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I'm trying to compute the state of a particular circuit using Dirac notation but I get stuck after a while .. you can find the circuit there. The numerical coefficients do not matter and I want to compute it for any rotation parameter. So far here's what I've done :

Computing the three states after rotation :

$R_X(\theta_1) |0⟩ = |\phi_1⟩= \cos(\frac{\theta_1}{2})|0⟩ - i \sin(\frac{\theta_1}{2}) |1⟩$

$R_Y(\theta_2) |0⟩ = |\phi_2⟩ = \cos(\frac{\theta_2}{2}) |0⟩ + \sin(\frac{\theta_2}{2}) |1⟩$

$R_Y(\theta_3) |0⟩ = |\phi_3⟩ = \cos(\frac{\theta_3}{2}) |0⟩ + \sin(\frac{\theta_3}{2}) |1⟩$

I then compute the joint state $|\phi_1 \phi_2⟩$ :

$|\phi_1 \phi_2⟩ = (\cos(\frac{\theta_1}{2}) |0⟩ - i \sin(\frac{\theta_1}{2}) |1⟩) \otimes (\cos(\frac{\theta_2}{2}) |0⟩ + \sin(\frac{\theta_2}{2}) |1⟩)$

$ = \cos(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) |00⟩ + \cos(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) |01⟩ - i \sin(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) |10⟩ - i \sin(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) |11⟩ $

I then compute the CNOT gate with control $|\phi_1⟩$ and target $|\phi_2⟩$ :

$ CNOT(|\phi_1 \phi_2⟩) = \cos(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) |00⟩ + \cos(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) |01⟩ - i \sin(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) |11⟩ - i \sin(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) |10⟩ $

But then I'm not sure how to compute the last CNOT gate.. I could do it using matrix computation but the whole point is being able to do this kind of computation quickly by hand.

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If you include the third qubit in your state, now, you have

$$(\cos(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) |00⟩ + \cos(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) |01⟩ - i \sin(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) |11⟩ - i \sin(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) |10⟩)\otimes(\cos(\frac{\theta_3}{2})|0⟩+\sin(\frac{\theta_3}{2})|1⟩)$$

Then you expand this:

$$\cos(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) \cos(\frac{\theta_3}{2})|000⟩ + \cos(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) \cos(\frac{\theta_3}{2})|010⟩ - i \sin(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) \cos(\frac{\theta_3}{2})|110⟩ - i \sin(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2})\cos(\frac{\theta_3}{2}) |100⟩$$

$$ + \cos(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) \sin(\frac{\theta_3}{2})|001⟩ + \cos(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) \sin(\frac{\theta_3}{2})|011⟩ - i \sin(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) \sin(\frac{\theta_3}{2})|111⟩ - i \sin(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2})\sin(\frac{\theta_3}{2}) |101⟩$$

Then do the CNOT on each basis state, changing the third bit iff the second bit is 1:

$$\cos(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) \cos(\frac{\theta_3}{2})|000⟩ + \cos(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) \cos(\frac{\theta_3}{2})|011⟩ - i \sin(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) \cos(\frac{\theta_3}{2})|111⟩ - i \sin(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2})\cos(\frac{\theta_3}{2}) |100⟩$$

$$ + \cos(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) \sin(\frac{\theta_3}{2})|001⟩ + \cos(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2}) \sin(\frac{\theta_3}{2})|010⟩ - i \sin(\frac{\theta_1}{2}) \cos(\frac{\theta_2}{2}) \sin(\frac{\theta_3}{2})|110⟩ - i \sin(\frac{\theta_1}{2}) \sin(\frac{\theta_2}{2})\sin(\frac{\theta_3}{2}) |101⟩$$

... and that's your final state! Kind of a mess as you can see; there's a reason people avoid computing these by hand. :)

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  • $\begingroup$ you were very brave for computing it ! I hoped there would be some kind of shortcut but I guess not really. Thanks a lot, I'll put your answer as accepted. $\endgroup$ Feb 25 '20 at 0:00

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