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Let there be a known a scheme (quantum circuit) of Controlled-G, where unitary gate G has G$^†$ such that G≠G$^†$ and GG$^†$=I (for example S and S$^†$, T and T$^†$, V and V$^†$, but not Pauli and H gates).

My question for the experts is:

How is correct that new scheme of Controlled-G$^†$ gate may be constructed from this known scheme of Controlled-G gate by reversing the order of used gates (U) and each U in this scheme changes to the corresponding U$^†$ (if U≠U$^†$ of course)?

For example, see below my OPENQASM-program (note that suffix 'dg' for gate name is used instead '$^†$'in OPENQASM), where it is used well-known scheme of Controlled-S gate and my scheme of Controlled-S$^†$ gate, constructed from this well-known scheme by the above method.
So far I have only received successful results of applying this method and have not found any obvious contradictions with the known theory [*], but suddenly I didn't take something into account.

My program in OPENQASM for example:

//Name of Experiment: Amy Matthew controlled-s and my controlled-sdg gates v7
OPENQASM 2.0; 
include "qelib1.inc";
qreg q[3];
creg c[3];
gate cs a,b {
// a is control, b is target
// see https//uwspace.uwaterloo.ca/bitstream/handle/10012/7818/AmyMatthew.pdf
// fig.4.6b
cx b,a;
tdg a;
cx b,a;
t a;
t b;
}
gate csdg a,b {
// a is control, b is target
// my controlled-sdg (I hope that is reverse of controlled-s)
tdg b;
tdg a;
cx b,a;
t a;
cx b,a;
}

h q[0];
cx q[0],q[1]
;
x q[2];
h q[2];
barrier q;
cs q[0],q[2];
cs q[1],q[2];
barrier q;
csdg q[0],q[2];
csdg q[1]

,q[2];

barrier q;
h q[2];
measure q -> c; 

enter image description here

[*]: Elementary gates for quantum computation Barenco et al. (1995)

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How is correct that new scheme of Controlled-$G^\dagger$ gate may be constructed from this known scheme of Controlled-G gate by reversing the order of used gates ($U$) and each $U$ in this scheme changes to the corresponding $U^\dagger$ (if $U≠U^\dagger$ of course)?

It's 100% correct:

Inverting a composed quantum gate is done with the algorithm you gave. You can check for example the implementation of the inverse() method in qiskit.CompositeGate:

def inverse(self):
    """Invert this gate."""
    # self.data is a list of the quantum gates composing the CompositeGate.
    self.data = [gate.inverse() for gate in reversed(self.data)]
    self.inverse_flag = not self.inverse_flag
    return self

Inverting a controlled-gate boils down to apply the controlled-[inverse of this gate].

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If I understand the question correctly, you're assuming that you have some gate $V$ that you've decomposed as $\prod_{i=1}^NU_i$ and you want to show that $V^\dagger$ is $\prod_{i=1}^NU_{N+1-i}^\dagger$ where the product is taken in the opposite order?

In that case, you just need to show that $VV^\dagger=\mathbb{I}$ given that $U_iU_i^\dagger=\mathbb{I}$. You can do this directly using the stated decompositions above: $$ VV^\dagger=(U_1U_n\ldots U_N)(U_N^\dagger\ldots U_2^\dagger U_1^\dagger)=U_1U_n\ldots U_{N-1}(U_NU_N^\dagger)U_{N-1}^\dagger\ldots U_2^\dagger U_1^\dagger $$ and so on.

This solution is completely general, and does not need to make any assumptions about the form of the original unitary $V$.

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  • $\begingroup$ Yes, that's right, thank you! But I'm little worried that output state of control qubit of V may be not equal input state this qubit. If so, then suddenly V† will not work correctly in some cases... And how do you think? $\endgroup$ – John Lancaster Jun 28 '18 at 14:31
  • $\begingroup$ You can't generally talk about the input and outputs of the control of a controlled gate as always needing to be the same - in general the two qubits will be entangled on output. Still, that doesn't affect the construction. $V^\dagger$ will work. If it helps, remember that quantum mechanics is linear. That means that if you can understand what it does on one set of basis states (such as one choice where the control is in the computational basis so that it doesn't change during the evolution), then linear combinations also work exactly as you would expect them to. $\endgroup$ – DaftWullie Jun 28 '18 at 14:52
  • $\begingroup$ You are absolutely right, I can't speak about the discrepancy between the input and output states of the control qubit in general, but I just wanted to illustrate the example of not entangled cases. Although the problem concerns entangled cases too. I'm trying clarify. Let there be some V issued for the correct implementation of Controlled-G, but actually V=WN, where W is correct implementation of Controlled-G and N is 'Not' gate for control qubit. Although VV†=I, but actually V† will be implementation of Negative Controlled-G†. $\endgroup$ – John Lancaster Jun 30 '18 at 15:11
  • $\begingroup$ I'm still not sure I understand what you're trying to get at. Are you saying: imagine we have $V=\prod_iU_i$ and I implement $(\prod_iU_i)N(\prod_iU_{N+1-i}^\dagger)$, how do I know that $(\prod_iU_{N+1-i}^\dagger)$ is still implementing the correct inverse (which, of course, does not mean that the overall action is $\mathbb{I}$)? $\endgroup$ – DaftWullie Jul 3 '18 at 13:54

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