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We are able to perform Schmidt decomposition for open-boundary MPSs with dimension of boundary bond $m_0 = 1$, $|\psi\rangle=\sum w_{a_l}|a_l\rangle_L|a_l\rangle_R$. Because we can make $|a_l\rangle_{L,R}$ orthorgonal.

However, I think it is hard to make a Schmidt decomposition for periodic chain with boundary dimension $m_0>1$, for we need to cut the chain twice to make if bipartite, and how can we keep the orthogonality of states during such kind of operations?

In other words, can we make a Schmidt decomposition with orthogonal bulk states as environment?

Example, we want to make a mixed ensemble consisted of $|a_l\rangle_L$ and $|a_r\rangle_R$ out of MPS $\Psi_{a_l,a_r}|a_l\rangle_L|a_la_r\rangle_C|a_r\rangle_R$ by tracing out the center part, we must make the center part orthogonal, which is not as easy as making the right or left ensembles orthogonal, the latter can be achieved by making MPS left or right canonical.

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  • $\begingroup$ Of course you can make a Schmidt decomposition. The problem is to truncate Schmidt coefficients by cutting individual bonds, not the Schmidt decomposition itself. $\endgroup$ – Norbert Schuch Jun 29 '16 at 20:20
  • $\begingroup$ @NorbertSchuch I am intended to get the mixed ensemble after tracing out one segment, not to perform truncation in DMRG, do you have any suggestion? PS: In fact, I've got one idea, but need some time to check whether it is correct : ) $\endgroup$ – 刘金国 Jun 30 '16 at 10:27
  • $\begingroup$ You have to be more specific. Tracing out is always possible. You should explain why it should be different/harder to do this with for a PBC MPS. (Maybe you have a specific representation of the mixed state in mind?) $\endgroup$ – Norbert Schuch Jun 30 '16 at 14:00
  • $\begingroup$ @NorbertSchuch , I added an example to this issue in the main text, and figured out a possible solution to it(see answers), It take some time to confirm that it works properly in my program... Do you have further suggestions on how most people cope with this problem? $\endgroup$ – 刘金国 Jul 2 '16 at 5:17
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One possible solution:

I figured out a possible solution to this problem.

1. Get the inner product of center block $M_{a_la_r,a_l'a_r'}=\Psi_{a_l,a_r}^{\sigma_c}\Psi^{\sigma_c}_{a_l',a_r'}$, here $\sigma_c$ is the physical indices for center block.

2. perform cholesky(if not able to perform it in rank revealing way, use eigen-value decomposition instead) decomposition, get something like $C_{a_la_r,k} C^\dagger_{k,a_l'a_r'}=M_{a_la_r,a_l'a_r'}$. Then $C^{-1}$ matrix is what diagonalize the central block, it's rank $r(C)\le min(d^{number\;of\;site},a_la_r)$. $C^{-1}$(not well defined in general?) is what normalize the center block noticing the normalization condition $C^{-1}MC^{-1\dagger}=1$.

3. Insert $CC^{-1}$ in to original MPS, $C^{-1}$ normalizes the enviroment, so we drop $C^{-1}M$(trace procedure) and replace the center block $\Psi\rightarrow C^\dagger$ and get the MPS representation of quantum mixture.

The drawback is that we can not always get the inverse of $C$ due to rank deficiency, but the result seems do not rely on this property, I believe there is a way to circumvent this issue in deduction. Still, I call for a smart way to do this, the above way is computational costly: $dm^5$ for m the states kept.

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