display | more...

### This is going to be terribly esoteric, isn't it?

Well... yes. What I'm going to describe here is the theory of adding angular momentum in Quantum Mechanics, and so knowlege of that write up is considered prerequisite, and that too was fairly esoteric. However, the theory is quite interesting, and for me writing this is more fun than doing real revision.

### The big idea

Suppose you have two angular momentum operators, J(1) and J(2). You can define their sum J = J(1)+J(2) as an operator acting on the product space of the two constituent operators, and you can show that this too is an angular momentum operator (since the commutator is a bilinear form, and J(1) and J(2) commute with each other when acting on the product space).
Big deal? Well, seeing as J is also an angular momentum, it has its own |j,m> eigenstates on the product space, and the connection between these and the angular momentum states of J(1) and J(2) is non trivial. Because of this, the theory of adding quantum mechanical angular momentum becomes... interesting. The Clebsch-Gordan Coefficients are what connect one set of states to the other, thus solving the problem.

### The general theory

One way to label the states of the product space (ie. that on which J acts) is in terms of the states of the J(i): these are
|j1,m1>|j2,m2> = |j1,m1;j2,m2>
Because we'll only be considering states of constant j1 and j2 at any one time, we'll supress these labels (to make the notation a little simpler), and simply call this state |m1,m2>A. The need for the subscript A will become clear in about 3 seconds' time (depending on how fast you read).
Now, the eigenstates of the operator J also span this space, so we can label these |j,m>B. The A and B distinguish between these two representations (since otherwise calling a particular state, say, |1,0> would be ambiguous). Now, using the completeness relation from Dirac's formalism, we may write:
|j,m>B = (sum on m1,m2) |m1,m2>A<m1,m2|j,m>B
and the numbers A<m1,m2|j,m>B are the Clebsch-Gordan coefficients.
To help calculate these, we can make use of the following two facts:
• Fact 1: j ranges from |j1-j2| up to j1+j2 in integer steps.
I have read about or been told of the following possible explanations for this:
• Explanation 1 (unconvincing): It works that way with classical angular momentum as a vector quantity, that is |a-b| <= |a+b| <= |a|+|b|. (And j is known to vary in integer steps already).
• Explanation 2 (dodgy): We need the bases of the space to have the same size, and this can be achieved if this condition holds. That is, because (sum on m from |j1-j2| to j1+j2) 2m+1 = (2j1+1)(2j2+1).
• Explanation 3 (hard to understand): It's one of the deeper results that comes out of the representions of SU2(C), the group that the angular momentum operators generate. Obviously this is the correct reason, but it would take too long to discuss here.
Just accept it for now.

• Fact 2: The coefficients vanish unless m = m1+m2.
Since J3 = J(1)3+J(2)3,
0 = A<m1,m2|J3-J(1)3-J(2)3|j,m>B = (m-m1-m2)A<m1,m2|j,m>B
et voila.
With this, we already can say that |j1+j2,j1+j2>B = |j1,j2>A, since the top state of the j=j1+j2 is a linear combination of states |m1,m2>A where m1+m2 = j1+j2, and there is only one such. Then, by noting that J- = J(1)-+J(2)- we can keep reducing the m value of the state until we've found all the states in the j=j1+j2 string (see example below if this is unclear).
After this, working out the other chains is straightforward: work out the top state by requiring it to be orthogonal to all the other |>B states already calculated, then use the J- operator to work out all the lower states in the chain.

### The simplest interesting example

Let's try adding the angular momentum of two spin 1/2 particles. Physically, this corresponds to working out the possible angular momenta of a state of two electrons bound together (as they are in phonons in a superconductor).
They're both spin 1/2, which means that J1=J1=1/2, and so the possible j values of the combined state are 0 and 1.
Right away, we have |1,1>B = |1/2,1/2>A. Then apply J-:
J-|1,1>B = (h-bar)Sqrt((1+1)(1-1+1)) |1,0>B =
(J(1)-+J(2)-) |1/2,1/2>A = (h-bar)Sqrt((1/2 + 1/2)(1/2 - 1/2 + 1)) |-1/2,1/2>A + (h-bar)Sqrt((1/2 + 1/2)(1/2 - 1/2 + 1)) |1/2,-1/2>A
ie. |1,0>B = Sqrt(1/2)( |1/2,-1/2>A + |-1/2,1/2>A ). Apply J- once more:
J-|1,0>B = (h-bar)Sqrt((1+0)(1-0+1)) |1,-1>B = (J(1)-+J(2)-)Sqrt(1/2)( |-1/2,1/2>A + |1/2,-1/2>A ) =
(h-bar)Sqrt((1/2)(1/2 + 1/2)(1/2 - 1/2 + 1))( |-1/2,-1/2>A + |-1/2,-1/2>A )
ie. |1,-1>B = |-1/2,-1/2>A, which was to be expected, since this is the only state with m1+m2 = -1.
The only other state that needs to be determined is |0,0>B. Remembering that m1+m1 = 0,
|0,0>B = a|1/2,-1/2>A+b|-1/2,1/2>A
Then by orthogonality with the other states a = -b = Sqrt(1/2). N.b. there is a choice of sign here, but it doesn't matter which you take.
So in summary,
|1,1>B = |1/2,1/2>A
|1,0>B = Sqrt(1/2)( |1/2,-1/2>A + |-1/2,1/2>A )
|1,-1>B = |-1/2,-1/2>A
|0,0>B = Sqrt(1/2)( |1/2,-1/2>A - |-1/2,1/2>A )

# Tell me something interesting

Have a look at the last equation above: it say that if you can arrange a paring of two electron to have total spin zero, then they exist in a superposed state where one is spin up (m = 1/2), and the other must be spin down (m = -1/2).
What's interesting about this is that once the electrons have been put into such a state, measuring the spin of one individual electron automatically forces the other to take the opposite state. Yes, notwithstanding such other physical laws such as relativity of simultaneity or that the speed of a signal travelling between them cannot exceed the speed of light. Here there seems to be an inconsistency between relativity and quantum mechanics, and it is the basis of the EPR paradox.

Log in or register to write something here or to contact authors.