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

[MisterX]

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

Homework Statement

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."

Homework 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

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?

 

[Goddar]

Try T_±2²=(x±iy)²...

 

[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*}<br /> T^{(2)}_{\pm 2} &amp;= T^{\left(1\right)}_{\pm 1} T^{\left(1\right)}_{\pm 1}\\<br /> T^{(2)}_{\pm 1} &amp;= \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)\\<br /> T^{(2)}_{0} &amp;= \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)<br /> \end{align*}

 

[Goddar]

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

 

[MisterX]

Yes, you can also read them out by analogy with the spherical harmonics..

How's that?

 

[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.

 

[Goddar]

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