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Consider a system composed of two parts $s$ (subsystem) and $R$ (reservoir), and let $\rho$ be the density matrix for some state of the combined system. Show that if subsystem $s$ is in the pure state, $\rho$ must have the form $P_s \otimes \rho_R$ , where $P_s$ is a projection operator on the Hilbert space of $s$.

From the book Quantum Mechanics: Fundamentals; Kurt Gottfried (chapter 2, problem 4).

I am having a difficult time in understanding this question.

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    $\begingroup$ Hello @histoman12. Can you show us what you have done so far? What is it that you don't understand? Thanks. $\endgroup$
    – joseph h
    Sep 26, 2020 at 4:34

2 Answers 2

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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 $\rho_S=\text{Tr}_R[\rho_{tot}]=|\varphi\rangle_S\langle\varphi|$ is pure, then $\rho_{tot}=\rho_S\otimes\rho_R=|\varphi\rangle_S\langle\varphi|\otimes\rho_R$. 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: $$ \rho_{tot}=\sum_k p_k |\psi_k\rangle\langle\psi_k|, $$ where $|\psi_k\rangle$ is a pure state of $S+R$ and $\sum_k p_k=1$. Then, each $|\psi_k\rangle$ can be expressed as $|\psi_k\rangle=\sum_j \sqrt{\lambda_j^{(k)}}|j_k\rangle_S\otimes|j_k\rangle_R$ by means of Schmidt decomposition, with $\lambda_j^{(k)}\in\mathbb{R^+}$. Let us now take the partial trace: $$ \rho_S=\sum_k p_k \text{Tr}_R[|\psi_k\rangle\langle\psi_k|]=\sum_{k,j} p_k \lambda_j^{(k)} |j_k\rangle_S\langle j_k|=\sum_{k} p_k \rho_S^{(k)}, $$ where $\rho_S^{(k)}=\sum_j\lambda_j^{(k)} |j_k\rangle_S\langle 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 $\rho_S^{(k)}=|\varphi\rangle_S\langle\varphi|$ for all $k$, so that $|\psi_k\rangle= |\varphi\rangle_S\otimes|j_k\rangle_R$ for a given $j$ with $\lambda_j^{(k)}\neq 0$, and $\rho_{tot}=|\varphi\rangle_S\langle\varphi|\otimes\sum_k p_k |j_k\rangle_R\langle j_k|$, proving the assertion.

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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: $$\rho_s=|\Psi_s\rangle \langle\Psi_s|$$. Then, if we compute $\rho_s$ from the total state $\rho$ we have to use the partial trace:

$$\rho_s=\text{Tr}_R[\rho]=|\Psi_s\rangle \langle\Psi_s|$$, 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 $\rho=\rho_s\otimes\rho_R$. 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 $P_s=|\Psi_s\rangle \langle\Psi_s|$.

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