# What is the energy functional for $\nu=5/2$ Moore-Read state?

I am trying to do some Monte Carlo simulations for Pfaffian state from Fractional Quantum Hall effect. I am wondering what is the energy functional for $\nu=5/2$ Moore-Read state?

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What do you mean by the "energy functional"? The energy as a function of the fermionic fields? Also, how do you do Monte Carlo for a Pfaffian system? Genuinely curious, because Pfaffian fermions have a horror of a sign problem--- if you could elaborate, perhaps somebody would know the answer. –  Ron Maimon Aug 19 '11 at 8:59

You know an explicit expression for the unnormalized Moore-Read wavefunction in (complex) position representation $$\psi_{\mathbf{MR}}(\{z_i\})= \mathbf{Pf}\left( \frac1{z_i-z_j} \right) \prod_{i<j} (z_i-z_j)^2 e^{-\sum_k |z_k|^2/4}\quad .$$ The energy of a wavefunction is given by the corresponding average value of the Hamiltonian. $$E_\psi=\langle H \rangle = \mathcal{N} \int dz_1\dots dz_n \left|\psi_{\mathbf{MR}}(\{z_i\})\right|^2 H(\{z_i\})$$ with the normalization constant $$\mathcal{N}= \left( \int dz_1\dots dz_n \left|\psi_{\mathbf{MR}}(\{z_i\})\right|^2 \right)^{-1} \quad .$$ In your case I guess the Hamiltonian only consists in a Coulomb term, thus its matrix elements in position space are $$H(\{z_i\})=\frac{e^2}{4\pi\epsilon} \sum_{i<j} \frac{1}{|z_i-z_j|}$$ since $||r_i-r_j||=|z_i-z_j|$.

The two integrals involved are computationally costly, but you can benefit from the Metropolis sampling method (http://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm), the most commonly used sampling method in Monte-Carlo algorithms I guess. It is very simple. In the latter you:

• randomly choose an initial configuration $\{z_i\}$ (i.e. set of position). Its probability is $\pi_0=|\psi\{z_i\}|^2$.

• generate a new configuration from the first one by randomly moving one of the particles. I note the new configuration probability $\pi_1$.

• the new configuration is accepted with probability 1 if it has a higher probability than the first one, and with probability $\pi_1$ otherwise.

• continue until the desired accuracy is reached or maximum number of attempt have be tried out.

Every accepted move contributes to the integral with weight 1/"number of sample configurations used". Explicitly if $I$ is the quantity I want to compute, $N$ the number of configurations $i$ I keep, and $f_i$ the value of the integrand for configuration $i$, $$I=\frac1N\sum_{i=1}^N f_i \quad .$$

Good luck.

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