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From what I understood the density operator $\rho$ is a mathematical tool which tells us about the probabilities of getting a particular output after measurement. I have two parties entangled with each other say $A$ and $B$ ( each having one qubit ) and the entangled state is initially give by $|\psi\rangle = \frac{|00\rangle+|11\rangle}{\sqrt{2}}$. Now if $A$ measures his qubit in the standard basis $\{|0\rangle,|1\rangle\}$,as a result of no signalling principle ( if no communication between the two takes place ) $B$'s reduced density operator should remain the same ( if I am interpreting the principle correctly ). But I have 3 scenarios where I am not able to prove this consequence of the no signalling principle ( In all the cases below $B$ does not know outcome of $A$'s measurement )

  1. $B$ knows that $A$ did a measurement and he did it in the basis $\{|0\rangle,|1\rangle\}$, then before and after measurement $B$'s reduced density of joint density operator for both, remains the same ie. $\frac{I}{2}$. Although I am able to see the principle holds, it holds for this particular initial entangled state $|\psi\rangle = \frac{|00\rangle+|11\rangle}{\sqrt{2}}$, I am unable to see how it would be proved for other entangled states in general.
  2. $B$ does not know that $A$ did a measurement. Lets say that after measurement the combined state was $|00\rangle\langle00|$, so $B$'s actual reduced density operator is $|0\rangle \langle0|$ after measurement, but for $B$ his reduced density operator is $\frac{I}{2}$. Now if $B$ does some measurement on his own qubit he expects the outcomes according to $\frac{I}{2}$ but are actually according to $|0\rangle \langle 0|$, won't there be any measurement where $B$ would be able to find out that $A$ did some measurement on his qubit( which violates no signalling principle ). I know this is a weaker case than $(1)$, but I am not able to find the mistake in my above reasoning.
  3. $B$ knows that $A$ did a measurement but does not know the specifics for the measurement. This is somewhat in between the scenarios $(1)$ and $(2)$.
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1 Answers1

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You have fallen into a very subtle trap, essentially the one I discussed here, but it's probably worth going into the specifics.

Lets say that after measurement the combined state was $|00\rangle\langle00|$, so $B$'s actual reduced density operator is $|0\rangle \langle0|$ after measurement

The problem with this is that you cannot consider the consequences of Alice measuring $|0\rangle$ without also considering the consequences of her measuring $|1\rangle$, and assigning them the appropriate weights. This means that

if $B$ does some measurement on his own qubit he expects the outcomes according to $\frac{I}{2}$ but are actually according to $|0\rangle \langle 0|$

... half the time. The other half of the time, Alice obtained $|1\rangle$, and Bob's density matrix is $|1\rangle\langle1|$, and his results will reflect this.

To deal with this, you can do two things:

  • You can appropriately combine those two density matrices to get Bob's observed state, which is $\tfrac 12(|0\rangle\langle0|+|1\rangle\langle1|)$ as expected.
  • You can consider only those results where Bob's density matrix is $|0\rangle\langle0|$, but only after he has had time to hear from Alice, via classical slower-than-light channels, which results he needs to account for and which ones he needs to disregard.

If Bob does not know that Alice did a measurement, then he certainly does not know the outcome, and you need to average as in the first point above.

It's unclear to me why you're confused about point 1, though (so, if I overshoot on level, it's on you to catch up, I'm afraid). Suppose Alice and Bob share some arbitrary bipartite state $\rho=|\psi\rangle\langle\psi|$. In the most general setting, Bob and Alice both perform generalized measurements modelled by POVMs $\{M_b\}$ and $\{M_a\}$ respectively, such that $\sum_bM_b=\sum_a M_a=1$. (These are the most general combinations of coupling to ancillas, unitary evolution, and projective measurements.) The outcome $a,b$ will occur with probability $$ P(a,b)=\mathrm{Tr}(M_aM_b\rho). $$ If Bob does not know the outcome of Alice's measurement, however, he needs to sum over all the possibilities for Alice's outcomes, so $$ P(b)=\sum_a P(a,b)=\sum_a\mathrm{Tr}(M_aM_b\rho)=\mathrm{Tr}(M_b\rho). $$ This is not only independent of $a$ (Alice's experimental outcomes cannot affect Bob's state without classical communication) but also of the choice of $\{M_a\}$ (Alice's choice of experiment also cannot affect Bob's state without classical communication). Thus the entangled channel cannot be used for communication by itself.

This is really textbook stuff though. If you want something more specific you will need to provide a more definite question.

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Emilio Pisanty
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