I've been going over Nielsen and Chuang's Quantum Computation and Quantum Information and I ran into Exercise 4.22, which says,

Prove that a $C^{2}(U)$ gate (for any single qubit unitary $U$) can be constructed using at most eight one-qubit gates, and six controlled-NOTs.

Given $U\in\text{U}(2)$, the operator $C^{2}(U)\in\text{U}(2^{3})$ is a controlled-$U$ gate with two control qubits and one target qubit. The exercise asks us to construct this purely out of one-qubit gates and CNOTs with the specified number of components. The intricate solution to this is given in this post.

This made me wonder a few questions:

  • Is there a circuit that improves on the number of one-qubit gates or number of CNOTs in any way? To be more specific, is there an equivalent circuit made of only one-qubit gates and CNOTs that has strictly less than $14$ components?
  • If the answer is no, is there a proof that $14$ components is the least number of components possible for this problem (assuming $U\in\text{U}(2)$ is arbitrary)?

1 Answer 1


It's not the smallest possible number. At least, not without making several additional restrictions.

You can do it in an expected cost of 4.5 two qubit gates and 8 single qubit gates if you use an ancilla, allow feedback, and count arbitrary-basis measurement and arbitrary-basis preparation as single qubit operations.

Here's the circuit that does it:

enter image description here

Note that I'm assuming all adjacent single qubit operations are fused together, and that includes the classically controlled X being conditionally fused with the final $U^{-1}$.

Note that I'm not counting classical controls towards the qubit count of an operation. Also, the measurement is 50/50 random so all the classically controlled gates are only present half of the time.


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