The Hamiltonian given by Rokhsar-Kivelson QDM is based on tensor products of the state vectors. Why is this the case?, is it because the lattice model is a mixed state? It'd be great if someone could go over the Hamiltonian on a step-by-step basis. Thanks.


1 Answer 1


It is not clear what is "tensor products of the state vectors". The QDM Hamiltonian, as any other Hamiltonians of spin models, is a Hermitian operator in a Hilbert space of spins (dimers in this case). However, this Hilbert space does not have a simple tensor product structure, since they consist of hardcore dimers covering a lattice, such that each site must be touched by only one dimer. To see the difference more explicitly, consider a model of spin-1/2's on a lattice of $N$ sites, and we know that the dimension of the Hilbert space is $2^N$, and the Hilbert space is simply $\mathcal{H}=\mathbb{C}^2\otimes\mathbb{C}^2\cdots$. For dimers, a famous classical result states that the number of ways to cover the square lattice with dimers grows asymptotically as $(1.3385...)^N$.

Now the RK Hamiltonian simply gives you the matrix elements. The Hamiltonian is local, so each term only affects/depends on a small patch (in the RK case, a square plaquette) of the lattice and that's why you see usually the Hamiltonian is presented by "pictures" relating dimer coverings which at most differ from each other by rearranging dimers in a single square plaquette. But once you know these matrix elements, it is not different from any other quantum Hamiltonian.

Not sure what you mean by "the lattice model is a mixed state". The lattice model is a model Hamiltonian, which is an operator and "mixed state" refers always to the state of a system as described by the density matrix. If you are confused about these things you probably should go back to more basic quantum mechanics before studying RK Hamiltonian.

  • $\begingroup$ The reason I used "mixed" state was due to the hamiltonian containing a ket followed by a bra. The pictures are given in terms of ket multiplied by a bra. I am not clear on why this is done? $\endgroup$
    – kb56
    Commented Aug 1, 2015 at 21:56
  • $\begingroup$ Ket followed by a bra has nothing to do with mixed states. Any operator $O$ is written this way: $O=\sum_{a,b}\langle a|O|b\rangle |a\rangle\langle b|$ where $|a\rangle$ (and $|b\rangle$) is a orthonomal complete basis of the Hilbert space. I'm not sure what else you can do to concretely specify an operator. $\endgroup$
    – Meng Cheng
    Commented Aug 1, 2015 at 22:02
  • $\begingroup$ So,the reason RK used a ket followed by a bra is to ensure that it ends up being an operator? $\endgroup$
    – kb56
    Commented Aug 1, 2015 at 22:15
  • $\begingroup$ It's a Hamiltonian. So it is also an operator. $\endgroup$
    – Meng Cheng
    Commented Aug 1, 2015 at 22:51
  • $\begingroup$ I was looking for an intuitive understanding but I figured it out.It's just an indirect way of writing the Hamiltonian operator via sum of outer products for the states. Thanks for being patient with me. $\endgroup$
    – kb56
    Commented Aug 1, 2015 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.