I am trying to read the original paper for the AKLT model,

Rigorous results on valence-bond ground states in antiferromagnets. I Affleck, T Kennedy, RH Lieb and H Tasaki. Phys. Rev. Lett. 59, 799 (1987).

However I am stuck at Eq. $(1)$:

we choose our Hamiltonian to be a sum of projection operators onto spin 2 for each neighboring pair: $$ \begin{align} H &= \sum_i P_2 (\mathbf{S_i} + \mathbf{S_{i+1}}) \\ &= \sum_i \left[\frac{1}{2}\mathbf{S_i}\cdot\mathbf{S_{i+1}} + \frac{1}{6}(\mathbf{S_i}\cdot\mathbf{S_{i+1}})^2 + \frac{1}{3}\right] \end{align}\tag{1} $$

In the equation, $H$ is the proposed Hamiltonian for which the authors intend to show that the ground state is the VBS ground state: the (unique) state with a single valence bond connecting each nearest-neighbor pair of spins, i.e. a type of spin-$1$ valence-bond-solid. Moreover, $\mathbf{S_i}$ and $\mathbf{S_{i+1}}$ are spin-$1$ operators, and $P_2$ is the projection operator onto spin-2 for the pair $(i,i+1)$.

I have several questions here.

  • First, how did the authors come up with the first line by observing the spin-1 valence-bond-solid state as below (Fig. 2 of the above paper)?

  • Why do they use a Hamiltonian which is "a sum of projection operators onto spin 2 for each neighboring pair"?

  • What does it mean exactly to project spin-$1$ pairs to spin $2$, and why do they want to project to spin $2$?

I feel there are quite a few steps skipped between here and standard QM textbooks. It would be great if somebody could recommend me some materials bridging them.

  • Secondly, I want to know how to go from the first line to the second line of equation $(1)$. However, I couldn't find the explicit form of $P_2$ either in the paper or by googling. Could somebody give me a hint?

Edit: I found a chapter of the unfinished book "Modern Statistical Mechanics" by Paul Fendley almost directly answers all my questions.

  • $\begingroup$ Is Sec. 2.3.1 of arxiv.org/abs/1306.5551 of any help? $\endgroup$ – Norbert Schuch Oct 16 '16 at 16:09
  • $\begingroup$ Thanks for the advice. The article you mentioned and some references it cites are very helpful. Also I found a chapter of the unfinished book "Modern Statistical Mechanics" by Paul Fendley almost directly answers all my questions. $\endgroup$ – entron Oct 18 '16 at 8:48
  • 3
    $\begingroup$ Related meta post: meta.physics.stackexchange.com/q/9308/2451 $\endgroup$ – Qmechanic Nov 1 '16 at 14:03

Let me try to answer my own questions to thank those who gave voice and support for undeleting this question. My main reference is the chapter 3 Basic quantum statistical mechanics of spin systems of the unfinished book "Modern Statistical Mechanics" by Paul Fendley.

The spin-1 valence-bond-solid state (VBS) in Fig. 2 can be imagined as the following:

  • Each spin-1 site is made of two spin-1/2 in triplet states (which has s=1)
  • Each of the imagined spin-1/2 forms a singlet state (valence-bond) with another spin-1/2 of the neighboring site.

In this sense, if we focus on two neighboring spin-1 sites, we can think them as 4 spin-1/2.

Given 4 spin-1/2, the only way to form a spin-2 is two pairs spin-1/2 form two spin-1 respectively, and then these two spin-1 forms a spin-2. If any pair of spins in this 4 spin-1/2 forms a spin-0 valence bond, then this 4 spin-1/2 no longer can form a spin-2 entity, which is the case for the spin-1 VBS state. Therefore, if we apply a projector to spin-2 on two neighboring sites in the spin-1 VBS we get 0 (annihilates the state). Consequently, the sum of such projectors on each pair of the neighboring sites, which is the proposed Hamiltonian, also annihilate the spin-1 VBS state. In other word, the spin-1 VBS state is an eigenstate of the hamiltonian with eigenvalue 0.

Considering the fact that a projector has two eigenvalues 0 and 1 and the sum of projectors has eigenvalues equal or bigger than 0. A 0 eigenvalue means spin-1 VBS is the ground state of the proposed Hamiltonian.

Note that in my question I thought $P_2$ is the projection operator onto spin-2 for the spin pair, but this is a mistake. Actually the whole notation $P_2(\mathbf{S_i} + \mathbf{S_{i+1}})$ is the projector operator. I don't like this confusing notation personally and would prefer something like $P_{2}^{\mathbf{S_i}, \mathbf{S_{i+1}}}$ or $P_{2}^{i, i+1}$, but I will use $P_2$ for short in the following.

Now I have answered the first 3 of my questions. My feeling is that it is more of a trick to get the Hamiltonian. The author could be inspired by previous work with Heisenberg model and Majumdar-Ghosh model as both Hamiltonian can also be expressed as the sum of projectors.

Now comes the last question which is to find out what $P_2$ is. $P_2$ acts on 2 spin-1 sites. The eigenstates of $X \equiv (\mathbf{S_i} + \mathbf{S_{i+1}})^2$, namely $|0\rangle$, $|1\rangle$, $|2\rangle$, are also the eigenstate of $P_2$. To be complete I listed the eigenvalues of $X$ and the three spin projectors in the following table:

| s | $X$| $P_0$| $P_1$ | $P_2$ |
| 0 | 0 | 1 | 0 | 0 |
| 1 | 2 | 0 | 1 | 0 |
| 2 | 6 | 0 | 0 | 1 |

We can express the projector as the function of $X$ so that it satisfy the above table:

$$ \begin{align} P_0 &= \frac{1}{12} (X-2)(X-6)\\ P_1 &= -\frac{1}{8} X(X-6)\\ P_2 &= \frac{1}{24} X(X-2) \end{align} $$

If we replace $X$ in the above equation with

$$ \begin{align} X &= (\mathbf{S_i} + \mathbf{S_{i+1}})(\mathbf{S_i} + \mathbf{S_{i+1}})\\ &= \mathbf{S_i^2} + \mathbf{S_{i+1}^2} + 2\mathbf{S_i} \cdot \mathbf{S_{i+1}} \\ &= 4 + 2\mathbf{S_i} \cdot\mathbf{S_{i+1}} \end{align} $$

then we get

$$ P_2 = \frac{1}{6} (\mathbf{S_i} \cdot\mathbf{S_{i+1}})^2 + \frac{1}{2} \mathbf{S_i} \cdot\mathbf{S_{i+1}} + \frac{1}{3} $$

  • $\begingroup$ Thanks for this answer to your own question :) It's simple yet effective especially the explanation of the projection operators ! $\endgroup$ – gertian May 4 '17 at 16:18

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