Spin State of Halperin \((mmn)\) States

This note aims to clarify many folklores that I have heard for a long time: that the Halperin \((112)\) state is a singlet; that the \((n+1,n+1,n)\) state that is a singlet; that all other \((mmn)\) states are not singlets. I thank Patrick J. Ledwith and Tomohiro Soejima for providing some of the crucial arguments discussed here. Main references are David Tong's Notes on quantum Hall effects and Jainendra Jain's book Composite Fermions.

The Halperin \((mmn)\) states

The Halperin \((mmn)\) state is defined as

\[\Psi(\{z_i\},\{w_i\}) = \prod_{i \left.<\right. j}^{N_z} (z_i-z_j)^m \prod_{i \left.<\right. j}^{N_w} (w_i-w_j)^m \prod_{i,j}(z_i-w_j)e^{-\phi(\{z_i\},\{w_i\})},\]

in which \(\phi(\{z_i\},\{w_i\}) = -(\sum_i |z_i|^2-\sum_i |w_i|^2)/4\ell^2\) describes the Gaussian factor of the lowest Landau level. In practice, this state should be interpreted as a two-component many-body state:

\[\ket{\Psi} = \int Dz_i Dw_i\Psi(\{z_i\},\{w_i\}) \prod_i c^\dagger_{\uparrow}(z_i) \prod_i c^\dagger_{\downarrow}(w_i) \ket{0}.\]

\(\ket{\Psi}\) is an eigenstate of \(S_z\) with eigenvalue \((N_\uparrow-N_\downarrow)/2\). It describes the Abelian Chern-Simons theory with

\[K = \begin{pmatrix} m & n \\ n & m \end{pmatrix}, \quad t = (1,1).\]

The ground state degeneracy on a torus is \(|\det K| = |m^2-n^2|\). (Note that \(K\) is assumed to be non-singular; the \((mmm)\) state describes a Laughlin \(\nu=1/m\) state that has degeneracy \(m\).) The filling at which this state can occur is

\[ \nu = t^T K^{-1} t = \frac{2m-2n}{m^2-n^2}=\frac{2}{m+n} \]

Due to Galilean symmetry, \(\sigma_{xy} = \nu\).

These are the basic features of the \((mmn)\) state. We will now set \(S_z=0\) and seek the answer to the question: which of these \((mmn)\) states are singlets? To do this we need an interesting condition called the Fock's cyclic condition (FCC).

Fock's cyclic condition

The FCC states the following condition. Given a many-body state \(\Psi(\{z_i\},\{w_j\})\), in which \(z_i\) describes the positions of the up-spin electrons and \(w_j\) describes the positions of the down-spin electrons, we say the state satisfies the FCC condition if and only if

\[ \sum_{i} (z_i, w_j)\Psi = \Psi \]

where \((z_i, w_j)\) means that one switches the position of \(z_i\) and \(w_j\) in the wavefunction. Although this might seem quite bizarre, it implies something deep: one can prove that, \(\Psi\) has total angular momentum \(S=S_z\) if and only if FCC holds; in other words, \(S^+ \Psi = 0\).

To prove this think about the action of \(S^+\) on \(\Psi\); this is going to split into \(S^+ = \sum_j S_j^+\). We will now show that \(S_1^+\Psi=0\) is equivalent to the FCC condition with \(j=1\); one can convince themselves that given fermionic exchange, the FCC statement shouldn't depend on \(j\).

With the many-body state (2) we write down the action of \(S_1^+\):

\[ S_1^+\ket{\Psi} = \int Dz_i Dw_j \Psi(\{z_i\},\{w_j\})\prod_i c^\dagger_{\uparrow}(z_i)c^\dagger_{\uparrow}(w_1)\prod_{j\neq 1} c^\dagger_{\downarrow}(w_j) \ket{0} \]

Now if this state is zero, all the wavefunction amplitudes must be zero. What is the wavefunction amplitude corresponding to the component \(\prod_i c^\dagger_{\uparrow}(z_i)c^\dagger_{\uparrow}(w_1)\prod_{j\neq 1} c^\dagger_{\downarrow}(w_j) \ket{0}\)? It is simply \(\Psi(\{z_i\},\{w_j\})\), anti-symmetrized over all the up-spin electron coordinates (the down-spins have already been anti-symmetrized.) It is then not hard to see that the antisymmetrization gives the wavefunction amplitude of \(\Psi - \sum_i (z_i, w_1)\Psi\).

Which are singlets?

We now have a powerful tool to check whether the state could be a singlet. Given that for a state to be a singlet, \(N_\uparrow\) must be equal to \(N_\downarrow\), or \(S_z\) must be zero, it is a singlet if and only if it satisfies the FCC.

We will now prove the only singlets are of type \((n+1,n+1,n)\). To see that \((n+1,n+1,n)\) is a singlet, note that it can be seen as symmetrically attaching \(n\) vortices to a fully filled (with both spins) LLL wavefunction. Given that symmetrically attaching vortices don't change the \(SU(2)\) symmetry (check for yourself!), and that a fully filled band is a singlet, the state must be a singlet. Proving that all others are not singlets is much more tricky.

