This may be a fairly basic question, but in Nielsen & Chuang, the following circuit is given for simulating $\exp\left(-i\Delta t Z_1 \otimes Z_2 \otimes Z_3\right)$: klocal simulation

which uses an ancilla qubit initialized to $|0\rangle$. But from messing around with different gate sequences, it seems like this can be done with

\begin{align} CNOT_{1,2}CNOT_{1,3}e^{-i\Delta t Z_3}CNOT_{1,3}CNOT_{1,2} \end{align} or even \begin{align} CNOT_{1,2}CNOT_{2,3}e^{-i\Delta t Z_3}CNOT_{2,3}CNOT_{1,2} \end{align}

which doesn't require the ancilla and has two fewer CNOTs. Is there any advantage to using an ancilla qubit in this circuit?


1 Answer 1


Yes, in this special case the circuit will simplify as you suggest. The advantage of the circuit that was given is that it generalises more easily, and works for any $H$ which has $\pm 1$ eigenvalues. Here's a general form of the circuit for your reference: enter image description here

This essentially comes down to an issue of how you can reversibly compute a one-bit function $f:\{0,1\}^n\mapsto\{0,1\}$ (the 0/1 answer conveys whether it's in the +1 or -1 eigenspace). You can guarantee to do this with $n+1$ bits. Sometimes you can do it with $n$ (basically if the function $f$ is balanced so that half the inputs give 0 answer and half give the 1 answer).


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