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Let's say I have a compound quantum system (CQS) in an (unknown to me) pure state $\left|\Psi\right>$. If an operator $\mathbf{A}$ acts only on variables of a subsystem (S) of CQS, then I can calculate expectation value of the corresponding observable as

$$ \bar{A} = \textrm{Tr} (\mathbf{\rho}_S \mathbf{A}) $$

where $\mathbf{\rho}_S$ is a partial density matrix of S (trace over other subsystems of CQS). I can do this for any observable (that acts only on S).

So, if I limit myself to S what is it that I cannot calculate using $\mathbf{\rho}_S$ that I'd be able to calculate if I knew $\left|\Psi\right>$? It seems such things should exist, otherwise I'd be able to construct the wave function for S (is that correct?)

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Below I explain how one would reconstruct a "wavefunction" from a density matrix known to come from a pure state. However, your situation is different. If $\rho$ is the density matrix of an entangled pure state of two systemsn $S$ and $S'$, then $\rho_S$ is the density matrix of a mixed state of $S$, see also this question and its answers. A mixed state does not have a unique wavefunction, so the question of whether you can reconstruct the wavefunction is meaningless. By design, if you are only interested in observables of $S$, then $\rho_S$ contains all necessary information for that - the partial trace is the reverse of the operation of "combining" quantum systems, and if you apply it to a non-entangled state of the combined system it just gives back the corresponding pure state of $S$, which obviously contains all necessary information about $S$.


You are able to construct the wave-function for a density matrix $\rho$ if you know that it comes from a pure state that has such a wavefunction. Or at least, you are able to construct the only physically meaningful content of the wavefunction:

You can compute how likely it is that the system is in a region $R\subset\mathbb{R}^n$ by computing the expectation value of the associated spectral projector $P_R$ (which acts on wavefunctions by multiplying them with the characteristic function of $R$; in non-rigorous language you may think of this as the operator that projects onto the space spanned by position eigenstates with eigenvalues in $R$), and this is what the wavefunction is - a probability density that tells you how likely it is the system is in a region $R$ by integrating the wavefunction over it. The values of the wavefunction at single points are physically irrelevant and therefore you should not expect to be able to recover them - the wavefunction is formally a representant of an equivalence class of square-integrable functions in $L^2(\mathbb{R}^n)$, not a single function whose values at individual points would carry any meaning - all that matters is what happens when you integrate it over regions of non-zero measure, and that information is recoverable from the density matrix.

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  • $\begingroup$ I don't quite understand your answer (esp. the part about associated spectral projector - what is it?), but the conclusion seems odd. Let's take an EPR-like state as an example. Do you mean that Alice is able to reconstruct the wave function of the whole system (Alice + Bob) just from her measurements and compute correlations between her and Bob's spin projections, etc? $\endgroup$ – xaxa Mar 22 '17 at 14:38
  • $\begingroup$ @xaxa No, because Alice can effectively only perform measurements on her part of the system, i.e. the reduced density matrix obtained by the partial trace over Bob's state space, and there is no wavefunction for that reduced matrix because the original state was entangled, see also this answer of mine. Note that my answer starts with the stipulation that you know that your matrix corresponds to a pure state. $\endgroup$ – ACuriousMind Mar 22 '17 at 14:41
  • $\begingroup$ I should probably update my question, because by $\rho_S$ I meant the reduced density matrix - partial trace over other subsystem's variables. $\endgroup$ – xaxa Mar 22 '17 at 14:48
  • $\begingroup$ @xaxa Oops, I completely overlooked that your question is about $\rho_S$! But, then I'm not sure what the question is - since the $\rho_S$ lives only on the space of (say, Alice), you can't compute expectation values of operators that act non-trivially on Bob's space at all, so it's directly obvious there's information in $\rho$ that's not in $\rho_S$. $\endgroup$ – ACuriousMind Mar 22 '17 at 14:50
  • $\begingroup$ The question is that if I'm only interested in Alice's observables, don't care about Bob's space, is there something I'm not able to calculate with $\rho_S$ that would otherwise be able to calculate if I knew $\left|\Psi\right>$? $\endgroup$ – xaxa Mar 22 '17 at 14:52
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So, if I limit myself to S what is it that I cannot calculate using $\mathbf{\rho}_S$ that I'd be able to calculate if I knew $\left|\Psi\right>$?

Locally to $S$? Nothing. As you've noted, the partial trace to $S$ gives you complete information about all local measurements.

It seems such things should exist, otherwise I'd be able to construct the wave function for S (is that correct?)

The problem here is that there might not be such a thing as a "wavefunction for $S$": even if the multipartite system is in a pure state $|\Psi\rangle$, if the state is entangled then there are states (and measurements on those states) that cannot be described by any wavefunction on the $S$ Hilbert space.

For an easy example, consider two entangled qubits on the Bell triplet state $$ \left|\Psi\right\rangle = \frac{|0\rangle|0\rangle+|1\rangle|1\rangle}{\sqrt{2}} $$ along with the local obervables $A_i=\sigma_i$, the Pauli matrices. Then the partial trace to the first qubit gives you the reduced density matrix $$ \rho_s = \frac{|0\rangle\langle0|+|1\rangle\langle1|}{2}, $$ which has zero expectation value for all three $A_i$s, which is impossible for a pure state. If you can describe $\rho_s$ in the form $|\psi\rangle\langle\psi|$, then by all means go for it, but in general this is not possible.

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