## Mixtures and subsystems

We have used the density operators to describe two distinct situations: the statistical properties of the mixtures of states, and the statistical properties of subsystems of composite systems.
In order to see the relationship between the two, consider a joint state of a bipartite system \mathcal{AB}, written in a product basis in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} as
\begin{aligned}
|\psi_{\mathcal{AB}}\rangle
&= \sum_{ij} c_{ij}|a_i\rangle\otimes|b_j\rangle
\\&= \sum_{j=1} |\widetilde\psi_j\rangle|b_j\rangle
\\&= \sum_{j=1} \sqrt{p_j}|\psi_j\rangle|b_j\rangle
\end{aligned}
\tag{9.7.1}
where |\widetilde\psi_j\rangle = \sum_i c_{ij}|a_i\rangle = \sqrt{p_j}|\psi_j\rangle, and the vectors |\psi_j\rangle are the normalised versions of the |\widetilde\psi_j\rangle.
Note that p_j=\langle\widetilde\psi_j|\widetilde\psi_j\rangle.

Then the partial trace over \mathcal{B} gives the reduced density operator of subsystem \mathcal{A}:
\begin{aligned}
\rho^{\mathcal{A}}
&=\operatorname{tr}_{\mathcal{B}} \sum_{ij} |\widetilde\psi_i\rangle\langle\widetilde\psi_j| \otimes |b_i\rangle\langle b_j|
\\&= \sum_{ij} |\widetilde\psi_i\rangle\langle\widetilde\psi_j| (\operatorname{tr}|b_i\rangle\langle b_j|)
\\&= \sum_{ij} |\widetilde\psi_i\rangle\langle\widetilde\psi_j| \langle b_j|b_i\rangle
\\&= \sum_{i} |\widetilde\psi_i\rangle\langle\widetilde\psi_i|
= \sum_{i} p_i |\psi_i\rangle\langle\psi_i|.
\end{aligned}

Now, let us see how \rho^{\mathcal{A}} can be understood in terms of mixtures.
Let us place subsystems \mathcal{A} and \mathcal{B} in separate labs, run by Alice and Bob, respectively.
When Bob measures part \mathcal{B} in the |b_j\rangle basis and obtains result k, which happens with the probability p_k, he prepares subsystem \mathcal{A} in the state |\psi_k\rangle:
\sum_{i=1} \sqrt{p_j}|\psi_i\rangle|b_i\rangle
\overset{\mathrm{outcome}\,k}{\longmapsto}
|\psi_k\rangle|b_k\rangle.
Bob does not communicate the outcome of his measurement.
Thus, from Alice’s perspective, Bob prepares a mixture of |\psi_1\rangle,\ldots,|\psi_m\rangle, with probabilities p_1,\ldots,p_m, which means that Alice, who knows the joint state but not the outcomes of Bob’s measurement, may associate density matrix \rho^\mathcal{A}=\sum_i p_i|\psi_i\rangle\langle\psi_i| with her subsystem \mathcal{A}.
This is the same \rho^{\mathcal{A}} which we obtained by the partial trace.

But suppose Bob chooses to measure his subsystem in some other basis.
Will it have any impact on Alice’s statistical predictions?
Measurement in the new basis will result in a different mixture, but Alice’s density operator will not change.
Suppose Bob chooses basis |d_i\rangle for his measurement.
Any two orthonormal bases are connected by some unitary transformation, and so we can write |b_i\rangle = U|d_i\rangle for some unitary U.
In terms of components, |b_i\rangle = \sum_j U_{ij}|d_j\rangle.
The joint state can now be expressed as
\begin{aligned}
|\psi_{\mathcal{AB}}\rangle
&= \sum_{i} |\widetilde\psi_i\rangle|b_i\rangle
\\&= \sum_{i} |\widetilde\psi_i\rangle \left( \sum_j U_{ij}|d_j\rangle \right)
\\&= \sum_j \underbrace{\left( \sum_i U_{ij}|\widetilde\psi_i\rangle \right)}_{|\widetilde\phi_j\rangle}|d_j\rangle
\\&= \sum_j|\widetilde\phi_j\rangle|d_j\rangle.
\end{aligned}

If Bob measures in the |d_i\rangle basis then he generates a new mixture of states |\phi_1\rangle,\ldots|\phi_m\rangle, which are the normalised versions of |\widetilde\phi_1\rangle,\ldots|\widetilde\phi_m\rangle, with each |\phi_k\rangle occurring with probability p_k=\langle\widetilde\phi_k|\widetilde\phi_k\rangle.
But this new mixture has exactly the same density operator as the previous one:
\begin{aligned}
\sum_j|\widetilde\phi_j\rangle\langle\widetilde\phi_j|
&= \sum_{ijl} U_{ij}|\widetilde\psi_i\rangle\langle\widetilde\psi_l|U^\star_{lj}
\\&= \sum_{il} \underbrace{\left(\sum_j U_{ij}U^\star_{lj}\right)}_{\delta_{il}}|\widetilde\psi_i\rangle\langle\widetilde\psi_l|
\\&= \sum_i|\widetilde\psi_j\rangle\langle\widetilde\psi_j|,
\end{aligned}
which is exactly \rho^\mathcal{A}.
So does it really matter whether Bob performs the measurement or not?

*It does not.*

After all, Alice and Bob may be miles away from each other, and if any of Bob’s actions were to result in something that is physically detectable at the Alice’s location that would amount to instantaneous communication between the two of them.

From the operational point of view it does not really matter whether the density operator represents our ignorance of the actual state (mixtures) or provides the only description we can have after discarding one part of an entangled state (partial trace).
In the former case, the system is in some definite pure state but we do not know which.
In contrast, when the density operator arises from tracing out irrelevant, or unavailable, degrees of freedom, the individual system cannot be thought to be in some definite state of which we are ignorant.
Philosophy aside, the fact that the two interpretations give exactly the same predictions is useful: switching back and forth between the two pictures often offers additional insights and may even simplify lengthy calculations.