### Preliminaries

1. This is quite a technical write up, and assumes knowledge of the Dirac formalism. If you are terrified by the messier ends of linear algebra then please look away now.
2. HTML is unfriendly to bra-ket notation, and to make things worse, the <,> symbols are overloaded here, as inequalities. To make a distinction, in the former case, they appear slightly larger, although you may need to look carefully to register the difference. Also, I've written Plank's-constant-divided-by-2pi as (h-bar), which looks slightly cumbersome. Apologies.

### Angular Momentum Operators

Angular momentum is treated as an observable, corresponding to an angular momentum operator (which must be hermitian). Such an operator, J = (J1,J2,J3)T is required to satisfy the commutation relations:
[ Ji,Jj ] = i(h-bar)εijkJk
##### Why?
There are two possible motivations for this, of which I can guarantee you will only approve of at most one:
1. (applied) The orbital angular momentum operator L = x × p (as in classical mechanics) satisfies these relations. By specifying the relations as a defining condition, we hope to examine a larger class of operators, which could give rise to some new Physics (and they do).
2. (pure-ish) The commutation relations are the same as those for the generators of the Lie algebra for the group SO3(R) of rotations of three-dimensional space. By reducing the group structure to a local form in this way, we hope to find the largest class of rotational symmetries of the system (and it is indeed larger than SO3(R)).

### Eigenvalues

The commutation relations given above supply us will all the necessary information to work out the eigenvalues of (some suitable commuting set of operators built out of) the Ji.

When we have a commuting set of operators, it will allow us to label the states of the system by their eigenvalues, which is our goal.

First, we need to decide what this commuting set will be: note that for each i, [ J2,Ji ] = 0 (J2 = J12+J22+J32), which can be proved with little effort. Since not two of the Jis commute with each other, we can only add one to the commuting set with J2 - conventionally it is J3. So, we call the states |a,b>, with
J2|a,b> = a|a,b>
J3|a,b> = b|a,b>
Now define J+ and J- by J± = J1 ± iJ2 (neither is hermitian, in fact they are hermitian adjoints of each other). The commutation relations can then be used to show that:
[ J3,J± ] = ±(h-bar)J±
[ J+,J- ] = 2(h-bar)J3
[ J2,J± ] = 0
And then,
J3J±|a,b> = (±(h-bar)J± + J±J3)|a,b> = (b ± (h-bar))J±|a,b>
so J±|a,b> lies in the b ± (h-bar) eigenspace of J3 ie. there exists some complex number C±(a,b) such that J±|a,b> = C|a,b ± (h-bar) > (Think about this for a moment. Yes, I am claiming that the eigenspace is one dimensional). That is to say, the J± act as creation/anihilation operators on the |a,b> states (for fixed a), rather like with the harmonic oscillator.

As with the harmonic oscillator, there are limitations imposed on the values of b. Note that J2-J32 = (1/2)(J+J- + J-J+) = (1/2)(J+J++ + J-J-+) is a positive definite operator, therefore

<;a,b|J2-J32|a,b> = (a-b2) >= 0
so a >= b2 which implies that there must be a limiting value of b, say btop, such that J+|a,btop> = 0 (where |a,btop> is a non-zero ket). Then,
0 = J-J+|a,btop> = (J2 - J32 - (h-bar)J3)|a,btop> = (a - btop2 - (h-bar)btop)|a,btop>;
Since |a,btop> is not a zero ket, a = btop(btop + (h-bar)). Repeating the whole argument for J-, there is a minimal b, bbtm such that J+|a,bbtm> = 0, and a = bbtm(bbtm - (h-bar)). Then we have that bbtm = -btop and there is some natural number n such that bbtm + n(h-bar) = btop.

