Y^2 - x^2 in the $\mid n\ell m \rangle$ basis - tensor Op.

1. Feb 10, 2014

MisterX

x^2 - y^2 in the $\mid n\ell m \rangle$ basis - tensor op.

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

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

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

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

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

Is there a general procedure for constructing $\ell = 2$ operators from $\ell = 1$ operators?

Last edited: Feb 10, 2014
2. Feb 10, 2014

Goddar

Try T±22=(x±iy)2...

3. Feb 11, 2014

MisterX

Yeah I found out - apparently it always works to use $T^{\left(2\right)}_{\pm 2} = T_{\pm 1}^{\left(1\right)}T_{\pm 1}^{\left(1\right)}$. The full set of operators is
\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*}

4. Feb 11, 2014

Goddar

Yes, you can also read them out by analogy with the spherical harmonics.. good luck with the rest!

5. Feb 11, 2014

MisterX

How's that?

6. Feb 11, 2014

Goddar

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.

7. Feb 11, 2014

Goddar

oops, in the last post r factors should be multiplied on the right, not divided..