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I'm aware, that if you measure just a subset of a circuit, which is entangled to another subset, it'll be in a mixed state. For example, if you measure q1 in the circuit

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its purity will be 0.5. But, if you measure the full circuit, it's purity will be 1.

Is it a general fact, that the "full" circuit as a theoretically isolated system will be always in a pure state?

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Yes. If you prepare a state, say $|0\rangle ^{\otimes n}$ (or any pure state in general) and only perform quantum gates (which are unitary operations) and qubits are perfectly isolated (there are no outside interactions with these qubits), then yes, the state will always be a pure state in the end.


This follows from the 3rd postulate of quantum mechanics.

Given a state $\rho(0)$, at time $t=0$, the state at a later time, $t>0$ is given by $$\begin{align} \rho(t) = U(t) \rho(0) U(t) ^ {\dagger} \,, \tag{1} \end{align}$$ where $U(t)$ is a unitary operator.


You can see that if the state of your full system, $\rho$ was a pure state initially, i.e., $$\rho(0) = |\psi(0)\rangle \langle \psi(0)|\,, \tag{2}$$ then state of your full system at time $t$, $$\begin{align}\rho(t) &= U(t) \rho(0) U(t) ^ {\dagger} \tag{3.1}\\ &= U(t)|\psi(0)\rangle \langle \psi(0)| U(t)^ {\dagger} \tag{3.2} \\ &= |\psi(t)\rangle \langle \psi(t)|\,.\tag{3.3}\end{align}$$ is also a pure state.

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