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Consider a four-qubit circuit with the following structure:

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where the boxes can be arbitrary two-qubit unitary operations, and the time- evolution proceeds from left to right.

That not every four-qubit unitary can be decomposed in such a way follows from a simple parameter-counting argument: $(2^4)^2\gg 3(2^2)^2$. However, consider the case in which we have a fixed input state, and want to generate a target output. From the point of view of the number of free parameters, this seems feasible: $2(2^4)< 3(2^2)^2$. Is it always possible?

In other words, given a fixed input $|\psi_0\rangle$ and a target $|\psi_t\rangle$, can we always find two-qubit unitaries $U_{12},U_{23},U_{34}$ (acting nontrivially only on the respective subspaces) such that $|\psi_t\rangle = (U_{12}\otimes U_{34})U_{23} |\psi_0\rangle$ ?

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No, it's impossible. It's easier to go backwards. Let $|\psi_t\rangle = \frac{1}{2}\big(|0000\rangle + |0101\rangle + |1010\rangle + |1111\rangle\big)$. The reduced density matrix $\rho_{12}$ of it on qubits $[1,2]$ will be maximally mixed state $\frac{1}{4}I$ (same on qubits $[3,4]$). Unitaries $U_{12}^{-1}$ and $U_{34}^{-1}$ won't change those $\rho_{12}$ and $\rho_{34}$, they will remain as $\frac{1}{4}I$ after the application. But reducing $\frac{1}{4}I$ to a qubit 1 will result in $\frac{1}{2}I$ and $U_{23}^{-1}$ can't change that. So, the reduced matrix of $|\psi_0\rangle$ on qubit 1 must be equal to $\frac{1}{2}I$, but it's not always the case, $|\psi_0\rangle=|0000\rangle$ is a counterexample.

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  • $\begingroup$ ah, nice argument. Working back from the solution, you are saying that if in the initial state the first qubit is uncorrelated with the rest, we can only end up having it correlated with the second qubit, therefore any output state which has correlations between first and third/fourth qubit, such as your $|\psi_t\rangle$, is unreachable. This also tells you that even adding an additional $U_{23}$ block at the end of the circuit isn't enough, as the first qubit still cannot correlate with the fourth. $\endgroup$
    – glS
    Apr 13, 2020 at 18:44
  • $\begingroup$ Not really, you can put SWAPs in place of $U_{12}$ or $U_{34}$ (or both) and q1 will become entangled with q3/q4 if q2 and q3 were entangled. Quantum correlations is a subtle thing :) I was thinking about entropy. I can't say right now what will be if we add another $U_{23}$ block. $\endgroup$
    – Danylo Y
    Apr 13, 2020 at 20:10
  • $\begingroup$ mmh how so? The first qubit still never gets to interact with third and fourth, because U23 does nothing on it if it starts separable. If the initial state is more general there is probably a similar argument about correlations that can be made but yes it's less intuitively clear $\endgroup$
    – glS
    Apr 13, 2020 at 20:50
  • $\begingroup$ Take $|\psi_0\rangle=|0000\rangle$, $U_{23}=CNOT$, $U_{12}=SWAP$, $U_{34}=I$. In the final state q1 and q3 will be entangled. We can just swap the entanglement. $\endgroup$
    – Danylo Y
    Apr 13, 2020 at 20:58
  • $\begingroup$ In what direction are you reading the circuit? It goes left to right thus the cnot acts first and does nothing (but with those unitaries this happens regardless, as they all act as the identity on that input) . But yes if you go from right to left then I agree with you that first and third (but not fourth) can get entangled $\endgroup$
    – glS
    Apr 13, 2020 at 21:15

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