Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Many body quantum states analyzed as probabilistic sequences

+ 9 like - 0 dislike
1074 views

Measurements of consecutive sites in a many body qudit system (e.q. a spin chain) can be interpreted as generating a probabilistic sequence of numbers $X_1 X_2 X_3 \ldots$, where $X_i\in \{0,1,\ldots,d-1 \}$.

Are there any studies on that approach, in particular - exploring predictability of such systems or constructing a Markov model of some order simulating it?

This post has been migrated from (A51.SE)
asked Jan 25, 2012 in Theoretical Physics by Piotr Migdal (1,260 points) [ no revision ]
retagged Mar 18, 2014 by dimension10
Maybe I'm missing something (I'm about to go to sleep). You take the spin-spin correlation functions and build (say) whatever order transition matrix you like, no?

This post has been migrated from (A51.SE)
@SHuntsman In the one ways (state -> sequence) it is straightforward. I am interested what can one deduce about the state (or Hamiltonian, if it an ground/eigenstate) knowing only the sequence.

This post has been migrated from (A51.SE)

2 Answers

+ 2 like - 0 dislike

Consider the 1-dimensional Ising model $$\mathcal{H}_N=-J\sum_{i=1}^{N} \sigma_i\sigma_{i+1}-H\sum_{i=1}^{N}\sigma_i\qquad\sigma_i=\pm 1$$
with periodic boundary conditions $\sigma_{N+1}=\sigma_1$ and magnetic field $H$. Let $\Omega$ be the set of the one-sided infinite sequences  $\omega=(\sigma_1,\sigma_2,\dots)\in\Omega$. Consider the map $x:\Omega\mapsto\{-1,1\}$, with coordinate function $x_n(\omega)=\sigma_n$ and the shift transformation $T:\Omega\mapsto\Omega$, with $\,x_n(T\omega)=x_{n+1}(\omega)=\sigma_{n+1}$. Let further ${\cal F}$ be a $\sigma$-algebra of subsets of $\Omega$, which is generated by the sets (cylinders) of the form $\{\omega\in\Omega:(\sigma_i,\dots,\sigma_{i+n-1})\in E\},$ while $E$ is a subset of $\{-1,1\}^n$. The $\sigma$-fields are  generated by what is called "thin" cylinders, that is, sets of the form $\{\omega\in\Omega: x_l(\omega)=\sigma_l, i\le l\le i+n-1\}$

Since the finite disjoint unions of thin cylinders form a field which generates ${\cal F}$,
a measure $P$ on ${\cal F}$ is uniquely determined by the values
$$p_n(s_i,\dots,s_{i+n-1})=P\{\omega\in\Omega:x_i(\omega)=s_i,\dots, x_{i+n-1}(\omega)=s_{i+n-1}\}$$
for $i\ge 1$. Now, for any finite $n, N$, $1\le n\le N$ and $1\le i\le N-n+1$, one can define the partial trace:
$$Z_{N}(i,n):=\sum_{\substack{\sigma_1,\dots,\sigma_{i-1},\\ \sigma_{i+n},\dots,\sigma_{N}}}
e^{-\beta\mathcal{H}(\sigma_1,\dots,\sigma_{N})}$$
and the ordinary trace:
$$Z_{N}:=\sum_{\sigma_1,\dots,\sigma_N} e^{-\beta\mathcal{H}(\sigma_1,\dots,\sigma_N)}.$$
Then, consider the fraction $$p^{(N)}_n(s_i,\dots,s_{i+n-1})=\frac{Z_{N}(i,n)}{Z_N}.$$
From the latter, we can derive a probability measure in the following sense
$$p_n(s_i,\dots,s_{i+n-1})=\lim_{N\rightarrow\infty}p^{(N)}_n(s_i,\dots,s_{i+n-1}),$$
and the associated transition probability is simply
$$p(s_1,\dots,s_n|s_{n+1})=\frac{p_{n+1}(s_1,\dots,s_{n+1})}{p_n(s_1,\dots,s_n)}.$$
By substitution of the traces and some further algebraic manipulations one can show that the latter is equal to a stochastic and irreducible transition matrix. This implies that the transition probability is associated with an ergodic Markov shift. 

answered Feb 25, 2017 by kaffeeauf (50 points) [ no revision ]
+ 1 like - 0 dislike

Consider first a deterministic problem in the form of a Poincaré surface of section of a system governed by a Hamiltonian, i.e. we take snapshots of the full state of the system after given fixed periods of time. We now have three options as to what is happening:

  1. The trajectory is quasi-periodic with none of the periods equal to an integer multiple of the snapshot time. Then we can asymptotically reconstruct the phase space trajectory of the particle.
  2. The trajectory is quasi-periodic with one of the periods equal to an integer multiple of our snapshot time, then we get a finite number of points along the phase space trajectory. However, this case is typically "of measure zero" for non-linear Hamiltonians.
  3. The trajectory is chaotic, then the portion of phase space where chaos "spills" is lost to any analytical reconstruction.

If we have case 1. in a large enough portion of the phase space, we will typically be able to approximate the motion by successive expansion via a set of action-angle coordinates. If it happens that case 2. spans a larger part of the phase space, we actually know that there the Hamiltonian looks like a harmonic oscillator! However, the reconstruction of the full classical Hamiltonian is impossible if chaos is present.

In practice, I would expect to gain useful insight only if you already have a very good idea of what your Hamiltonian looks like, and you just need to determine a few free parameters.


Now to the quantum case. The construction of a Markov process from the quantum+measurement dynamics is obvious, every step is constructed by the deterministic $e^{-iHt}$ evolution plus a probabilistic collapse.

One way which comes to my mind for the reconstruction of the Hamiltonian is to consider the fact that the expectation values of variables fulfill the classical equations of motion

$$\langle f(\hat{P},\hat{Q})\rangle = \{f(p,q),H(p,q)\}$$

What you measure are quantities of the sort of $\langle Q \rangle,\,\langle Q^2 \rangle - \langle Q \rangle^2,...$ with initial $\langle Q \rangle$ being the value just measured and all the other momenta zero. I.e., there is an entangled Poincaré surface of section hidden in your Markovian chain. This seems to be a good enough basis for the reconstruction of the Hamiltonian, at least up to the limitations mentioned above.

answered Feb 22, 2017 by Void (1,645 points) [ revision history ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...