1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    [tex]\begin{align*}
    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)
    \end{align*}[/tex]
     
  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..
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Y^2 - x^2 in the [itex]\mid n\ell m \rangle[/itex] basis - tensor Op.
  1. Spin 1/2 Basis Change (Replies: 3)

  2. Why is <n|n-2> = 0? (Replies: 1)

Loading...