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If we have a system that consists of 2 subsystems A and B. Lets say that the quantum state of the total system $|\Psi\rangle_{AB}$ is a pure state, is it possible that the individual states of the subsystem can be:

  1. Quantum states of them (that means each state of each subsystem can be expressed as a linear superposition of the basis ket of the respective Hilbert space) and not only eigenstates?

  2. Mixed states? (main thing that I want to know)

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  • $\begingroup$ Example, subsystem A has a basis ket, which are in the most common case eigenvectors of the hamiltonian and you can express every other quantum state as a linear combination of them. $\endgroup$
    – imbAF
    Oct 2, 2021 at 21:18
  • $\begingroup$ no, whether, even though the state of the total system is a pure state, the states of the subsystems, whose tensor product gives us the total state of the system, whether these states can be mix states, or no? $\endgroup$
    – imbAF
    Oct 2, 2021 at 21:20
  • $\begingroup$ No but still that doesn't answer my question. $|\Psi\rangle_{AB}=|\Psi\rangle_{A}\times |\Psi\rangle_{B}$. If $|\Psi\rangle_{AB}$ is a pure state (total system state) can the subsystem state's be mixed, or no? $\endgroup$
    – imbAF
    Oct 2, 2021 at 21:25
  • $\begingroup$ or : total state=pure => states of the subsystems=pure & total state=mixed => states of the subsystems=mixed ? $\endgroup$
    – imbAF
    Oct 2, 2021 at 21:26
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – imbAF
    Oct 2, 2021 at 21:26

1 Answer 1

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OP specified the question in the comments, which we'll interpret as follows: Consider a bipartite system $\mathscr H = \mathscr H_1 \otimes \mathscr H_2$ and a density operator $\rho$ on $\mathscr H$. If $\rho$ is pure, does this imply that its reduced density matrices $\rho_1$ and $\rho_2$ are pure? Further, does a mixed $\rho$ imply that $\rho_1$ and $\rho_2$ are mixed?

The answer to both questions is, in general, no. For the case of a pure $\rho$ it is easy to verify (cf. Schmidt decomposition) that $\rho_1$ and $\rho_2$ have the same non-zero eigenvalues and are pure if and only if $\rho$ is a product state.

For example, if $\rho$ corresponds to a maximally entangled Bell state of a two-qubit system, then its reduced density matrices are maximally mixed.

To answer the second question, let us consider a mixed density matrix $\rho_1$ on $\mathscr H_1$ and a pure density matrix $\rho_2$ on $\mathscr H_2$. Then $\rho \equiv \rho_1 \otimes \rho_2$ is a density matrix on $\mathscr H$. We compute: $$\mathrm{Tr} \,\rho^2 = \mathrm{Tr}\, \rho_1^2 \, \mathrm{Tr} \,\rho_2^2 = \mathrm{Tr}\, \rho_1^2<1 \quad ,$$

which shows that $\rho$ is mixed. Obviously, $\rho_1$ and $\rho_2$ are the reduced density matrices of $\rho$. Thus, it is not necessary for the reduced density matrices of a mixed state to be mixed as well.

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