Preemptively, we will call our chosen \(w_j\) to be risen \(z_0\). We will use the sign convention that the wavefunction is written as

\[ \Psi = \prod_{i\left.<\right. j} (z_i-z_j)^m(w_i-w_j)^m \prod_{i, j} (z_j-w_i)^n \prod_i (z_0-w_i)^m (z_0-z_i)^n \]

where the Gaussian factor has been premptively omitted. We will use the convention that Latin indices \(i,j,\dots\) will not run through \(0\).

Then, given that

\[ (z_0, z_k)\Psi = (-1)^n\prod_{i\neq k}\left(\frac{z_0-z_i}{z_k-z_i}\right)^{m-n}\prod_{j}\left(\frac{z_k-w_j}{z_0-w_j}\right)^{m-n} \Psi \]

Define \(L=m-n\). Thus, what we need to prove is that

\[ F(\{z_i\},\{w_j\})=1-(-1)^n\sum_k \prod_{i\neq k}\left(\frac{z_0-z_i}{z_k-z_i}\right)^{L}\prod_{j}\left(\frac{z_k-w_j}{z_0-w_j}\right)^{L} \neq 0 \]

for \(L\neq 1\). Now the proof is divided into several cases.

\(L=0\)

This case is trivial; as long as the particle number is large, \(F\) is a large nonzero constant.

\(L \left.<\right. 0\)

This case is also easy. We note that \(F\) is a holomorphic function in all the variables. Let's consider the variable \(q_{kj}=z_k-w_j\). Given this is a linear transformation the function had better be a holomorphic function of \(q_{kj}\) too. Now we only need to look at the function as a monovariable holomorphic function, with all other variables as parameters: i.e. \(F=F(q_{kj};\dots)\). How do we prove that this is nonzero? We will study its singular behavior. We note that the leading singularity of \(q_{kj}\to 0\) come only from one term (note \(L<0\)!), and it could not be cancelled by anything else; this leads to the conclusion that \(F\neq 0\).

\(L \left. > \right. 0\)

This is more difficult. We will consider the leading singularity when \(z_{ij}=z_i-z_j\) goes to zero.. Now the leading contributions come from two terms, which can be possibly cancelled:

\[ \begin{aligned} F(z_{ij},\dots) &\propto \frac{\prod_{k\neq i,j}(z_0-z_k)^L\prod_{k\neq i,j, \, l}(z_k-w_l)^L}{\prod_l (z_0-w_l)^L}\\ &\left(\frac{(z_0-z_j)^L\prod_l (z_j+z_{ij}-w_l)^L}{\prod_{k\neq i,j}(z_j+z_{ij}-z_k)^L}+(-1)^L \frac{(z_0-z_{j}-z_{ij})^L\prod_l (z_j-w_l)^L}{\prod_{k\neq i,j}(z_j-z_k)^L}\right) \frac{1}{z_{ij}^L} + \dots \end{aligned} \]

Let's define

\[ F_1=\frac{(z_0-z_j)^L\prod_l (z_j+z_{ij}-w_l)^L}{\prod_{k\neq i,j}(z_j+z_{ij}-z_k)^L}, \quad F_2 = \frac{(z_0-z_{j}-z_{ij})^L\prod_l (z_j-w_l)^L}{\prod_{k\neq i,j}(z_j-z_k)^L}. \]

One can convince themselves that \(F_1=F_2\) when \(z_{ij}=0\). Hence, if \(L\) is even, \(F_1\) and \(F_2\) wouldn't cancel each other; the leading singularity survives. So the only possibility is from \(L\) positive and odd.

Now let's prove that

\[ F_1-F_2 = F_3 z_{ij} + O(z_{ij})^2 \]

where \(F_3\) is a nonzero holomorphic function. Explicit derivatives show that

\[ F_3 = L\frac{(z_0-z_j)^L\prod_l (z_j-w_l)^L}{\prod_{k\neq i,j}(z_j-z_k)^L}\left(\sum_l\frac{1}{z_j-w_l}-\sum_{k\neq i,j}\frac{1}{z_j-z_k} + \frac{1}{z_0-z_j}\right) \]

which, again according to the singularity counting trick, is nonzero.

Thus the leading singularity of \(F\) in \(z_{ij}\) is \(z_{ij}^{-(L-1)}\). If \(L=1\), regular terms that we have ignored can come in and potentially cancel them out; however, when \(L>1\), no other terms can cancel this singularity. Hence the FCC is not satisfied, and these states cannot be singlets.

In conclusion: the only singlet is the \((n+1, n+1, n)\) state.

So what's the mysterious \((112)\) state?

As it turns out this is really a misnomer. I think it's better thought of a \((\bar{1}\bar{1}2)\) state – at filling \(2/3\), one can check that filling two hole Landau levels (which can be spin up plus spin down LLLs) of the two flux composite fermion gives the correct filling. This state should be what was found in numerics.