# Construct a two-qubit quantum gate with given action using the gates ${CNOT, H, T}$

Using the gates $${CNOT, H, T}$$, construct a 2-qubit gate that acts as follows on the computational basis

$$|0⟩⊗|0⟩ = |0⟩⊗|0⟩$$

$$|0⟩⊗|1⟩ = e^{pi*i/4}|0⟩⊗|1⟩$$

$$|1⟩⊗|0⟩ = e^{pi*i/4}|1⟩⊗|0⟩$$

$$|1⟩⊗|1⟩ = |1⟩⊗|1⟩$$

I'm not sure how to solve this - I'm sure that if someone shows me the solution (ie the actual circuit) I'd understand it, but I really want to know how to generally come up with solutions to problems where we're meant to create circuits as well(I've been using trial and error so far).

• The last qubit is always unchanged (equivalent to passing it through a CNOT with $$|0⟩$$ as the first input qubit).

• We definitely need to use the $$T$$ gate for the terms which have $$e^{pi*i/4}$$

The $$T$$ gate applies a phase of $$e^{\pi i / 4}$$ to $$|1\rangle$$, i.e., it has the following effect: $$T(\alpha|0\rangle + \beta|1\rangle) = \alpha|0\rangle + e^{\pi i / 4}\beta|1\rangle$$. We want this effect to be applied when one of the qubits is in $$|1\rangle$$, but not both. Therefore, we need a way to distinction these two states.

We can use the $$\mathrm{CNOT}$$ gate to differentiate these two states from the rest. To do this, apply a $$\mathrm{CNOT}$$ gate with the first qubit as control and the second as target. This will leave $$|00\rangle$$ and $$|01\rangle$$ unaffected, $$|10\rangle$$ turns into $$|11\rangle$$ and $$|11\rangle$$ into $$|10\rangle$$. In other words, it will set the second qubit to $$|1\rangle$$ if and only if one qubit but not both were in the $$|1\rangle$$ state.

The states we want to apply the $$\pi / 4$$ phase to are now mapped to the states with $$|1\rangle$$ in the second qubit. Thus, we can apply the $$T$$ gate to the second qubit, changing $$|01\rangle$$ into $$e^{\pi i / 4}|01\rangle$$ and $$|11\rangle$$ into $$e^{\pi i / 4}|11\rangle$$.

Now we only have to map back to our original states, which can be easily done with a $$\mathrm{CNOT}$$ gate just as the first one.

Put into a Qiskit circuit, this look as follows.

from qiskit import QuantumCircuit
from qiskit.quantum_info import Operator

qc = QuantumCircuit(2)

qc.cx(0, 1)
qc.t(1)
qc.cx(0, 1)

print(qc)
print(np.round(Operator(qc).data, 5))


Which gives the circuit and unitary:

q_0: ──■─────────■──
┌─┴─┐┌───┐┌─┴─┐
q_1: ┤ X ├┤ T ├┤ X ├
└───┘└───┘└───┘
[[1.     +0.j      0.     +0.j      0.     +0.j      0.     +0.j     ]
[0.     +0.j      0.70711+0.70711j 0.     +0.j      0.     +0.j     ]
[0.     +0.j      0.     +0.j      0.70711+0.70711j 0.     +0.j     ]
[0.     +0.j      0.     +0.j      0.     +0.j      1.     +0.j     ]]

• Why do we apply $T$ to the second qubit? We need $|0⟩⊗|1⟩=e^{pi∗i/4}|0⟩⊗|1⟩$ where the $e^{pi*I/4}$ term is on the first qubit not the second Apr 29, 2022 at 6:53
• @SVMteamsTool $e^{\pi i / 4}$ is a global phase in $e^{\pi i / 4} |01\rangle$, meaning that $e^{\pi i / 4} |0\rangle \otimes |1\rangle = |0\rangle \otimes e^{\pi i / 4} |1\rangle$. Try writing both states as vectors to see they are equivalent. Apr 29, 2022 at 7:06