The textbooks say, the relative angular momentum of any two particles i and j should be no less than n, n is the power of (zi-zj)^n in the equation below

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I don't see this is an obvious argument. Can anybody clarify why?

Also, which one is a better definition of relative angular momentum between particle i and j?

$<r_i \times p_i-r_j\times p_j>$

For 2D complex coordinates, the argument is:
$\left \langle \Psi|\bar{z} \bar{\partial}-z\partial | \Psi \right \rangle\geqslant n$


1 Answer 1


First, I think the 2nd definition you provided here is the correct one.

Second, to see that the relative angular momentum between each pair in this Laughlin wavefunction is no less than $n\hbar$, we can change the variables $Z_{ij}=\frac{z_i+z_j}{\sqrt{2}},z_{ij}=\frac{z_i-z_j}{\sqrt{2}}$ in the Laughlin wavefunction $\Psi$, keeping other variables $z_{l},\ne i,j$ unchanged. Then we can expand out $\Psi$ as follows \begin{align} \Psi & =\sum_{m^{\prime}} \,\{\sum_m f_{m,m^{\prime}}(..,z_{i-1},z_{i+1},...,z_{j-1},z_{j+1},...)\, Z_{ij}^m e^{-|Z_{ij}|^2}\}\, z_{ij}^{m^{\prime}} \; e^{-|z_{ij}|^2} \\ & \equiv \sum_{m^{\prime}} g_{m^{\prime}}(..,z_{i-1},z_{i+1},...,z_{j-1},z_{j+1},...,Z_{ij}) \;z_{ij}^{m^{\prime}} \; e^{-|z_{ij}|^2} \end{align} From the function form of $\Psi$, it is quite clear that $m^{\prime}\ge n$. Since each term in the 2nd line above is an eigenfunction of the relative angular momentum operator $\hat{L}_{ij}=-i\hbar[z_{ij}\partial_{z_{ij}}-\bar{z}_{ij}\partial_{\bar{z}_{ij}}]$ with an eigenvalue $m^{\prime}\hbar \ge n\hbar$, we immediately see that the expectation value $<\Psi|\hat{L}_{ij}|\Psi>\ge n\hbar$.


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