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Suppose $\rho_{AB}$ denotes the density matrix of a bipartite system. Reduced density matrix of A ($\rho_A$) is obtained by tracing out B $$\rho_A\equiv\sum_{i}\langle i_B |\rho_{AB}|i_B\rangle$$ where $\{|i_B\rangle \}$ is the basis of the subsystem B. It is said that $\rho_A$ is the physical state of the subsystem A. What justifies this claim?

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The partial trace over $B$ of the quantum state of a bipartite system $AB$ corresponds to discarding $B$: that is, the reduced density matrix $\rho_A=\mathrm{Tr}_B(\rho_{AB})$ is the complete description of the state of the system for any and all measurements that are completely local to $A$.

This can be made precise by considering an arbitrary hermitian measurement operator $\mathcal O_A$ (which includes, among other things, eigenprojectors corresponding to the measurement of some other observable), whose expectation value is $$ \langle \mathcal O_A\rangle = \mathrm{Tr}\mathopen{}\left( \hat{\rho}_{AB} \ \hat{\mathcal O}_A\otimes \mathbb I\right)\mathclose{}. $$ Here the trace can be decomposed as $$ \langle \mathcal O_A\rangle = \mathrm{Tr}_A\mathopen{}\left( \mathrm{Tr}_B\mathopen{}\left( \hat{\rho}_{AB} \ \hat{\mathcal O}_A\otimes \mathbb I\right)\mathclose{}\right)\mathclose{}, $$ and since $\hat{\mathcal O}_A$ does not act on the $B$ sector, it can be factored out of the $B$ trace, giving $$ \langle \mathcal O_A\rangle = \mathrm{Tr}_A\mathopen{}\left( \mathrm{Tr}_B\mathopen{}\left( \hat{\rho}_{AB}\right)\mathclose{} \hat{\mathcal O}_A\right)\mathclose{}, $$ or in other words, $$ \langle \mathcal O_A\rangle = \mathrm{Tr}_A\mathopen{}\left( \hat{\rho}_{A} \hat{\mathcal O}_A\right)\mathclose{}. $$ Thus, if you want to predict the results of any possible experiment that only involves $A$, then you need no more, and no less, than $\hat{\rho}_{AB}$.

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  • $\begingroup$ It might help to distinguish $\hat{\mathcal O}_A$ and $\hat{\mathcal O}_A\otimes I$. $\endgroup$ – Norbert Schuch Sep 18 '17 at 9:28
  • $\begingroup$ @Norbert Yeah, I wondered whether to do that or not. It's probably a good idea. $\endgroup$ – Emilio Pisanty Sep 18 '17 at 9:33
  • $\begingroup$ Well, IMO the essential point in the whole story is to define what it means to measure part of a system. This can e.g. be explained by demanding that for a product state, the result on A is independent of the state of B. $\endgroup$ – Norbert Schuch Sep 18 '17 at 9:50
  • $\begingroup$ @NorbertSchuch I suspect you can write this in a much more clear and concise way than I could. $\endgroup$ – Emilio Pisanty Sep 18 '17 at 9:52
  • $\begingroup$ I wouldn't know why :-) $\endgroup$ – Norbert Schuch Sep 18 '17 at 11:36

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