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Suppose we have a $3$-qubit system at time $t_0$ in the state $$\vert{\psi(t_0)}\rangle= \vert{q_0}\rangle \otimes \vert{q_1} \rangle \otimes \vert{q_2}\rangle. $$ 1 We want to check if, for instance, the first qubit at time $t_1$ is equals to itself at time $t_0$:

$$\vert{\psi(t_1)} \rangle = \vert{q_0} \rangle \otimes \vert{\phi} \rangle, $$ where $\vert{\phi} \rangle$ is the state of the qubits $q_1, q_2$ after the operations.

My first attempt is to calculate the reduced density matrix using the partial trace over the two other qubits, to get the reduced density matrix at $t_0$ and $t_1$:

$$\rho^0(t_0) = \operatorname{Tr}_{12}(\rho(t_0)),\\ \rho^0(t_1) = \operatorname{Tr}_{12}(\rho(t_1)).$$ So my question is: is there an equivalence relation between the equality of the reduced matrices and the two subsystems being in the same state?

$$\rho^0(t_0) = \rho^0(t_1) \Leftrightarrow \vert{q_0}\rangle = \text{the state of the subsystem at }t_1 $$

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Density matrix completely describes the state of a system (or a subsystem). So if two density matrices are equal then the two states are equal (and vice versa).

But you should not forget that

  1. The state of a subsystem can be mixed (e.g. look at the Bell state $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$). Density matrices uniquely describe mixed states as well, so calculating the partial trace is the most general method for checking the equality.
  2. In general, the state of the whole system isn't completely determined by the states of subsystems. So, we can't deduce the equality of the whole states if states of subsystems are equal.
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  • $\begingroup$ if you are really sure about your two first lines, so that is my answer. thank you to provide me with a mathematical proof. $\endgroup$ – El-Mo Jun 24 at 18:03
  • $\begingroup$ My words can't be counted as a proof :), though this fact indeed can be proved from the postulates and the Born rule. $\endgroup$ – Danylo Y Jun 24 at 20:59
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It depends what you mean in wanting to say that the states "are the same". You probably mean

all observable consequences on just that subsystem (i.e. measurements in arbitrary bases etc, without using the other qubits) are identical.

In which case, it is sufficient that the reduced density matrices are the same - the whole point of the density matrix is to encapsulate all observables of that particular subsystem.

Incidentally, the whole point of needing to use the reduced density matrix is that, in general, states are not separable, and hence you cannot write the state of the first qubit as $|q_0\rangle$. The $\rho$ is the best description that you can give.

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