# 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..