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The toffioli gate flips the target bit when both of the control bits are $\vert 1 \rangle $.

Would it be possible to instead have a gate which flips a target bit when both control bits are 'equal'?

Possible meanings of equal:

  1. Control bits are identical: $\vert c1 \rangle = \alpha \vert 0 \rangle + \beta \vert 1 \rangle = \vert c2 \rangle$.

  2. Control bits have the same sign/phase: doesn't flip if $\vert c1 \rangle = -\vert c2 \rangle$. Flips otherwise.

I am thinking about this in the context of qubits which may or may not have been marked by the oracle in Grover's algorithm if that helps.

Many thanks

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You can decompose that operation into CNOTs and NOTs.

enter image description here

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I propose a $3$-input "Agnew" gate acting on $\{0,1\}^3$ and producing $3$ outputs as follows:

1 2 T || 1 2 T
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0 0 0 || 0 0 1
0 0 1 || 0 0 0
0 1 0 || 0 1 0
0 1 1 || 0 1 1
1 0 0 || 1 0 0
1 0 1 || 1 0 1
1 1 0 || 1 1 1
1 1 1 || 1 1 0

Here, 1 and 2 are the two control (qu)bits, while T is the target (qu)bit.

This can be written in matrix form as:

$$\mathsf{AGNEW}= \left( \begin{array}{cccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ \end{array} \right)$$

This of course can be written in many other ways.

This is a permutation matrix, and is unitary. Thus, it is reversible and can be achieved with a quantum computer.

However, this is a purely classical gate, and the comments in the question about control qubits having the same phase do not immediately fit in to this gate.

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  • $\begingroup$ Would there be a way to make the global phase stuff fit in non-immediately or is that impossible? $\endgroup$ May 8 '20 at 13:01
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  1. Control bits have the same sign/phase: doesn't flip if $\vert c1\rangle=−\vert c2\rangle$. Flips otherwise.

is difficult.

If the state is $-\vert 000\rangle$, there's no way to decide

$\vert c1\rangle, \vert c2\rangle, \vert t\rangle = \vert 0\rangle, \vert 0\rangle, -\vert 0\rangle$ or

$\vert c1\rangle, \vert c2\rangle, \vert t\rangle = \vert 0\rangle, -\vert 0\rangle, \vert 0\rangle$.

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