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Question: How one obtains $$\langle \psi(z_1) \cdots \psi(z_N) \sigma(w_1) \sigma(w_2) \rangle \sim \mathrm{Pf}\left( \frac{f(z_i,z_j; w_1, w_2)}{z_i-z_j}\right) \prod_{i=1}^N (z_i-w_1)^{-1/2} (z_i-w_2)^{-1/2},$$ where $\mathrm{Pf}$ is the Pfaffian, $\psi$ is the neutral fermion and $\sigma$ is the spin field in the Ising conformal field theory, and $$f(z_i,z_j;w_1,w_2) = (z_i-w_1)(z_j-w_2) + (z_j-w_1)(z_i-w_2).$$ The result appears in the seminal paper of Read and Rezayi [see Eqs. (2.17) and (2.18)], discussing a new fractional quantum Hall phase (now called Read-Rezayi state). The paper states that the above correlator represents a two-quasihole wavefunction of Moore-Read state.

Note: I can evaluate the fermion correlators $$\langle \psi(z_1) \cdots \psi(z_N) \rangle = \mathrm{Pf}\left( \frac{1}{z_i-z_j} \right)$$ using Wick's theorem. What I cannot understand is how one deals with the spin field $\sigma$.

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1 Answer 1

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We proceed by using the bosonization formula. The square of the correlator is, after the bosonization, $$\langle \psi(z_1) \cdots \psi(z_N) \sigma(w_1) \sigma(w_2) \rangle ^2 = \frac12 \langle b(z_1) \cdots b(z_N) \left( V_{\frac12}(w_1) V_{-\frac12}(w_2) + (-1)^N V_{-\frac12}(w_1) V_{\frac12}(w_2) \right) \rangle,$$ where $b(z) = -\partial \phi(z)$ is the U(1) current, and $V_\alpha(z) \sim e^{i \alpha \phi(z)}$ is the vertex operator. We used the fact that two Ising anyons should fuse to identity channel if $N$ is even, and to $\psi$ channel if $N$ is odd.

The left-hand side of the above equation should be proportional to, using the statement given in the question, $$\det\left( \frac{f(z_i,z_j; w_1, w_2)}{z_i-z_j}\right) \prod_{i=1}^N \frac{1}{z_i-w_1} \frac{1}{z_i-w_2},$$ where we used the fact that $\mathrm{Pf}\,(A)^2 = \det A$ for a matrix $A$. It remains to evaluate the right-hand side, but this can be done by using the Ward identity.

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