# How do you implement a negative controlled gate using the regular controlled gate?

I have been reading a paper about perfect error correction codes, and when the circuit is described, the author uses some negative controlled gates, that is:

The gate is applied if the control is $$|0\rangle$$ and trivial operation is applied when the control is $$|1\rangle$$.

Consequently, it is the reverse operation that a regular controlled gate. The author represents such gates with an empty circle instead of the filled dot used for regular controlled gates.

I am wondering how could one implement one of such negative controlled gates by using a standard controlled gate, that is, add some other gates to the controlled unitary so that its operation is reversed.

I am wondering how could one implement one of such negative controlled gates by using a standard controlled gate, that is, add some other gates to the controlled unitary so that its operation is reversed.

This is addressed in Nielsen and Chuang's section 4.3 (~ p. 184, 10th edition).

Basically,$$c_{|0\rangle}U_{AB} \equiv (X_A\otimes I_B)c_{|1\rangle}U_{AB}(X_A\otimes I_B),$$

i.e. the $$X$$ gate does a basis transformation $$|0\rangle_A\to|1\rangle_A$$ for you such that you can apply the required unitary and then transform back to the original basis using another $$X$$.

In the controlled gates we have been considering, conditional dynamics on the target qubit occurs if the control bits are set to one. Of course, there is nothing special about one, and it is often useful to consider dynamics which occur conditional on the control bit being set to zero. For instance, suppose we wish to implement a two qubit gate in which the second (‘target’) qubit is flipped, conditional on the first (‘control’) qubit being set to zero. In Figure 4.11 we introduce a circuit notation for this gate, together with an equivalent circuit in terms of the gates we have already introduced. Generically we shall use the open circle notation to indicate conditioning on the qubit being set to zero, while a closed circle indicates conditioning on the qubit being set to one.

Let me work out here the more general case, in which you have a controlled operation but in which the "turned-off" and "turned-on" states are not necessarily $$\ket0$$ and $$\ket1$$.


Define now the "inverse-controlled" operation $$\tilde{\mathcal U}$$ defined as $$\tilde{\mathcal U}\equiv \mathbb P_{\phi_2}\otimes U_1+\mathbb P_{\phi_1}\otimes U_2.$$ The question is whether you can implement $$\tilde{\mathcal U}$$ by using only $$\mathcal U$$ and local operations.

More specifically, it is easy to see that we can do this by only using local operations on the controlling qubit. To see this, let $$V$$ be a single-qubit unitary, and consider the composition of $$\mathcal U$$ with two $$V$$ on the first qubit:

$$(V\otimes I)\mathcal U(V^\dagger\otimes I)= (V\mathbb P_{\phi_1} V^\dagger)\otimes U_1+ (V\mathbb P_{\phi_2} V^\dagger)\otimes U_2.$$ We thus simply need a $$V$$ such that $$V\mathbb P_{\phi_i}V^\dagger=\mathbb P_{\phi_{3-i}}$$. This is equivalent to asking for a $$V$$ such that \begin{align} V\ket{\phi_1} &=\ket{\phi_2} \\ V\ket{\phi_2} &=\ket{\phi_1}. \end{align} In the case $$\ket{\phi_1}=\ket0, \ket{\phi_2}=\ket1$$ you recover $$V=X$$, while more in general you want an operation $$V$$ whose SVD reads $$V=\ketbra{\phi_1}{\phi_2}+\ketbra{\phi_2}{\phi_1}.$$