9
$\begingroup$

I am quite interested in the understanding of the relation between p_ip wave superconductor(SC) and the Moore-Read(MR) state.

They share many similar properties, for example, p+ip SC has majorana as vortex excitation, MR has nonabelian quasi hole excitation. More interestingly, their wavefuncions are similar to some extent. All of them can be found in Read and Green's paper: http://arxiv.org/pdf/cond-mat/9906453v3.pdf. (where you can find the wavefunction for p+ip SC in eqn.12 and above, MR wavefunction in eqn.5.1 in http://www.physics.rutgers.edu/~gmoore/MooreReadNonabelions.pdf).

These two wave functions are similar only when electrons are separated far from each other, as shown in eqn 16 in Read Green's paper.

Does anyone know how to explain the relation between these two states? Why they share similar property at long distance?

Thanks in advance.

$\endgroup$
1
2
$\begingroup$

I'm not an expert in quantum Hall physics, and this may not be a comprehensive answer, but let me point out an interesting connection. In arXiv:1712.09904 the Read-Green model is restudied at the special point $\mu=\frac{m\Delta^2}{2}$ where the Hamiltonian can be written in a factorized form \begin{eqnarray}\label{eq:K_bb} \hat{K}&\equiv&\int_{S}\left[ \frac{\nabla \psi_{z}^{\dagger }\cdot \nabla \psi _{z}}{2m}-(\Delta \psi_{z}\partial_{\bar{z}}\psi_{z}+\mathrm{H.c.})+\mu \psi_{z}\psi_{z}^{\dagger }\right] d^{2}z, \\ &=&\int_{S}\frac{1}{2m}(2\partial_{z}\psi_{z}^{\dagger }-m\Delta^*\psi _{z})(2\partial_{\bar{z}}\psi_{z}-m\Delta\psi_{z}^{\dagger })d^{2}z, \end{eqnarray} where $S$ denotes a region in the 2D plane with complex coordinates $z=x+iy$, $\partial_{z}=(\partial_{x}-i\partial_{y})/2$, and $\psi_{z}$ is the fermionic annihilation operator at position $z$. Then it can be easily checked that the state \begin{equation} |G\rangle =\exp \left[ \frac{m\Delta}{4\pi}\int_{S}\frac{1}{z-z^{\prime }}\psi _{z}^{\dagger }\psi_{z^{\prime }}^{\dagger }d^{2}zd^{2}z^{\prime }\right] |0\rangle.\end{equation} is the exact ground state of $\hat{K}$ by noticing that $|G\rangle$ is annihilated by $(2\partial_{\bar{z}}\psi_{z}-m\Delta\psi_{z}^{\dagger })$ and that $\hat{K}$ is positive semi-definite. Then we notice that if we project $|G\rangle$ to the $N=2M$ particle subspace we immediately get a Pfaffian state \begin{eqnarray} \hat{P}_{N=2M}|G\rangle &=&\frac{1}{M!} \left[ \frac{m\Delta}{4\pi}\int_{S}\frac{1}{z-z^{\prime }}\psi _{z}^{\dagger }\psi_{z^{\prime }}^{\dagger }d^{2}zd^{2}z^{\prime }\right]^M |0\rangle\\ &=& \mathrm{const.}\int_S \mathrm{Pf}\left[\frac{1}{z_i-z_j}\right]_{1\leqslant i,j\leqslant 2M} \psi^\dagger_{z_1}\ldots \psi^\dagger_{z_{2M}}d^2z_1\ldots d^2 z_{2M}|0\rangle, \end{eqnarray} (see the definition of Pfaffian by wedge product). By comparison, the Moore-Read wavefunction without quasiholes is \begin{equation} \psi(z_1,\ldots,z_{2M})=\mathrm{Pf}\left[\frac{1}{z_i-z_j}\right]_{1\leqslant i,j\leqslant 2M} \left[ \prod_{1 \leqslant i < j \leqslant 2M}\left( z_i-z_j \right)^2 \right] \prod^{2M}_{k=1}\exp\left( - \mid z_k \mid^2 \right). \end{equation} As you can see, the two states are almost the same except that the $p_x+ip_y$ ground state don't have the Laughlin factor. The physical interpretation for this is that the Moore-Read state can be viewed as a $p_x+ip_y$ superconducting state of composite fermions in the background of a Laughlin state. At $\nu=5/2$ FQHE every electron absorb 2 quantized vortices forming a composite fermion, and the weak attactive interaction between those composite fermions causes a Cooper instability to form a $p_x+ip_y$ superconductor.

More interesting things happen when vortices/quasihole are introduced. The exact ground state (in the even particle number sector) of $\hat{K}$ with 2 vortices at position $\eta_1,\eta_2$ is \begin{equation} |G_{\eta_1,\eta_2}\rangle =\exp \left\{ \frac{m\Delta}{4\pi}\int_{S}\frac{1}{z-z^{\prime }}\left[\sqrt{\frac{(z-\eta_2)(z'-\eta_1)}{(z-\eta_1)(z'-\eta_2)}}+(z\leftrightarrow z')\right]\psi _{z}^{\dagger }\psi_{z^{\prime }}^{\dagger }d^{2}zd^{2}z^{\prime }\right\} |0\rangle, \end{equation} which is also annihilated by $(2\partial_{\bar{z}}\psi_{z}-m\Delta\psi_{z}^{\dagger })$. The extra square root factor is necessary because in our gauge convention fermion fields $\psi^\dagger_z$ are anti-periodic around a vortex. The projected state $\hat{P}_{N=2M}|G_{\eta_1,\eta_2}\rangle$ has a Pfaffian wavefunction \begin{equation} \hat{P}_{N=2M}|G_{\eta_1,\eta_2}\rangle\propto\int_S\mathrm{Pf}\left\{\frac{1}{z_i-z_j} \left[\sqrt{\frac{(z_i-\eta_2)(z_j-\eta_1)}{(z_i-\eta_1)(z_j-\eta_2)}}+(z_i\leftrightarrow z_j)\right] \right\}_{1\leqslant i,j\leqslant 2M}\psi^\dagger_{z_1}\ldots \psi^\dagger_{z_{2M}}d^2z_1\ldots d^2 z_{2M}|0\rangle, \end{equation} while the Moore-Read wavefunction with 2 quasiholes is \begin{equation} \psi^{\text{MR}}_{\eta_1,\eta_2}(z_1,\ldots,z_{2M})=\mathrm{Pf}\left\{\frac{(z_i-\eta_1)(z_j-\eta_2)+(z_i\leftrightarrow z_j)}{z_i-z_j} \right\}_{1\leqslant i,j\leqslant 2M}\times(\text{Laughlin factor}). \end{equation} As we see, beyond the Laughlin factor, the $p+ip$ ground state has an extra factor $1/\sqrt{(z_i-\eta_1)(z_i-\eta_2)(z_j-\eta_1)(z_j-\eta_2)}$ inside the Pfaffian, indicating a difference in the short-range physics between the two states. Nevertheless, it was explicitly checked in arXiv:1712.09904 that the two states has exactly the same topological degeneracy and non-Abelian statistics, as the analytic continuation of both wavefunctions give exactly the same non-Abelian unitary rotation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.