Suppose we have some quantum system with sub-systems A and B. It could be, for example, two qubits or groups of qubits. Is it fair to say that tracing out the sub-system A is always akin discarding any information about the sub-system A? So, if our information comes only from the measurement of the sub-system B, the measured probabilities will correspond to the same reduced density matrix regardless of what happens to sub-system A: whether it has been measured by someone else without our knowledge, or not?

If this intuitive understanding is true, is there any mathematical proof you could point me to that a reduced density matrix is independent from what happens to the traced-out sub-system--for any mixed state, any combination of separable and entangled states, and any operators that could affect the traced-out sub-system? I feel this should be simple to prove, but couldn't find anything explicit.

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    $\begingroup$ Tracing out a sub-system is equivalent to ignoring all observables that are associated exclusively with that sub-system. I mean, it really is completely equivalent: if we don't compute any expectation values involving that sub-system's observables, then we might as well have taken a partial trace. Or not. Makes no difference. By the way, the ignore-these-observables idea is more flexible than the partial-trace idea, because the ignore-these-observables idea still makes sense even when no tensor-product structure is given. Does this thought address the question? $\endgroup$ – Chiral Anomaly May 21 at 4:03
  • $\begingroup$ It would be helpful if you specify what one is allowed to use (and what not), regarding quantum mechanical evolution, measurement, etc. (Stinespring etc). Otherwise a self-contained answer would be a whole introductory chapter to a quantum information lecture. $\endgroup$ – Norbert Schuch May 21 at 10:04
  • $\begingroup$ Maybe purification of a quantum state is what you're looking for. $\endgroup$ – Leviathan May 21 at 21:23
  • $\begingroup$ @NorbertSchuch Good question. Let's narrow it down to using only measurement. $\endgroup$ – triclope May 22 at 2:06

Detailed version

Let $\rho$ be a density matrix describing a state shared by Alice and Bob. We can generically write it as $$\newcommand{\ketbra}[2]{\lvert#1\rangle\!\langle#2\rvert}\rho=\sum_{ijkl}\rho_{ijkl}\ketbra ij\otimes\ketbra kl,$$ where here $\ketbra ij$ refer to Alice's degrees of freedom, while $\ketbra kl$ to Bob's.

The reduced state $\rho_A$ of Alice is obtained by tracing out Bob's degrees of freedom, and reads $$\rho_A\equiv\operatorname{Tr}_B\rho=\sum_{ij}\underbrace{\left(\sum_k\rho_{ijkk}\right)}_{\equiv(\rho_A)_{ij}}\ketbra ij.$$ The state that results from Bob doing something on his side can be written as $$\rho' \equiv(\mathbb1\otimes\mathcal E)\rho=\sum_{ijkl}\rho_{ijkl}\ketbra ij\otimes\mathcal E(\ketbra kl),$$ where $\mathcal E$ is a generic (completely positive, trace-preserving) map. We can also get an explicit expression for the matrix elements of $\rho'$ by writing $$\rho'=\sum_{ijkl}\rho_{ijkl}\ketbra ij\otimes\mathcal E(\ketbra kl) =\sum_{ijmn}\underbrace{\left(\sum_{kl}\rho_{ijkl}\langle m\rvert\mathcal E(\ketbra kl)\lvert n\rangle\right)}_{\rho'_{ijmn}}\,\,\ketbra ij\otimes\ketbra mn.$$ In other words, we have $\rho'_{ijmn}=\sum_{kl}\rho_{ijkl}\langle m\rvert\mathcal E(\ketbra kl)\lvert n\rangle.$

How is Alice's share of the state affected by Bob's acting on his? If no classical information is exchanged (that is, if Alice is not given any information about Bob's observations), then her share of the state is described again by the reduced density matrix, this time of $\rho'$: $\rho'_A\equiv\operatorname{Tr}_B\rho'$. This reads $$\rho'_A=\sum_{ij}\left(\sum_m \rho'_{ijmm}\right)\ketbra ij,$$ where $$\sum_m\rho'_{ijmm}=\sum_{klm}\rho_{ijkl}\langle m\rvert\mathcal E(\ketbra kl)\lvert m\rangle= \sum_k\rho_{ijkk},$$ where in the last step we used that any completely positive trace preserving linear map must satisfy $\operatorname{Tr}[\mathcal E(\ketbra kl)]=\delta_{kl}$, which you can easily see passing through Kraus' decomposition: $$\operatorname{Tr}[\mathcal E(\ketbra kl)] =\operatorname{Tr}\left[\sum_\ell A^\ell\ketbra kl A^{\ell \dagger}\right] = \operatorname{Tr}\left[\sum_\ell A^{\ell \dagger} A^\ell\ketbra kl \right] = \operatorname{Tr}[\ketbra kl]. $$ (This also immediately follows from the property of any quantum channel to be trace-preserving). You can therefore conclude that $\operatorname{Tr}_B(\rho)=\operatorname{Tr}_B(\rho')$, which means that no matter what Bob is doing on his side, Alice will not see any measurable difference on her side. There can of course be (classical and nonclassical) correlations between the measurement results, but these correlations are only observable by exchanging classical information between Alice and Bob.

Note that this is totally general, as any operation (including measurements and such) performed by Bob can be written as some quantum map $\mathcal E$ acting on his system.

Short version

Working directly on the matrix elements of the states and the channel, using Einstein's convention for repeated indices, and using numbers instead of latin characters for the indices, we have $$(\rho'_A)_{12}=\{\operatorname{Tr}_B[(\mathbb1\otimes\mathcal E)\rho]\}_{12} =[(\mathbb1\otimes\mathcal E)\rho]_{1233} =\rho_{1245}\mathcal E_{3345}=\rho_{1233}\equiv(\rho_A)_{12},$$ where I'm defining the matrix elements of $\mathcal E$ as $\mathcal E_{1234}\equiv \langle1\rvert \mathcal E(\ketbra34)\lvert 2\rangle$.


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