Can you make a Toffoli-like gate which flips if control bits are equal?

Question

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

Answer 1

You can decompose that operation into CNOTs and NOTs.

Answer 2

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

1 2 T || 1 2 T
==============
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.

Answer 3

2. 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

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