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

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.

• Hello @histoman12. Can you show us what you have done so far? What is it that you don't understand? Thanks. Sep 26, 2020 at 4:34

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.
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 $$P_s=|\Psi_s\rangle \langle\Psi_s|$$.