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I am a bit confused by the following scenario:

Suppose we are given an entangled state $| \psi \rangle$ that has $2$ qubits (we can generalize to $n$ qubits).

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Suppose we had a circuit like above. Because $| \psi \rangle$ is entangled, we cannot really CNOT each qubit, or Hadamard a single qubit. So how would we predict an outcome of such a circuit in general, mathematically?

And how could we extend this to more complex gates, like a CSWAP?

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  • $\begingroup$ why do you think that $|\psi\rangle$ being entangled means you cannot apply the CNOT or Hadamard on each qubit? $\endgroup$
    – glS ♦
    Mar 31 at 11:53

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Suppose $|\psi\rangle = (|00\rangle+|11\rangle)/\sqrt{2}$. The effect of the first CNOT is to flip the first qubit, therefore the entagled state becomes $|\psi'\rangle = (|10\rangle+|01\rangle)/\sqrt{2}$. Mathematically, you consider the whole state $|1\rangle|\psi\rangle = ..$ and apply the CNOT matrix to the first and second qubit (starting from the top).

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  • $\begingroup$ What if $| \psi \rangle = \frac{1}{\sqrt{3}}( | 00 \rangle + | 01 \rangle + | 11 \rangle )$ being applied in a CNOT? In this case, there are two kets that have a “0” qubit in them. $\endgroup$
    – Andrew Baker
    Mar 31 at 6:16
  • $\begingroup$ In that case you get $|\psi'\rangle = \frac{1}{\sqrt{3}}(|10\rangle + |11\rangle + |01\rangle)$. It is linear algebra, so you just apply the gate to each of the superposed elements of your state. $\endgroup$
    – Michele Amoretti
    Mar 31 at 7:00

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