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This question is motivated by understanding decoherence processes. Consider a bipartite quantum system $S$ composed of two subsystems, $S_{soi}$ (soi = system of interest) and $S_{env}$ (env = environment) and represented by $\mathcal{H} = \mathcal{H}_{soi} \otimes \mathcal{H}_{env}$.

With this stated, I now ask about the following (proposed?) theorem:

Theorem. Suppose the composite system is in an entangled state $|\psi\rangle \in \mathcal{H}$. Then the reduced density matrix for $S_{soi}$, $\rho_{soi}$, is not in a pure form.

How do I prove this? Of course I expect (know) this is true on physical grounds; we expect the reduced density matrix of an entangled system to be impure since the entire notion of entanglement is to suggest that, since the system is entangled with its environment, no individual quantum state can be attributed to the system itself.

Suppose I go for a direct proof in which case I write the following:

Let $\{ |\psi_i\rangle_{soi}|\psi_j\rangle_{env} \}$ (with $i,j$ ranging over appropriate values based on the size of the respective spaces) be the product basis for the system. Then we can expand the system state as $$|\psi \rangle = \sum_{i,j} c_{ij} |\psi_i\rangle_{soi}|\psi_j\rangle_{env}$$ where we must have at least 2 distinct $i$ and 2 distinct $j$ such that $c_{ij} \neq 0$ or else we don't have an entangled state. But I can't see how to proceed from here because the "at least 2 distinct $i$ and 2 distinct $j$ such that $c_{ij} \neq 0$" condition is sort of weird to work with. I can't see how to formulate this analytically so that I can take the partial trace of the corresponding density operator...

Any help would be greatly appreciated. Note that this is NOT a homework problem. This theorem sprung to mind while following the discussion in Schlosshauer's textbook on decoherence.

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Formally, the question asks whether given two Hilbert spaces $H_1$ and $H_2$, the partial trace $\rho_1 = \mathrm{tr}_2(\rho_\psi)$ of the density matrix of a pure state $\lvert\psi\rangle\in H_1\otimes H_2$ has $\mathrm{tr}(\rho_1^2) < 1$ when $\lvert \psi\rangle$ is entangled.

The trick here is choosing the right basis. By Schmidt decomposition, for any $\rho_\psi$ we can pick orthonormal sets (not bases if the dimensions of $H_1$ and $H_2$ are different!) $\lvert \psi_i\rangle_1$ and $\lvert \phi_i\rangle_2$ such that $$ \lvert\psi\rangle = \sum_i c_i \lvert \psi_i\rangle_1\otimes\lvert \phi_i\rangle_2$$ with $c_i\in\mathbb{R}^+, \sum_i c_i^2 = 1$. Then the density matrix is $$ \rho_\psi = \sum_{i,j} c_i c_j \left(\lvert \psi_i\rangle_1 \langle\psi_j\rvert_1 \otimes \lvert \phi_i\rangle_2\langle \phi_j\rvert_2 \right)$$ and the partial trace is $$ \mathrm{tr}_2(\rho_\psi) = \sum_i c_i^2 \lvert \psi_i\rangle_1\langle \psi_i\rvert_1.$$ The square of the partial trace is hence $$ \mathrm{tr}_2(\rho_\psi)^2 = \sum_i c_i^4 \lvert \psi_i\rangle_1 \langle\psi_i\rvert_1$$ and so $\mathrm{tr}(\rho_1^2) = \sum_i c_i^4$. Since all the $c_i$ are between 0 and 1 and we know that $\sum_i c_i^2 = 1$, it follows that $\rho_i^2$ is pure if and only if exactly one of the $c_i = 1$ and all the other $c_i$ are zero, since otherwise $\sum_i c_i^4 < 1$.

Looking back to the definition, exactly one $c_i$ being non-zero means that $\lvert \psi\rangle$ was not entangled since it was just $\lvert \psi_i\rangle_1\otimes\lvert \phi_i\rangle_2$, so indeed the reduced density matrix of a state is mixed if and only if that state was entangled.

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  • $\begingroup$ Reading your answer right now. How come you deleted the commentary re: the pathological cases though? $\endgroup$
    – EE18
    Mar 11, 2023 at 0:06
  • $\begingroup$ @EE18 I realized it was unnecessary because by definition of an entangled state, there just aren't any entangled states for the case where one of the $H_i$ has dimension 1, so the statement "every entangled state leads to a mixed state on the subsystems" is vacuously true. $\endgroup$
    – ACuriousMind
    Mar 11, 2023 at 0:09
  • $\begingroup$ Noted! A few other questions: (1) You write $\sum_i c_i^2 = 1$. This is not intrinsic to the Schmidt decomposition right? This is because the particular vector were decomposing is normalized? (2) Obviously the answer is that things are basis independent, but it's not a priori obvious to me; in particular, you've shown that $\rho_i$ is pure if and only if there is only one term in the Schmidt decomposition. A state is unentangled if there exists a factorization. Thus how do we know that there isn't a factorization which we can't see in the Schmidt decomposition? $\endgroup$
    – EE18
    Mar 11, 2023 at 0:24
  • $\begingroup$ Re: (2) I suppose if you can just comment on the lemma "One term in Schmidt decomposition $\iff$ state unentangled". Maybe it should be obvious to me but I can't see it. The $\implies$ is clear, but I can't see the converse. Maybe this has something to do with the uniqueness of the decomposition but I'm not familiar enough with the theorem to know. $\endgroup$
    – EE18
    Mar 11, 2023 at 0:29
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    $\begingroup$ @EE18 I'm not sure what the problem here is. An unentangled state is exactly of the form $c_i = 1$ for one $i$ and $c_i = 0$ for the rest (regardless of whether or not this is a "Schmidt decomposition"). So my answer shows that such unentangled states have pure subsystem states. Nothing in my answer depends on the exact algorithm for Schmidt decomposition, all we need is that there exist always some orthonormal sets such that I can write $\lvert \psi\rangle$ as in my first equation. Uniqueness etc. is completely irrelevant. $\endgroup$
    – ACuriousMind
    Mar 11, 2023 at 1:11

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