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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.

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    $\begingroup$ Usually we wouldn't call a Valence Bond Solid an RVB, right? $\endgroup$ – Ruben Verresen May 9 '16 at 14:10
  • $\begingroup$ @Ruben Verresen Yes. VBS breaks translation symmetry while RVB does not. I use the term here just for convenience. $\endgroup$ – Kai Li May 11 '16 at 9:33
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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.

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  • $\begingroup$ @ 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 $D=(N-1)!!$? $\endgroup$ – Kai Li Dec 9 '13 at 18:51
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    $\begingroup$ @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. $\endgroup$ – Everett You Dec 10 '13 at 6:33
  • $\begingroup$ @ Everett You I see. I misunderstood that all the $(N-1)!!$ VBS are linearly independent without carefully thinking. $\endgroup$ – Kai Li Dec 10 '13 at 8:41
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The short answer is yes.

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.

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  • $\begingroup$ @ 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. $\endgroup$ – Kai Li Dec 13 '13 at 10:34
  • $\begingroup$ @ 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 ? $\endgroup$ – Kai Li Dec 14 '13 at 16:17
  • $\begingroup$ @K-boy No, not at the moment. But I can think about it. $\endgroup$ – Isidore Seville Dec 14 '13 at 19:34
  • $\begingroup$ @ Isidore Seville Do you know how to compute the dimension of space $W$ or $V_0$? $\endgroup$ – Kai Li Dec 16 '13 at 15:10
  • $\begingroup$ @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. $\endgroup$ – Isidore Seville Dec 16 '13 at 18:37

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