# Thread: SU(2) spin 1/2 functions

1. ## SU(2) spin 1/2 functions

It's been a while since I posted a Physics question for y'all to ponder so here's another one.

One of the deficiencies in learning Physics is that no one seems to want to explicitly write down SU(2) invariant functions for spin 1/2 systems. Usually one finds references to defined vectors (2-spinors, actually, but they work just as if they were vectors) in the appropriate Hilbert space and the matrix representation is used. For some reason, the integer spin representations ARE given as invariant functions (the spherical harmonics functions as a well-known example).

To be specific I am looking for one of two things:
1) A differential equation for a spin 1/2 representation to find the invariant functions.
2) The spin 1/2 functions themselves.

(Oddly enough, I'd be happier with the differential equation.)

Some details. I know it is possible to go back and forth between matrix equations and differential equations in certain cases. (I've seen it done, but I don't know the details of how to do it in a general case.) I'll give the details of the matrix representation of a spin 1/2 system hoping that it will be helpful.

The following uses the Dirac "bra-ket" convention. $\displaystyle | \psi >$ (a "ket") represents a vector in a complex Hilbert space and $\displaystyle < \psi |$ (a "bra") represents the dual "vector" in the space dual to the aforementioned Hilbert space. $\displaystyle < \psi | \phi >$ (a "bra-ket") represents the scalar product between a bra and a ket.

The usual Physics setup for a spin 1/2 system is the choose the system to be oriented in the z direction: $\displaystyle |+> \equiv \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ represents a state where the spin is +1/2 in the z direction (dropping the usual h-bar unit) and $\displaystyle |-> \equiv \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ represents a state where the spin is -1/2 in the z direction. We can show that {|+>, |->} is an orthonormal basis in the Hilbert space in use. ({<+|, <-|} is a basis in the dual space.) We have 5 operators in use, which I'll give you since there are arbitrary choices involved in their representation:
$\displaystyle S_x=\frac{1}{2} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$
$\displaystyle S_y=\frac{1}{2} \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}$
$\displaystyle S_z=\frac{1}{2} \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$
And we have two "step" matrices
$\displaystyle S_+ \equiv S_x + i S_y = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$
$\displaystyle S_- \equiv S_x - i S_y = \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}$

The use of these are as follows. $\displaystyle S_{\pm}$ changes the spin of a basis vector by +/- 1/2. (ie. $\displaystyle S_- |+> = |->$ and $\displaystyle S_+ |->=|+>$. All other applications of $\displaystyle S_{\pm}$ to a basis vector gives zero.) The others measure the average spin component of an arbitrary vector in the Hilbert space in the given spatial direction by the following formula:
$\displaystyle < \psi > \equiv < \psi | S | \psi >$ (is called the "expectation value.")

If this isn't the kind of information needed to give a function representation let me know and I'll give you what information you would need. I realize this may be an impossible problem to solve, so I'll simply thank any and everyone for their efforts in advance.

-Dan

2. By the way, as I said the integer spin functions are known. For example, (in terms of spherical harmonics) a SU(2) spin 2 function representation of a spin 2 object with a z component of spin 1 would be:
$\displaystyle Y_2^1(\theta, \phi)=-\sqrt{\frac{15}{8 \pi}}(sin \theta \, cos \theta )e^{i \phi}$.

All I know about a spin 1/2 function representation of an object with a z component of spin +/- 1/2 is that it cannot be represented using the spherical harmonics. (Using two methods we can derive two different differential equations to obtain solutions which contradict each other.) Being that the integer spin representations and the spin 1/2 representations of SU(2) are irreducible, you might have expected that.

-Dan