# Why reduced density operator being same is necessary sufficient for no signalling?

Problem Statement :
Two parties $A$ ( Alice ) and $B$ ( Bob ) ( in order ) share an entangled pair $\frac{1}{\sqrt{3}}(|00\rangle+|01\rangle +|11\rangle)$. Bob does a measurement in basis $\{ |0\rangle,|1\rangle\}$ and gets the outcome of his qubit as $|0\rangle$. Bob wants to guess if Alice had performed a measurement on his qubit in the basis $\{ |0\rangle,|1\rangle\}$ or not. If Bob is able to guess even partially ( with a probability $\neq \frac{1}{2}$ ) then it means signalling takes place ( if I am not wrong ).

My doubt related to the problem above:
Let $i$ be the event that Alice performed the measurement and $m$ the event that Bob's measurement outcome for his qubit is $|0\rangle$. So I want to calculate is $p(i|m)$. $$p(i|m) = \frac{p(i)p(m|i)}{p(m)}$$ Bob calculates this probability so according to him $p(i)=\frac{1}{2}$ ( if this was not the case he might know something already ), $p(m)$ from Bob's perspective is $\frac{1}{3}$. Now I don't understand what $p(m|i)$ should be ( why should it be $\frac{1}{3}$, it comes out to $\frac{2}{3}$ for me but can't find the mistake ). For no signalling to hold valid $p(i|m)=\frac{1}{2}$.

My doubt in general :
Basically my doubt arises because partially density operator being the same is the condition for no signalling ( talking from Bob's perspective ). But partial density operator just tells with what probability to expect the outcome, but once a particular outcome is observed I might able to partially deduce something. I mean why do we base validation of no signalling on the probabilities of outcomes of a measurement ( given by reduced density ) rather than validating it by inability to deduce the desired information even partially from the outcomes of my measurements ( talking by Bob's perspective ).

• Can you explain what you mean by "guessing"? What can you deduce from a particular outcome? – CuriousOne May 20 '15 at 23:26
• By guessing I mean that if I am certain that an event has occurred with a probability $p > \frac{1}{2}$ then I say the event has occurred if $p < \frac{1}{2}$ I say it has not occurred. But if $p=\frac{1}{2}$ I can't say anything. – sashas May 20 '15 at 23:30
• That's what I don't understand... how can you be certain about that? – CuriousOne May 20 '15 at 23:32
• I am not sure. It's like if I have to bet on something I will bet in favor if p>1/2 for me and against p<1/2 and I wouldn't be sure what to bet on if p=1/2. I might be wrong sorry for vague terminologies. – sashas May 20 '15 at 23:38
• Nature doesn't care about your bets, though. Whether information is transmitted by entanglement or not is a physical fact that you can't influence by the way you try to analyze the data. You may lose information by doing it wrong, but you can't create any by trying to be smart. That doesn't even work in a casino where the probabilistic outcomes are generated by a classical mechanism. – CuriousOne May 21 '15 at 0:28

There is a problem with your statement about the condition of no signalling is $p = \frac{1}{2}$. We can expand your problem into three parts: how to understand the condition conceptually, how to use the condition in your problem and how the condition is reflected in the density matrix formalism. For the sake of convenience, I will denote $|\psi\rangle = \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |11\rangle)$.

To start with, the fact that Bob can guess with $p \neq \frac{1}{2}$ does not imply any signalling between Alice and Bob, for the concept of 'impartiality' depends on the quantum state that you are dealing with. It is actually quite easy to see it with a classical example: suppose Alice has a fair coin (with $p_H = \frac{1}{2}$) and Bob has a biased coin (with $p_H \neq \frac{1}{2}$) and they both flip the coin independently. There is no communication or signaling between Alice and Bob, but it is obvious that for Bob he should find that his coin has $p_H \neq \frac{1}{2}$, which really is attributed to the fact that the coin is biased.

The case is similar in your problem, as for Bob to discriminate between the case where Alice measures in $\{|0\rangle, |1\rangle\}$ basis or not, he needs two possible results that give him different probabilities (or frequencies) corresponding to each case. However, it can be demonstrated that he cannot do that, as for example if he measures his qubit in $\{|0\rangle, |1\rangle\}$ basis, he will get $|0\rangle$ with probability:

$P_0 = \langle \psi | 0 \rangle \langle 0 | \psi \rangle = \frac{1}{3}$.

This would be the baseline he can use to compare with the situation where Alice steers the system by measuring her qubit in $\{|0\rangle, |1\rangle\}$ basis. With probability $\frac{2}{3}$ Alice would collapse the state to $\frac{1}{2} (|0\rangle + |1\rangle)$, and with probability $\frac{1}{3}$ Alice would collapse the state to $|1\rangle$. Therefore the probability of Bob measuring the state to be in $|0\rangle$ given Alice had done a measurement is

$P_0' = \frac{2}{3} \times \frac{1}{2} + \frac{1}{3} \times 0 = \frac{1}{3}$.

In other words, Bob will not be able to tell whether Alice had made a measurement by simply measuring his qubit, for both give the same probability $\frac{1}{3}$. So despite the system has some probability $p \neq \frac{1}{2}$, it is the fact that Bob obtains the same probability for both cases rather than the probability being $\frac{1}{2}$ that suggests Bob cannot decide what Alice had done, i.e. the condition for no signalling.

One can obviously extend this question by thinking about whether there are other ways to conduct signalling - say for instance, would Alice's measurement in other basis such as $\{|+\rangle, |-\rangle\}$ make an observable difference in Bob's measurement? The answer is summed up pretty nicely by the reduced density matrix formalism: Bob's measurement on his qubit is completely determined by $\text{Tr}_A (|\psi\rangle \langle\psi|)$. On the other hand, for Alice to conduct any measurement on some basis on her qubit, the state collapses to

$|\psi\rangle \langle \psi| \rightarrow \Pi_a |\psi\rangle \langle \psi| \Pi_a^\dagger + \Pi_b |\psi\rangle \langle \psi| \Pi_b^\dagger$,

where $\Pi$ are projectors in the basis denoted by $a$ and $b$. When Bob measures his state, using the partial trace rule once again, his measurement is completely determined by

$\text{Tr}_A (\Pi_a |\psi\rangle \langle \psi| \Pi_a^\dagger + \Pi_b |\psi\rangle \langle \psi| \Pi_b^\dagger) = \text{Tr}_A((\Pi_a + \Pi_b) |\psi \rangle \langle \psi|) = \text{Tr}_A |\psi\rangle \langle\psi|$,

since the projectors sums up to identity operator. Here we can see the reduced density matrix for Bob's qubit is identical in both situations, and therefore Bob can do pretty much nothing to tell whether Alice had made a measurement or not, or on which basis her measurement is conducted.

• I figured it out, but thanks a lot for explaining so nicely. I appreciate it. – sashas Jun 3 '15 at 4:13