When reading Prof. David Tong's notes on FQHE, I got a question regarding the Haldane pseudopotential. We have the unique analytic solution of the lowest Landau level of two particles with a potential which only depends on the distance between the two particles: $V(r_1,r_2)=V(|r_1-r_2|)$ as: $$\psi_{mM}(z_1,z_2)=(z_1-z_2)^m(z_1+z_2)^Me^{-\frac{1}{4}(|z_1|^2+|z_2|^2)}$$ with $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$. $m,M$ are non-negative integers. The Haldane pseudopotential is defined as: $$v_m=\frac{\langle mM|V|mM\rangle}{\langle mM|mM\rangle}\tag{3.11}$$ and the claim is that $v_m$ is not dependent on $M$. I'm having a little trouble to see why this is true? Is there any way that cancels out the $M$ dependent term in the integral: $$\int d\vec{r}_1\int d\vec{r}_2\psi_{mM}(\vec{r}_1,\vec{r}_2)^* V(|\vec{r}_1-\vec{r}_2|) \psi_{mM}(\vec{r}_1,\vec{r}_2)$$ or is there any other way to compute the expectation value?
2 Answers
Let's do a change of variables, from $\vec{r}_1,\vec{r}_2$ to $\vec{\Delta}=\vec{r}_1-\vec{r}_2$ and $\vec{\Sigma}=\vec{r}_1+\vec{r}_2$. Note that the Jacobian associated with this transformation is $2$.
Furthermore, we note that $\psi$ has the following nice factorization property: \begin{eqnarray} \psi_{mM}(\vec{r}_1,\vec{r}_2) &=& |\vec{r}_1-\vec{r}_2|^m |\vec{r}_1 + \vec{r}_2|^M \exp\left(-\frac{1}{4} \left(|\vec{r}_1|^2+|\vec{r}_2|^2\right)\right) \\ &=& |\vec{\Delta}|^m|\vec{\Sigma}|^M \exp\left(-\frac{1}{8} \left(|\vec{\Sigma}|^2 + |\vec{\Delta}|^2\right)\right) \\ &=& \phi_m(\vec{\Delta}) \phi_M(\vec{\Sigma}) \end{eqnarray} where to go from line 1 to line 2 we used \begin{equation} |\vec{\Sigma}|^2+|\vec{\Delta}|^2=2(|\vec{r}_1|^2+|\vec{r}_2^2|+|\vec{r}_1\cdot\vec{r}_2|-|\vec{r_1}\cdot\vec{r}_2|)=2(|\vec{r}_1|^2+|\vec{r}_2|^2) \end{equation} and where we defined $\phi_j(x) \equiv |x|^j e^{-|x|^2/8}$.
Then we can write the numerator as \begin{eqnarray} \langle mM | V | mM \rangle &=& \int d \vec{r}_1 \int d\vec{r}_2 \psi^\star_{mM}(\vec{r}_1,\vec{r}_2) V(|\vec{r}_1-\vec{r}_2|) \psi_{mM}(\vec{r_1}\vec{r}_2) \\ &=& 2 \int d \vec{\Sigma} \int d \vec{\Delta} |\psi_{mM}(\vec{\Sigma},\vec{\Delta})|^2 V(|\vec{\Delta}|) \\ &=&2 \left[\int d \vec{\Sigma} |\phi_M(\vec{\Sigma})|^2\right] \left[\int d \vec{\Delta} |\phi_m(\vec{\Delta})|^2 V(|\Delta|)\right] \end{eqnarray} Similarly we can write the denominator as \begin{equation} \langle mM | mM \rangle = 2 \left[\int d \vec{\Sigma} |\phi_M(\vec{\Sigma})|^2\right] \left[\int d \vec{\Delta} |\phi_m(\vec{\Delta})|^2 \right] \end{equation} The first term in brackets, that depends on $M$, cancels when we divide the numerator and denominator.
Andrew presented a really nice solution here. However, I struggle to reproduce what he calls a nice factorization property. Specifically, how to arrive from $(z_1-z_2)^m(z_1+z_2)^M$ to $|\vec{r}_1-\vec{r}_2|^m|\vec{r}_1+\vec{r}_2|^M$.
I can't tell if that step is incorrect or my math is a little rusty. Anyway, I found that the following is true $(z_1-z_2)^m[(z_1-z_2)^m]^*(z_1+z_2)^M[(z_1+z_2)^M]^* = |\vec{r}_1-\vec{r}_2|^{2m}|\vec{r}_1+\vec{r}_2|^{2M}$, by simply using trigonometric notation, i.e. $z = re^{i\phi}$, and doing some basic algebra. So, the solution is still correct. Just wanted to add this into consideration, if anyone else might get confused with that first step.
