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Spherical Tensor operators for half-integers

  1. Jul 14, 2012 #1

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

    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 [itex]k[/itex] instead of integer. And if not, why?

    Thank you,
  2. jcsd
  3. Jul 14, 2012 #2


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    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.
  4. Jul 14, 2012 #3
    Great! Thanks Bill. The integer ones seem to have explicit forms relating back to Cartesian components, for instance,
    etc... Do such forms exist for the half integer spherical operators? For instance is it possible to write [itex]T\left( \frac{7}{2},q\right) [/itex] in terms of Cartesian components [itex](a_{0},a_1 ,a_2,...)[/itex]?

    Thanks again.
  5. Jul 15, 2012 #4


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    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).
  6. Jul 15, 2012 #5
    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 [itex]C[/itex] some associated coefficients? Would these just be the CB coefficients, or does it not even make sense?
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