# Spherical Tensor operators for half-integers

1. Jul 14, 2012

### jfy4

Hi,

There are, for example, lists of spherical tensor operators for $l=\text{integer}$ steps, e.g. $l=0,1,2,...$.

$$T_{k}^{q}(J)\rightarrow T_{0}^{0}=1, \quad T_{1}^{\pm 1}=\mp \sqrt{\frac{1}{2}}J_{\pm},\quad T_{1}^0=J_z$$
and this continues forever. I was wondering if there are operators in this spherical tensor form for half integers steps of $k$ instead of integer. And if not, why?

Thank you,

2. Jul 14, 2012

### Bill_K

Sure, the definition is: an irreducible tensor operator T(k) is a set of 2k+1 operators T(k q), q = -k,... , k which transforms under an infinitesimal rotation as

[J±, T(k q)] = [(k ∓ q)(k ± q + 1)]1/2
[J0, T(k, q)] = q T(k q)

This works for both integer and half-integer k.

3. Jul 14, 2012

### jfy4

Great! Thanks Bill. The integer ones seem to have explicit forms relating back to Cartesian components, for instance,
$$J_{+}=a_{1}+ia_{2}$$
etc... Do such forms exist for the half integer spherical operators? For instance is it possible to write $T\left( \frac{7}{2},q\right)$ in terms of Cartesian components $(a_{0},a_1 ,a_2,...)$?

Thanks again.

4. Jul 15, 2012

### Bill_K

OK, now it gets interesting. What you're asking is the relationship between the components of a tensor in the spherical basis as opposed to the Cartesian basis. The Cartesian basis starts with the 3 representation and the three basis vectors ex, ey, ez, and for ℓ = 1 the relationship is easy to write down: e± = ex ± i ey, e0 = ez. (I'm ignoring √2's)

The representations for ℓ = integer > 1 are irreducible and contained in 33 ⊗ ... ⊗ 3. Which means you take the tensor product of ℓ copies of ex, ey, ez with itself and project out the highest irreducible part. So for example for ℓ = 2,
e0 = 2 exex - eyey - ezez
e±1 = ez(ex ± i ey) + (ex ± i ey)ez
e±2 = (ex ± i ey)(ex ± i ey)
From these relations you can read off the Cartesian components Txx, Txy, etc in terms of the spherical components T0, T±1, T±2.

However, when you come to a half-integer representation, ℓ = n + ½, it is not contained in any product of 3's. What you must do is to take in addition a spinor quantity, transforming as the 2 representation, and consider products of this with n copies of 3. Thus the building blocks are the "spin spherical harmonics", i.e. the spherical decomposition of a two-spinor field χ(x, y, z).

5. Jul 15, 2012

### jfy4

Thanks Bill. I understand these things live on a sphere, but have you ever seen their contraction written with a metric like
$$C_{l\, l',m\,m'}A^{l,m}\sigma^{l',m'}=C_{00,m\, m'}A^{0,m}\sigma^{0,m'}+C_{\frac{1}{2}\frac{1}{2},m\, m'}A^{\frac{1}{2},m}\sigma^{\frac{1}{2},m'}+ C_{11,m\, m'}A^{1,m}\sigma^{1,m'}+...$$
with $C$ some associated coefficients? Would these just be the CB coefficients, or does it not even make sense?