Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,549 answers , 19,356 comments
1,470 users with positive rep
410 active unimported users
More ...

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

+ 1 like - 0 dislike
37 views

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
asked Dec 9, 2013 in Theoretical Physics by Kai Li (975 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

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 Everett You (660 points) [ no revision ]
@ 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
+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-07 16:29 (UCT), posted by SE-user Isidore Seville
answered Dec 13, 2013 by Isidore Seville (10 points) [ no revision ]
@ 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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...