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The AKLT Hamiltonian and the chain is described in Wikipedia, and also the page 17 of this year Nobel Prize advanced information

I have questions concerning the info released by nobelprize.org, and wonder whether they make some incorrect statement in my question 2 below.

The Hamiltonian is given by $$ \hat H = \sum_j \left(\frac{1}{2}\left(\vec{S}_j \cdot \vec{S}_{j+1} + \frac{1}{3} \left(\vec{S}_j \cdot \vec{S}_{j+1}\right)^2\right) + 1/3\right)=\sum_j P_2\left[\vec{S}_j + \vec{S}_{j+1}\right] $$ The version from Nobel prize info has an additional 1/2 factor and plus shifting the ground state energy by 1/3.

question 1. What is the easiest way to prove that the following spin-0 singlet pairing between the two nearest neighbor spin-1/2 states (the spin 1/2 is obtained from splitting the spin-1 on a site to two spin-1/2, defined in the oval projection below) described by the Wikipedia on AKLT is the lowest energy ground state?

enter image description here

question 2. I am slightly confused by the description in Nobel prize info, it says that the Hamiltonian is equivalent to the $P_2$ projector $\sum_j P_2\left[\vec{S}_j + \vec{S}_{j+1}\right]$, also it says that: "$P_2$ projects on the subspace corresponding to spin 2 on two adjacent lattice sites." What does it mean to "project to the the subspace corresponding to spin 2 on two adjacent lattice sites?" Is this statement from nobelprize.org incorrect?

Comment: My tentative thought is that if I analyze $\frac{1}{2}\left(\vec{S}_j \cdot \vec{S}_{j+1} + \frac{1}{3} \left(\vec{S}_j \cdot \vec{S}_{j+1}\right)^2\right) + 1/3$ on two neighbor sites $j$ and $j+1$, I find that the total spin-0 and the total spin-1 sectors for the two spins $\vec{S}_j + \vec{S}_{j+1}$ seems to have the lowest energy, while the total spin-2 has a higher energy, thus the total spin-2 sector has the disfavored energy penalty. Thus, shouldn't we project out the total spin-2 (get rid of the total spin-2) and project to the total spin-0 and total spin-1? Isn't that the remaining four states in the total spin-0 and total spin-1 correspond to the precise degree of freedom of the four fold degeneracy of zero modes on the open chain?

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    $\begingroup$ You are aware that the Hamiltonian you give (twice!) differs from the one given in eq. (18) on pg 17 of the linked nobel price description? Also, please try to focus your question. $\endgroup$ – Norbert Schuch Oct 31 '16 at 3:10
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    $\begingroup$ @mysteriousness The additive constant is 1/3, not 1/6. With 1/3, you do not get a projector any more, so your question becomes much more obscure. I think the fact that you claim sth. is given in the nobel text and it is not, together with the fact that you assert it might be wrong, justifies a downvote. Also, there is lots of text which makes it very hard to parse the actual question, and on the other hand very few to guide someone answering what you do understand (or it is very hidden). Finally, do not add a space after @ : Otherwise, I do not get ping-ed. $\endgroup$ – Norbert Schuch Oct 31 '16 at 22:29
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    $\begingroup$ Duplicate of physics.stackexchange.com/q/286601 $\endgroup$ – Norbert Schuch Nov 1 '16 at 21:50
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    $\begingroup$ Note that the duplicate-question mechanism is for pointing people to answered questions. Neither this question nor the proposed duplicate currently has any answers, upvoted or otherwise. The ground-state question here is distinct, as far as I can tell, so I think it's appropriate to leave both questions open. $\endgroup$ – rob Nov 3 '16 at 3:08
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    $\begingroup$ @NorbertSchuch I suppose a well-written answer would clarify the issue, hint, hint :) $\endgroup$ – rob Nov 3 '16 at 19:47
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Question 1:
If you consider two consecutive sites in your pictures you will find that the left- and rightmost spin both have spin $\tfrac12$, while the two spins in the middle are in a singlet, i.e., have spin $0$. Together, the two sites thus have spin $$ \tfrac12\otimes 0\otimes \tfrac12 = 0\oplus 1 $$ This spin is not changed by the projection onto the spin 1 space in the AKLT construction. Thus, the two consecutive spin $1$ sites have a space $$ 1\otimes 1 = 0\oplus 1 \oplus 2\ , $$ but by the above argument the spin $2$ never shows up. Thus, a Hamiltonian which gives energy zero to nearest neighbors with spin $0$ or $1$, but positive energy for spin $2$, has this state as its ground state. As it turns out, this state is the unique ground state.

Question 2:
What does it mean to "project to the the subspace corresponding to spin 2 on two adjacent lattice sites?" Each site is a spin $1$. Two sites together thus have spin $1\otimes 1 = 0\oplus 1 \oplus 2$, i.e., we can decompose their Hilbert space as an (orthogonal) direct sum of a spin $0$, spin $1$, and spin $2$ subspace. The projector onto spin $2$ is exactly the projector onto the spin $2$ subspace.

Is this statement from Nobel prize.org incorrect? No.

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  • $\begingroup$ But I thought is is projected to spin 0 (singlet) and spin 1 (triplet), instead of spin-2. I need some clarification on the "project to" or "project out." I thought is it project to spin 0 (singlet) and spin 1 (triplet), and project out spin-2. $\endgroup$ – user32229 Nov 7 '16 at 15:08
  • $\begingroup$ The circle in the picture you copied is a projection onto spin 1. The Hamiltonian is a projector onto spin 2: This means that the states with spin 0 and 1 have low energy (namely zero), while the states with spin 2 have high energy (namely 1). $\endgroup$ – Norbert Schuch Nov 7 '16 at 15:12
  • $\begingroup$ I see. Thanks, just a little thing here I misunderstood. $\endgroup$ – user32229 Nov 7 '16 at 15:14

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