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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,873 answers , 20,701 comments
1,470 users with positive rep
502 active unimported users
More ...

Average entropy of quantum system in bipartite pure state for finite temperature

+ 4 like - 0 dislike
84 views

[I got halfway through writing this when I found the paper that answers the question in (essentially) the affirmative. I'll post it anyways in case anyone is interested.]

Background: If a random vector $\psi$ (pure quantum state) is drawn from a $NM$-dimensional vector space for finite $N$ according to the Haar measure, then the entanglement entropy $$S(\rho_N) = \mathrm{Tr}[\rho_N \mathrm{log} \rho_N], \qquad \rho_N = \mathrm{Tr}_M[\psi\psi^\dagger]$$ across a tensor decomposition into $N$ and $M$ dimensional vector spaces (subsystems) is highly likely to be almost the maximum $$S_{\mathrm{max}} = \mathrm{log}_2(\mathrm{min}(N,M)) \,\, \mathrm{bits},$$ for any such choice of decomposition. More precisely, if we fix $\alpha=M/N$ and let $N\to \infty$, then the fraction of the Haar volume of states that have entanglement entropy more than an exponentially small (in $N$) amount away from the maximum is suppressed exponentially (in $N$). In physics this was known as Page's conjecture (later proved), and for mathematicians it is a simple consequence of the concentration of measure phenomenon.

For any given Hermitian operator $H$ on the vector space acting as a Hamiltonian, we can assign an expected value of the energy $\langle H \rangle_\psi = (\psi^\dagger, H \psi)$ to a given vector $\psi$. The Haar measure, then, can be considered a Gibbs probability distribution $p_\psi \propto e^{-\beta\langle H \rangle_\psi}$ in the infinite temperature limit ($\beta \to 0$), i.e. all states are equally likely. The Page conjecture is exactly what you'd expect if a subsystem of a global pure state is maximally mixed, given its size.

My question: is there a way to extend this statement about typical entanglement entropies to cases where the probability distribution is one for finite temperature? More physically, can we prove that the typical entanglement entropy of a system in a pure global state is exponentially close to

$$S_\beta = \mathrm{Tr}[\rho^{(\beta)}_N \mathrm{log} \rho^{(\beta)}_N], \qquad \rho^{(\beta)}_N = \mathrm{Tr}_M[\psi\psi^\dagger]$$

with high probability, according to the Gibbs distribution $p_\psi \propto e^{-\beta\langle H \rangle_\psi}$?


This post imported from StackExchange MathOverflow at 2015-03-17 12:03 (UTC), posted by SE-user Jess Riedel

asked Oct 15, 2014 in Theoretical Physics by Jess Riedel (220 points) [ revision history ]
edited Mar 17, 2015 by Dilaton

1 Answer

+ 2 like - 0 dislike

The answer by Popescu et al is basically "yes", and is so close to my question that I suspect I read this paper and forgot it.

The precise statement has been proved not with a Gibbs distribution on global pure states, but with a constraint (i.e. constant weight positive on some subspace, and 0 outside it; "constant" being determined by the Haar measure). When the constraint is global energy, this turns out to be effectively very close to a Gibbs distribution because the density of states is exponential in the energy, so that the distribution of energy is highly peaked around the corresponding temperature. Their statement is significantly stronger than mine in that they prove the reduced state of the system is actually very close to the canonical thermal state, not just that it has the same entropy.

The statement, eq. (2) in the article, is that for arbitrarily small $\epsilon$ there exist $$\eta = \epsilon +\frac{1}{2}\sqrt{\frac{d_s}{d_E^{\mathrm{eff}}}} \quad , \quad \eta' = 4e^{-C d_R \epsilon^2}$$ such that $$\frac{V[\{ \vert \phi\rangle \in \mathcal{H}_R \vert D(\rho_S(\phi),\Omega_S)\ge\eta\}]}{V[\{ \vert \phi\rangle \in \mathcal{H}_R\}]} \le \eta'.$$ Here, $D$ is the trace distance, $\mathcal{H}_R$ is the subspace of the global Hilbert space satisfying the constraint, $\rho_S$ is the reduced state of the system, $d_S$ and $d_R$ are the dimensions of the system and the constrained subspace, and $$d_E^{\mathrm{eff}} = \frac{1}{\mathrm{Tr} \Omega_E^2} \ge \frac{d_R}{d_S}$$ is the effective size of the environment. In these expression, $\Omega_S = \langle \rho_S \rangle_{\mathcal{H}_R}$ and $\Omega_E = \langle \rho_E \rangle_{\mathcal{H}_R}$ are the Haar-average reduced states conditional on the constraint.

This post imported from StackExchange MathOverflow at 2015-03-17 12:03 (UTC), posted by SE-user Jess Riedel
answered Oct 15, 2014 by Jess Riedel (220 points) [ no revision ]

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$\hbar$ysicsOv$\varnothing$rflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...