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Spherical Tensor operators for half-integers

  1. Jul 14, 2012 #1
    Hi,

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

    [tex]
    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
    [/tex]
    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

    Bill_K

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    Science Advisor

    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,
    [tex]
    J_{+}=a_{1}+ia_{2}
    [/tex]
    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

    Bill_K

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    Science Advisor

    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
    [tex]
    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'}+...
    [/tex]
    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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