So much for the derivation; in practice the eigenvalue labels j and m are used instead, with j = (1/2)n = btop/(h-bar) and b = m(h-bar). Then the restriction bbtm <= v <= btop becomes m = -j, -(j-1), -(j-2) ... j-2,j-1,j (taking 2j+1 possible values), and

J2|j,m> = (h-bar)2 j(j+1)|j,m>
J3|j,m> = (h-bar) m|j,m>
Nearly done, but still need to find out the C±(j,m) factors mentioned above. Note that J+|j,m> = C+(j,m)|j,m+1> so |c+(j,m)|2 = <j,m|J++J+|j,m> = <j,m|J2 - J32 - (h-bar)J3|j,m> = (h-bar)2(j(j+1)-m(m+1)) = (h-bar)2((j-m)(j+m+1)). By convention, the Cs are real and positive, so just take the square root of this. By similar means, C-(j,m) = (h-bar)2((j+m)(j-m+1)). In summary,
J±|j,m> = Sqrt((h-bar)(j-±m)(j±m+1))|j,m±1>

In fact, the above should be strangely familiar
to anyone who knows about representations of sl(2C)

### How am I meant to make sense of any of that?

Not easily, as this whole topic is very abstract from the outset. One mental picture you can adopt is that there are a number of 'strings', correpsonding to different values of j, and along each string there are (2j+1) 'beads' each corresponding to a different value of m.
Then, applying J2 to a state will tell you which string it is on (as the eigenvalues is a function of j), and applying J3 to the state tells you its position on the string (its eigenvalue being essentially m). Applying J+ and J- to states will move them respectively up and down the string (up to a multiplicative factor).
It's easy to show that the sum of two angular momentum operators yeilds another angular momentum operator, but the way in which the states add is far from trivial: the strings become in some sense entangled with each other. You can work out what the amalgamated states are using the Clebsch-Gordan Coefficients, when it's been written up.

### Rotation operators

A rotation of three dimensional space can be specified by a unit vecor n (axis of rotation) and an angle of rotation θ. Such a rotation induces a rotation on the space of states which can be expressed in terms of the spin operator as follows:
U(n,θ) = exp(-iJ·nθ/(h-bar))
As for why, the question really belongs in a study of Lie algebras. In short, the commutation relation given at the top has been 'integrated up' to produce this group element. I stated the result as I am about to use it below...

### Spin

One advantage of all this nightmareish algebra is that the class of angular momentum operators is indeed larger than just the orbital angular momentum (L): it also includes the spin angular momentum, S. All elementary particles have a definite spin, this is the eigenvalue j asscoiated with S. Spin is found experimentally to be either a non-negative integer (0,1,2,...) or a non-negative half odd integer ((1/2),(3/2),...) (particles with integer spin are called bosons, and the others are called fermions). Given this j value, the particle then has 2j+1 possible spin states it can occupy - eg. the electron has spin 1/2, so can either be in a spin up | 1/2, 1/2> or spin down | 1/2, -1/2> state.
##### Rotating fermions
Now here's the interesting part: what happens if you rotate a state about the z-axis? Well, U(z,θ) = exp(-iJ3θ/(h-bar)) and then
U(z,θ)|j,m> = exp(-iJ3θ/(h-bar))|j,m> = exp(-imθ)|j,m> = (cos(mθ)-isin(mθ))|j,m>
which doesn't seem to be stating anything profound until you plug in θ = 2π. Then you get U|j,m> = -|j,m>, since the sin term vanishes and the cos term is -1 (since m is half an odd integer). That is to say, if you rotate the whole apparatus around in a circle, the fermion will not return to its original state! (A result that Stephen Hawking quoted in A Brief History of Time which left many people puzzled) Not only is the result bizzare, but it has been confimed by experiment.

### Wave functions

Previously, given some state |ψ> you could simply cast it onto a position bra <x| in order to obtain the particle wave function ψ(x) = <x>. When spin is taken into account, extra degrees of freedom are present, so |ψ> needs to be cast onto a larger space of states: this is formed by taking a product of the spaces containing <x| and <j,m| to produce the bra <x,j,m|, giving a 'vector valued' wave function ψjm(x) = <x,j,m|ψ>
The ordinary Schrodinger wave equation tells you nothing about how to deal with a vector valued wave functions; for this you need The Dirac Equation (any takers?).

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