0 we can go
through the same procedure to get:
α = m_{min}^{2} - hm_{min}
If there are n steps between m_{max} and m_{min} then:
2m_{max} = nh ∴ m_{max} = nh/2
Now α = m_{max}^{2} + hm_{max} so,
α = n^{2}h^{2}/4 + nh^{2}/2
= nh/2(nh/2 + h)
If we let j = n/2 and we get:
α = h^{2}j(j + 1)
Therefore,
J^{2}|j,m> = h^{2}j(j + 1)|j,m>
<j,m|J_{+}^{†}J_{+}|j,m> = <j,m|J_{-}J_{+}|j,m>
_{ } = <j,m|J^{2} - J_{z}^{2} - hJ_{z}|j,m>
_{ } = <j,m|J^{2} - J_{z}^{2} - hJ_{z}|j,m>
_{ } = <j,m|h^{2}j(j + 1) - h^{2}m^{2} - h^{2}m|j,m>
_{ } = (j(j + 1) - h^{2}m^{2} - h^{2}m)<j,m|j,m>
_{ } = h^{2}(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_{-} = (J_{x} + iJ_{y})(J_{x} - iJ_{y})
_{ } = J^{2} - J_{z}^{2} - i[J_{x}J_{y}]
_{ } = J^{2} - J_{z}^{2} + hJ_{z}
Which leads to:
C_{-}^{2} = h^{2}(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 J_{2} = J ⊗ 1 + 1 ⊗ J for 2 momenta and
J_{3} = J_{2} ⊗ 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 |
- - - - - -
S_{x} = (1/2)(S_{+} + S_{-}) gives:
- -
S_{x} = h/2| 0 1 |
_{ } | 1 0 |
- -
S_{y} = (1/2i)(S_{+} - S_{-}) gives:
- -
S_{y} = h/2| 0 -i |
_{ } | i 0 |
- -
For S_{z} we use:
S_{z}|s,m> = m|s,m> and find the following:
S_{z}|1/2,-1/2> = -h/2|1/2,-1/2>
- - - - - -
S_{z} = h/2| 1 0 || 1 | = h/2| 1 |
_{ } | 0 -1 || 0 | | 0 |
- - - - - -
- - - - - -
S_{z} = h/2| 1 0 || 0 | = h/2| 0 |
_{ } | 0 -1 || 1 | | 1 |
- - - - - -
These are the familiar Pauli matrices.
Finally,
- -
S^{2} = 3h^{2}/4| 1 0 |
^{ } | 0 1 |
- -
Compare this with S^{2}|1/2,1/2> = h^{2}(1/2)(1/2 + 1)|1/2,1/2>
= 3h^{2}/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> = h^{2}√[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> = h^{2}√[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 |
- - - - - -
L_{x} = (1/2)(L_{+} + L_{-}) gives:
- -
_{ } | 0 1 0 |
L_{x} = h/√2| 1 0 1 |
_{ } | 0 1 0 |
- -
L_{y} = (1/2i)(L_{+} - L_{-}) gives:
- -
_{ } | 0 1 0 |
L_{y} = h/i√2| -1 0 1 |
_{ } | 0 -1 0 |
- -
For L_{z} we use:
L_{z}|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 |
L_{z} = h| 0 0 0 |
_{ } | 0 0 -1 |
- -
Finally,
- -
^{ } | 1 0 0 |
L^{2} = 2h^{2}| 0 1 0 |
^{ } | 0 0 1 |
- -
Compare this with L^{2}|1,1> = h^{2}1(1 + 1)|1,1>
= 2h^{2}
NOTE: THESE L_{x}, L_{y} AND L_{z} NATRICES SHOULD
NOT BE CONFUSED WITH THE MATRICES FOR SO(3).
Irreducible Representations
---------------------------
The Clebschâ€“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 J_{Z}:
- -
_{ } | 3/2 0 0 0 0 0 |
_{ } | 0 1/2 0 0 0 0 |
J_{Z} = | 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 |
UJ_{z}U^{†} = | 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 = j_{1} ± j_{2}.
Along the second row is the value of m = m_{1} ± m_{1}.
The columns give the eigenstates of the overall
momenta in terms of the eigenstates of the individual
momenta. Therefore,
|j,m> = ΣC(|j_{1},m_{1}> ⊗ |j_{2},m_{2}>)
= A(|j_{1},m_{1}> ⊗ |j_{2},m_{2}>) + B(|j_{1},m_{1}> ⊗ |j_{2},m_{2}>)
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,
|j_{1},m_{1}> ⊗ |j_{2},m_{2}> = √(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)