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 ...

  Partial trace of matrix product state

+ 0 like - 0 dislike
653 views

I have come accross a formula that puzzles me a bit in the proof of lemma 23 (page 32) of [this paper][1].

The authors start from a (translationally-invariant) matrix product state:
$$\lvert\psi\rangle := \sum_{i_1, \ldots, i_n}\textrm{tr}(A_{i_1} \ldots A_{i_n})\lvert i_1 \ldots i_n\rangle$$

where the $A_i$ are matrices of dimension $D$ satisfying $\sum_{i}A_i^{\dagger}A_i = \sum_{i}A_iA_i^{\dagger} = \mathbf{1}_{D \times D}$ (that is, they constitute Kraus operators for a unital channel). Then, they claim that the reduced state of the first $l$ qudits looks like:

$$\rho_l = \sum_{i_1, \ldots, i_l}\textrm{tr}\left(A_{i_1} \ldots A_{i_l}\frac{\mathbf{1}_{D \times D}}{D}A_{j_1}^{\dagger} \ldots A_{j_l}^{\dagger}\right)\lvert i_1 \rangle\langle j_1\rvert \otimes \ldots \otimes \lvert i_l \rangle\langle j_l \rvert$$

Yet, the first expression that looks obvious to me would rather be:
$$\rho_l = \sum_{i_1, \ldots, i_l}\left(\sum_{i_{n + 1}, \ldots, i_n}\textrm{tr}\left(A_{i_1}\ldots A_{i_{l}}A_{i_{l + 1}}\ldots A_{i_n}\right)\textrm{tr}\left(A_{i_n}^{\dagger}\ldots A_{i_{l + 1}}^{\dagger}A_{i_l}^{\dagger}\ldots A_{i_1}^{\dagger}\right)\right)\lvert i_1 \rangle\langle j_1\rvert \otimes \ldots \otimes \lvert i_l \rangle\langle j_l \rvert$$

It looks as if by some magic the authors converted the sum of product of traces above to the trace of a product! Any idea where that would come from? At first, I thought it had to do with the $A$ being randomized (see paragraph on top of page 32), but by giving it a closer look, it seems the statements of lemma 23 do not include an expectation value with respect to the Haar measure, as opposed e.g to lemma 24, so I am not sure what to think.

  [1]: https://arxiv.org/abs/1206.2947

asked Nov 14, 2019 in Theoretical Physics by eduardgil (0 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$\hbar\varnothing$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
...