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Y^2 - x^2 in the [itex]\mid n\ell m \rangle[/itex] basis - tensor Op.

  1. Feb 10, 2014 #1
    x^2 - y^2 in the [itex]\mid n\ell m \rangle[/itex] basis - tensor op.

    1. The problem statement, all variables and given/known data
    I must determine the matrix elements of [itex]x^2 - y^2[/itex] in the [itex]\mid n\ell m \rangle[/itex] basis.
    "...use the fact that [itex]x^2 - y^2[/itex] is a sum of spherical components of a rank two tensor, together with the explicit form of the Winger-Eckart theorem."
    "Show that [exp. proportional to [itex]x^2 - y^2[/itex]] is a sum of ITOs with [itex]\ell =2[/itex]."
    2. Relevant equations
    Winger Eckart Theorem:
    [itex]\langle \ell_1 m_1|T^\ell_m|\ell_2 m_2\rangle = \underbrace{\langle \ell_2m_2;\ell m|\ell_1 m_1 \rangle}_{\text{C.G. coeff.}} \langle j||T^\ell||j'\rangle[/itex]

    The operators must satisfy these properties:
    [itex]\left[L_z, T^{\ell}_m\right] =mT^{\ell}_m[/itex]
    [itex]\sum_{i=1}^3\left[ L_i,\left[L_i, T^{\ell}_m\right]\right] =\ell\left(\ell + 1 \right) T^{\ell}_m[/itex]
    [itex]L_{\pm} = L_{1} \pm iL_{2}\;\;\;\;\;\;\;\; \left[L_\pm, T^{\ell}_m\right] =\sqrt{\ell\left( \ell + 1\right)-m\left(m\pm 1\right)}\,T^{\ell}_{m+1} [/itex]

    I have some operators with a lower value of [itex]\ell[/itex]:
    [itex]T^1_{1} = -x -iy[/itex] is an operator with [itex]\ell = 1, m = 1[/itex].
    [itex]T^1_{0} = z[/itex] is an operator with [itex]\ell = 1, m = 0[/itex].
    [itex]T^1_{-1} = x -iy[/itex] is an operator with [itex]\ell = 1, m = -1[/itex].

    [itex]\left[L_i, x_j\right] = \sum_k \epsilon_{ijk}x_k[/itex]

    3. The attempt at a solution
    What I need to do is find the family of five operators with [itex]\ell = 2[/itex], and express [itex]x^2-y^2[/itex] as superposition of those operators. All I need is one of these operators and I can construct the rest. I guessed that [itex]z^2[/itex] might be an operator with [itex]\ell = 2, m = 0[/itex] but I found that [itex]\sum_{i=1}^3\left[ L_i,\left[L_i, z^2\right]\right] =4z^2 - 2x^2 - 2y^2 \neq \ell \left(\ell + 1 \right)z^2 = 6z^2 [/itex].

    Is there a general procedure for constructing [itex]\ell = 2[/itex] operators from [itex]\ell = 1[/itex] operators?
    Last edited: Feb 10, 2014
  2. jcsd
  3. Feb 10, 2014 #2
    Try T±22=(x±iy)2...
  4. Feb 11, 2014 #3
    Yeah I found out - apparently it always works to use [itex]T^{\left(2\right)}_{\pm 2} = T_{\pm 1}^{\left(1\right)}T_{\pm 1}^{\left(1\right)}[/itex]. The full set of operators is
    T^{(2)}_{\pm 2} &= T^{\left(1\right)}_{\pm 1} T^{\left(1\right)}_{\pm 1}\\
    T^{(2)}_{\pm 1} &= \frac{1}{\sqrt{2}}\left( T^{\left(1\right)}_{\pm 1} T^{\left(1\right)}_0 + T^{\left(1\right)}_0 T^{\left(1\right)}_{\pm 1} \right)\\
    T^{(2)}_{0} &= \frac{1}{\sqrt{6}}\left( T^{\left(1\right)}_{+1} T^{\left(1\right)}_{-1} + T^{\left(1\right)}_{-1} T^{\left(1\right)}_{+1} + 2T^{\left(1\right)}_0 T^{\left(1\right)}_0 \right)
  5. Feb 11, 2014 #4
    Yes, you can also read them out by analogy with the spherical harmonics.. good luck with the rest!
  6. Feb 11, 2014 #5
    How's that?
  7. Feb 11, 2014 #6
    well, look at the spherical harmonics remembering that z = cosθ/r, x2+y2=sin2θ/r2, etc... if you discard the numerical factors, you'll see the pattern.
  8. Feb 11, 2014 #7
    oops, in the last post r factors should be multiplied on the right, not divided..
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