Detailed derivation and explanation of the AKLT Hamiltonian

Question

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.

Answer 1

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

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

Answer 2

In my opinion, the statement that the VBS state is annihilated by the total spin-2 projector on neighbouring sites needs more explanation, because it is somehow vague to say "spin pair (1, 2) and spin pair (3, 4) are in triplet state, while spin pair (2, 3) is in singlet state". (By spin I mean spin-1/2 unless otherwise stated.) There can be two operational understanding of this statement:

• Put the initial state such that (1, 2), (3, 4) are both triplet, then project (2, 3) onto the singlet space.
• Put the initial state such that (2, 3) is singlet, then project (1, 2), (3, 4) onto the triplet space.

The first statement is not true because the resulting state is not necessarily in triplet space for (1, 2) and (3, 4), and thus is not a valid representation of the spin-1 chain. To see this, denote the 2-spin projector onto singlet and triplet subspace as $\Pi_s$ and $\Pi_t$ respectively, and one can just verify that $$ \Pi_{s,(1,2)}\Pi_{s,(3,4)}\Pi_{s,(2,3)}\Pi_{t,(1,2)}\Pi_{t,(3,4)} \neq 0. $$ (The key observation is that the state that (1, 2), (3, 4) form singlets and the state that (1,4), (2, 3) form singlets are not orthogonal, but I will not expand this here. Nevertheless, you can verify the inequality by a direct calculation.)

Therefore, we should take the second statement. However, it is then NOT true that in the final state, (2, 3) form a singlet. One can see this by verifying $$ \Pi_{t,(2,3)}\Pi_{t,(1,2)}\Pi_{t,(3,4)}\Pi_{s,(2,3)} \neq 0. $$ As a result, the statement "the two sites cannot have total spin 2 because there is a singlet" cannot be understood simply as that (2, 3) form a singlet. In our 4-spin example, it is true that the final state has no overlap with the spin-2 subspace, yet the reason takes a longer argument. The argument goes as,

• First, (2, 3) are initialised to a singlet, making the initial state having no overlap with the spin-2 subspace.
• Projectors onto the triplet subspaces on (1, 2) and (3, 4) commute with the global rotation operators, leaving the total angular momentum on the 4 spins invariant.
• The final state has no overlap with the spin-2 subspace.

We can apply this argument on a 1-dimensional chain with $N$ sites, where each site has two spins, denoted L and R. The spin-1 VBS state is constructed by initialising pairs $(i_R, (i+1)_L)$ to singlets, and projecting pairs $(i_L, i_R)$ onto triplets. For every two neighbouring sites, say $i$ and $i+1$, initially $(i_R, (i+1)_L)$ is in singlet, making the 4 spins having no overlap with the spin-2 subspace. Then the statement that the final state has no overlap with the spin-2 subspace follows from the triplet projectors commuting with a global rotation on the 4 spins.

This argument may seem unnecessary to some intuition inclined readers. However, it matters when we consider more layers of projectors. Suppose we have 3 layers of triplet projectors forming a brick wall structure. That is, this first layer project onto triplets on $(i_L, i_R)$, the second layer on $(i_R, (i+1)_L)$, and the third layer the same as the first. Then the final state is NOT annihilated by a spin-2 projector on two neighbouring sites. The reason is that the second layer breaks angular momentum conservation for the 4 spins on the two sites.