# Is the spin-singlet state also a Resonating-Valence-Bond(RVB) state?

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The spin-singlet state of a lattice spin-1/2 system is defined as $S_x\Psi=S_y\Psi=S_z\Psi=0$, where $S_\alpha=\sum S_i^\alpha(\alpha=x,y,z)$ are the total spin operators, in other words, a spin-singlet state is a spin state with gobal $SU(2)$ spin-rotation symmetry. On the other hand, a RVB state is the superposition of various configurations of singlet-product states (no need for equal weight superposition here), thus, a RVB state is of course a spin-singlet state according to the above definition.

For the reversed statement, is the spin-singlet state also a RVB state? How to prove it? For the extreme 2-spin system, it's direct to show the equivalence between the spin-singlet state and the RVB state.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user K-boy

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Yes, a spin-singlet state is also an RVB state. The valence bound states (singlet-product states) over-complete the Hilbert space of spin-singlet states.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user Everett You
answered Dec 9, 2013 by (785 points)
@ Everett You Overcomplete? Let $N$(even) be the number of lattice sites, then there will be $(N-1)!!$ different valence bound states(VBS), right? Since each VBS is a spin-singlet state, the dimension $D$ of the Hilbert space of spin-singlet states must $D\geqslant (N-1)!!$, do you mean that $D\leqslant (N-1)!!$ at the same time and hence $D=(N-1)!!$?

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user K-boy
@K-boy Yes, there will be $(N-1)!!$ different VBS states. But these states are not orthogonal to each other. For example, if you have 4 spins, there will be 3 VBS configurations, but the 3 state vectors are 120 degree to each other in a 2 dim plane, so there are only 2 linearly independent (or orthogonal) singlet state, i.e. the Hilbert space dim is 2. In general, for $N$(even) spins, the number of singlet states is $N!/((N/2)!(N/2+1)!)$, known as the Catalan numbers (oeis.org/A000108), which is far less than $(N-1)!!$ VBS states. So the VBS basis is over-complete.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user Everett You
@ Everett You I see. I misunderstood that all the $(N-1)!!$ VBS are linearly independent without carefully thinking.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user K-boy
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One can convince oneself this is indeed the case by doing the dimensional counting as it was done by Everett You. However, it is by no means a proof. The problem is that the valence bond states are not linearly independent. Even though there are much more valence bond states than the number of singlets made from $N$ spin-one-half spins, it is still possible that the dimension of the vector space generated by these valence bond states is smaller than the dimension of singlet space.

One can formulate a proof along the following lines.

Let $V$ be the linear space carries the $s=1/2$ representation of $SU(2)$. Then, the Hilbert space of $N$ spins, $V^{\otimes N}$, not only carry a reducible representation of $SU(2)$ but also a reducible representation of permutation group $S_N$. The key observation is that the singlet space $V_0$, i.e. the space of all singlet states formed by $N$ spins, carries an $\textit{irreducible representation}$ of $S_N$. (This is a trivial example of Schur-Weyl duality)

Now, let us label the spins by $1,2,\cdots N$, and $N$ is an even number. Consider the following singlet state:

$$|\psi\rangle=|1,2\rangle\otimes|3,4\rangle\otimes\cdots|N-1,N\rangle.$$

Here $|i,j\rangle$ stands for the singlet formed by spins $i$ and $j$. Apparently, this is a valence bond state. Furthermore, acting a permutation $\pi\in S_N$ on $|\psi\rangle$ gives rise to another valence bond state:

$$U(\pi)|\psi\rangle=|\pi(1),\pi(2)\rangle\otimes|\pi(3),\pi(4)\rangle\otimes\cdots|\pi(N-1),\pi(N)\rangle.$$

Thus, we can construct a vector space $W$ generated by $U(\pi)|\psi\rangle$, $\pi\in S_N$, and it carries a representation of $S_N$. As all valence bond states can be obtained in this way, $W$ is actually the space generated by valence bond states.

Now I claim $W=V_0$. To see this, we notice any state $U(\pi)|\psi\rangle$ is a singlet, and therefore is in $V_0$. In other words, $W\subseteq V_0$. Furthermore, $W$ carries a representation of $S_N$ and $V_0$ is irreducible, which implies $W=V_0$ or $W=\emptyset$. As $W\neq\emptyset$ by construction, $W=V_0$.

Recall that $W$ is the space generated by all valence bond states and $V_0$ is the space of all singlets that can be formed by $N$ spins. Since $W=V_0$, we see any singlet state made of $N$ spins can be written as a linear superposition of valence bond states.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user Isidore Seville
answered Dec 13, 2013 by (10 points)
@ Isidore Seville, thanks for your detailed answer. Yes, you present a proof here, please give me a little more time to understand your proof and I will make some comments later.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user K-boy
@ Isidore Seville, I'm sorry that I am not familiar with representation theory but I believe that your proof is correct. Do we have a proof without the language of representation theory ?

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user K-boy
@K-boy No, not at the moment. But I can think about it.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user Isidore Seville
@ Isidore Seville Do you know how to compute the dimension of space $W$ or $V_0$?

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user K-boy
@K-boy Yes. Not only you can compute the dimension of $V_0$, you can also compute the dimension of $V_S$ for any total spin quantum number. A detailed answer is too long to fit in as a comment, though.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user Isidore Seville

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