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

Please welcome our new moderators!

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

122 submissions , 103 unreviewed
3,497 questions , 1,172 unanswered
4,545 answers , 19,342 comments
1,470 users with positive rep
408 active unimported users
More ...

Accurate quantum state estimation via "Keeping the experimentalist honest"

+ 9 like - 0 dislike
18 views

Bob has a black-box, with the label "V-Wade", which he has been promised prepares a qubit which he would like to know the state of. He asks Alice, who happens also to be an experimental physicist, to determine the state of his qubit. Alice reports $\sigma$ but Bob would like to know her honest opinion $\rho$ for the state of the qubit. To ensure her honesty, Bob performs a measurement $\{E_i\}$ and will pay Alice $R(q_i)$ if he obtains outcome $E_i$, where $q_i={\rm Tr}(\sigma E_i)$. Denote the honest probabilities of Alice by $p_i={\rm Tr}(\rho E_i)$. Then her honesty is ensured if $$ \sum_i p_i R(p_i)\leq\sum_i p_i R(q_i). $$ The Aczel-Pfanzagl theorem holds and thus $R(p)=C\log p + B$. Thus, Alice's expected loss is (up to a constant $C$) $$ \sum_i p_i(\log{p_i}-\log{q_i})=\sum_i {\rm Tr}(\rho E_i)\log\left[\frac{{\rm Tr}(\rho E_i)}{{\rm Tr}(\sigma E_i)}\right]. $$ Blume-Kohout and Hayden showed that if Bob performs a measurement in the diagonal basis of Alice's reported state, then her expected loss is the quantum relative entropy $$ D(\rho\|\sigma)={\rm Tr}(\rho\log\rho)-{\rm Tr}(\rho\log\sigma). $$ Clearly in this example Alice is constrained to be honest since the minimum of $D(\rho\|\sigma)$ is uniquely obtained at $\sigma$. This is not true for any measurement Bob can make (take the trivial measurement for example). So, we naturally have the question: which measurements can Bob make to ensure Alice's honesty? That is, which measurement schemes are characterized by $$ \sum_i {\rm Tr}(\rho E_i)\log\left[\frac{{\rm Tr}(\rho E_i)}{{\rm Tr}(\sigma E_i)}\right]=0\Leftrightarrow \sigma=\rho? $$

Note that $\{E_i\}$ can depend on $\sigma$ (what Alice reports) but not $\rho$ (her true beliefs).

Partial answer: Projective measurement in the eigenbasis of $\sigma$ $\Rightarrow$ yes, Blume-Kohout/Hayden showed that this is the unique scheme constraining Alice to be honest for a projective measurement.

Informationally complete $\Rightarrow$ yes, clearly this constrains Alice to be honest since the measurement will uniquely specify the state (moreover, the measurement can be chosen independent of $\sigma$).

Trivial measurement $\Rightarrow$ no, Alice can say whatever she wants without impunity.

This post has been migrated from (A51.SE)
asked Nov 14, 2011 in Theoretical Physics by Chris Ferrie (645 points) [ no revision ]
retagged Mar 7, 2014 by dimension10

1 Answer

+ 7 like - 0 dislike

Trivially for any set of measurements $\{E_i\}$ where $\rho$ and $\sigma$ have equal expectation value for each $E_i$, $$\sum_i\mbox{Tr}(\rho E_i) \log \left[ \frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}\right] = \sum_i\mbox{Tr}(\rho E_i) \times 0 = 0.$$

Note that the log-sum inequality theorem says that $$\sum_i a_i \log\left(\frac{a_i}{b_i}\right) \leq \big(\sum_i a_i\big)\times \log \left(\big(\sum_i a_i\big)/\big(\sum_i b_i\big)\right),$$ with equality only if $\frac{a_i}{b_i}$ is constant for all $i$. This implies that $$\sum_i\mbox{Tr}(\rho E_i) \log \left[ \frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}\right] = 0$$ if and only if $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}$ is constant. Since the sum of the traces over $i$ is 1 for both density matrices this implies that you must have $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)} = 1$.

If, however, the projections onto each subspace selected by $E_i$ aren't equal, then there exists some $i$ such that $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)} \neq 1$.

Hence for Alice to have zero expected loss, then $\mbox{Tr}(\rho E_i) = \mbox{Tr}(\sigma E_i) ~~~\forall i$. If expectation value for each $E_i$ uniquely identifies a density matrix, then necessarily $\rho = \sigma$ if Alice has zero expected loss. A sufficient condition for this is when $E_i$ project onto $(\dim \mathcal{H})^2 -1$ linearly independent density matrices, where $\mathcal{H}$ is the Hilbert space of the system.

This post has been migrated from (A51.SE)
answered Nov 14, 2011 by Joe Fitzsimons (3,555 points) [ no revision ]
Most voted comments show all comments
@ChrisFerrie: Sorry, I meant if the measurements can depend on $\sigma$ not $\rho$. I've edited my above comments to reflect this.

This post has been migrated from (A51.SE)
Oh good, thanks. This is a very important point and I think you should add it to your answer.

This post has been migrated from (A51.SE)
This argument is for the case where the set of measurements is fixed and independent of $\sigma$. If the scheme need only work for certain classes of $\sigma$ then this imposes correlations between entries in the density matrix which reduces the number of linearly independent measurement operators required to uniquely identify it. An example of this is where $\sigma = |+\rangle\langle +|$, where purity implies that the state has expectation value 0 for $Z$ and $Y$ measurements, and if $\mbox{Tr}(\rho X) = \mbox{Tr}(\sigma X)$ then $\mbox{Tr}(\rho Y) = \mbox{Tr}(\sigma Y)$, etc.

This post has been migrated from (A51.SE)
I made a few changes to clarify which measurement schemes are allowed: they are independent on $\rho$ but not necessarily fixed as they can depend on what state $\sigma$ Alice reports. What is surprising is that if Bob is going to perform a PVM, he must do it in the eigenbasis of $\sigma$. The proof used in the paper is not intuitive to me. Your line of reasoning is more appealing but I'd have to think a bit harder to continue it to reach the same conclusion about PVMs.

This post has been migrated from (A51.SE)
@ChrisFerrie: No, that is not the same as the sufficient condition I gave since I only required linear independence and not orthogonality and only for a sufficiently large subset of $\{E_i\}$, so it includes a bunch of weak measurements. However that was only an example. The necessary and sufficient condition is that all expectation values are the same. This includes a huge range of measurements, such as measuring a randomly chosen Pauli operator, which yields only a single bit of information.

This post has been migrated from (A51.SE)
Most recent comments show all comments
Also, if you take any complete or over-complete basis for tomography you can make the measurements and make them arbitrarily weak, you still satisfy the criterion (though Alice's expected loss tends towards zero as the measurement tends towards the identity).

This post has been migrated from (A51.SE)
And hence you need only measure in the $X$ basis, even though this does not have sufficiently many linearly independent outcomes to uniquely identify an arbitrary density matrix.

This post has been migrated from (A51.SE)

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:
$\varnothing\hbar$ysicsOverflow
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
...