# Pure state of a system and mixed states of subsystems

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)

• 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. Oct 2, 2021 at 21:18
• 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? Oct 2, 2021 at 21:20
• 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? Oct 2, 2021 at 21:25
• or : total state=pure => states of the subsystems=pure & total state=mixed => states of the subsystems=mixed ? Oct 2, 2021 at 21:26
• Oct 2, 2021 at 21:26

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.