If subsystem $s$ is in the pure state $ρ$ must have the form $P_s ⊗ ρ_R$

Question

Consider a system composed of two parts s (subsystem) and R (reservoir), and let ρ be the density matrix for some state of the combined system. Show that if subsystem s is in the pure state, ρ must have the form Ps ⊗ ρR , where Ps is a projection operator on the Hilbert space of s.
from the book Quantum Mechanics: Fundamentals; Kurt Gottfried (chap 2 prob 4).
I am having a difficult time in understanding this question.

RMPsp · Accepted Answer

The problem is asking for the necessary condition over $rho$ that yields a pure state for the subsystem. This necessary condition must be the one that the problem proposes, let's see it.
Being $rho_s$ the state of the subsystem, we want it to be a pure state:
$$rho_s=|Psi_srangle langlePsi_s|$$.
Then, if we compute $rho_s$ from the total state $rho$ we have to use the partial trace:
$$rho_s=text{Tr}_R[rho]=|Psi_srangle langlePsi_s|$$,
where $text{Tr}_R$ is the partial trace over the reservoir. The only way of getting the pure state for the subsystem is that subsystem and reservoir are not entangle ,i.e., $rho$ is a product state like $rho=rho_sotimesrho_R$. Once we have the product state, the partial trace removes "cleanly" the reservoir. And since we are looking for a pure state in a density matrix form, it must be a projector, since projectors are written exactly as the thing we are looking for, i.e. as $P_s=|Psi_srangle langlePsi_s|$.

Goffredo_Gretzky · Answer

I think the point at issue may be restated as: prove that, if $rho_{tot}$ is the state of a composed system ($S+R$) and $rho_S=text{Tr}_R[rho_{tot}]=|varphirangle_Slanglevarphi|$ is pure, then $rho_{tot}=rho_Sotimesrho_R=|varphirangle_Slanglevarphi|otimesrho_R$. In other words, if the state of the system is pure, then the overall state must be factorized.
Let us understand why. The overall state can be written as:
$$
rho_{tot}=sum_k p_k |psi_kranglelanglepsi_k|,
$$
where $|psi_krangle$ is a pure state of $S+R$ and $sum_k p_k=1$. Then, each $|psi_krangle$ can be expressed as $|psi_krangle=sum_j sqrt{lambda_j^{(k)}}|j_krangle_Sotimes|j_krangle_R$ by means of Schmidt decomposition, with $lambda_j^{(k)}inmathbb{R^+}$. Let us now take the partial trace:
$$
rho_S=sum_k p_k text{Tr}_R[|psi_kranglelanglepsi_k|]=sum_{k,j} p_k lambda_j^{(k)} |j_krangle_Slangle j_k|=sum_{k} p_k rho_S^{(k)},
$$
where $rho_S^{(k)}=sum_jlambda_j^{(k)} |j_krangle_Slangle j_k| $ are physical states of the system. $rho_S$ is a pure state, so it cannot be written as a non-trivial convex combination of different physical states. Therefore, either $exists k':$ $p_{k'}=1$ and all the rest are zero, so that the $rho_{tot}$ is pure and trivially factorized, or $rho_S^{(k)}$ must be independent of $k$. In the latter case, straightforwardly we observe $rho_S^{(k)}=|varphirangle_Slanglevarphi|$ for all $k$, so that $|psi_krangle= |varphirangle_Sotimes|j_krangle_R$ for a given $j$ with $lambda_j^{(k)}neq 0$, and $rho_{tot}=|varphirangle_Slanglevarphi|otimessum_k p_k |j_krangle_Rlangle j_k|$, proving the assertion.

If subsystem $s$ is in the pure state $ρ$ must have the form $P_s ⊗ ρ_R$

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