# How can we implement controlled-T gate using CNOT and H, S and T gates?

In general, is there any way to implement a controlled version of an arbitrary gate U if we are given only CNOT and U gate?

• An observation that might be helpful: any arbitrary controlled-U gate on two qubits can be written as: $\text{controlled-}U = |0 \rangle \langle 0 | \otimes \mathbb{I} + |1 \rangle \langle 1 | \otimes U$. But that doesn't automatically tell us how to express the full controlled-U gate in terms of elementary gates...but I'd start by writing U in terms of basic gates and then go from there. Aug 1, 2020 at 5:51

You can use Toffolis and an ancilla to remove the need for the control on the T, then decompose the Toffolis into T+H+CNOT. Since the second Toffoli is uncomputing the ancilla, it can be replaced by a measurement based uncomputation. We can implement $$CT$$ using the following circuit: This solutions uses an extra gate which isn't available above $$R_I(\pi/8) = \sqrt{T}$$

Explanation:

We know that $$T = \sqrt{S} = Z^{\frac{1}{4}}$$. Since $$S,T,Z$$ are all diagonal matrices hence their controlled versions will also be diagonal matrices $$CZ,CS,CT$$. Thus $$CT=CZ^{\frac{1}{4}}$$.

If we can implement the fourth root of $$CZ$$ using the above gates we will have a $$Controlled-T$$ gate.

We can implement a Controlled $$Z$$ gate using $$CNOT$$ and $$H$$ gates $$CZ = |0\rangle\langle 0|\otimes I + |1\rangle\langle 1|\otimes Z \\ = |0\rangle\langle 0|\otimes HIH + |1\rangle\langle 1|\otimes HXH \\ = (I\otimes H)(|0\rangle\langle 0|\otimes I + |1\rangle\langle 1|\otimes X)(I\otimes H)\\ = (I\otimes H)CNOT(I\otimes H)$$

We can implement $$CNOT^{2t}$$ gates using the circuit show in Craig Gidney's Answer We need $$CNOT^{1/4}$$ hence $$t=\frac{1}{8}$$. This requires $$Z^{\frac{1}{8}}$$ i.e $$R_{I}(\pi/8)$$ and its Adjoint gate.

Let $$U = (I\otimes H)CNOT^{\frac{1}{4}}(I\otimes H)$$, then $$U^4= UUUU \\ = (I\otimes H)CNOT^{\frac{1}{4}}(I\otimes H) \times (I\otimes H)CNOT^{\frac{1}{4}}(I\otimes H) \times (I\otimes H)CNOT^{\frac{1}{4}}(I\otimes H) \times (I\otimes H)CNOT^{\frac{1}{4}}(I\otimes H) \\ = (I\otimes H)CNOT^{\frac{1}{4}}CNOT^{\frac{1}{4}}CNOT^{\frac{1}{4}}CNOT^{\frac{1}{4}}(I\otimes H) \\ = (I\otimes H)CNOT(I\otimes H) \\ = CZ$$ Thus $$U = CZ^{\frac{1}{4}} = CT$$

In general for implementing $$Controlled-Z^{\frac{1}{2^n}}$$ gate we would need $$H$$, $$CNOT$$ and $$Z^{\frac{1}{2^{n+1}}}$$(along with its adjoint version) gate.

In general you need more than $$U$$ and CNOT to implement a controlled-$$U$$. One approach to constructing a controlled-$$U$$ gate, for arbitrary $$U \in \mathbf{U}(2)$$, from single qubit and CNOT gates begins by parameterizing $$U$$ in terms of $$U(\alpha,\beta,\gamma,\delta)$$ according to $$U = e^{i\alpha}\begin{bmatrix} \exp\left({-i\frac{\beta+\delta}{2}}\right)\cos \frac{\gamma}{2} & -\exp\left({-i\frac{\beta-\delta}{2}}\right)\sin \frac{\gamma}{2} \\ \exp\left({i \frac{\beta-\delta}{2}}\right)\sin \frac{\gamma}{2} & \exp\left({i\frac{\beta+\delta}{2}}\right)\cos \frac{\gamma}{2} \end{bmatrix}.$$

With this parameterization, $$U$$ can be decomposed in terms of the Pauli matrices $$\sigma_x, \, \sigma_y, \, \sigma_z$$ as

$$U=e^{i\alpha}A \, \sigma_x B \, \sigma_x C, \;\;\; A \equiv e^{-\frac{i}{2}\beta \sigma_z}e^{-\frac{i}{4}\gamma \sigma_y}$$ $$B \equiv e^{\frac{i}{4}\gamma \sigma_y}e^{\frac{i}{4}(\beta+\delta) \sigma_z}, \;\;\; C\equiv e^{\frac{i}{4}(\beta-\delta) \sigma_z}.$$

The usefulness of this decomposition is that $$U$$ can be expressed as above, while $$ABC=I$$. Defining one additional gate $$D=\begin{bmatrix} 1 & 0 \\ 0 & e^{i \alpha} \end{bmatrix}$$ allows circuit implementation of an arbitrary controlled-$$U$$ using four single qubit gates and two CNOT gates as In the specific case noted in the title, $$U=T$$, the parameterization is $$\alpha=\beta=\delta=\frac{\pi}{8}$$, and $$\gamma=0$$. This gives $$A=e^{-\frac{i \pi}{16} \sigma_z} \cong \sqrt{T}$$, $$B=e^{\frac{i \pi}{16} \sigma_z} \cong \sqrt{T^\dagger}$$, $$C=I$$, $$D=\sqrt{T}$$ (where $$\cong$$ indicates equivalence up to global phase). It seems that you need access to $$\sqrt{T}$$ gates to implement the desired controlled-$$T$$.