# Angular momentum of pair of particles

## Homework Statement

There are two noninteracting, spinless, distinguishable particles an isotropic oscillator potential. One particle is in the state $|1\rangle =\frac{1}{\sqrt{2}}(|1,1,-1\rangle + |1,1,0\rangle )$ and the other is in the state $|2\rangle = \frac{1}{\sqrt{2}}(|1,1,0\rangle - |1,1,1\rangle )$ where $|a,b,c\rangle$ corresponds to $n=a,l=b,m=c$. What is the expectation value of $L^{2}$?

## The Attempt at a Solution

We know that the angular momentum of the system is equal to the sum of the individual angular momenta ($L=L_{1}+L_{2}$), and since the particles are independent, $|\varphi \rangle =|1\rangle |2\rangle$ where $|\varphi \rangle$ is the state of the entire system. We also know that $L^{2}=L_{x}^{2}+L_{y}^{2}+L_{z}^{2}$. Now taking $\langle L^2 \rangle = \langle \varphi | L^2 | \varphi \rangle$, we end up with $\langle L^2 \rangle = \langle L_{1}^2 \rangle + \langle L_{2}^2 \rangle + 2 \langle L_{1}L_{2} \rangle$. The $\langle L_{1}^2 \rangle$ and $\langle L_{2}^2 \rangle$ are easy to calculate, but I don't see how to calculate $\langle L_{1}L_{2} \rangle$. Perhaps there's a way to show that $\langle L_{1}\rangle=\langle L_{2} \rangle$ so that this would reduce to $\langle L^2 \rangle = 4 \langle L_{x}^2 \rangle$, but I don't see a way to do that.

vela
Staff Emeritus
Since you have the states, you could just explicitly calculate what the expectation value of $$L_1\cdot L_2 = {L_1}_x{L_2}_x + {L_1}_y{L_2}y + {L_1}_z{L_2}_z$$ is by expressing Lx and Ly in terms of the ladder operators, L+ and L-.