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Consider a two-qubit Clifford gate that maps $|00\rangle$ to $|00\rangle$, up to a phase. Can it map at least one of the other computational basis states $|10\rangle, |01\rangle, |11\rangle$ into a nontrivial superposition of computational basis states (i.e. a linear combination of at least two states)? Or do no such Clifford gates exist?

Similarly, for a three-qubit Clifford gate that maps $|000\rangle$ to $|000\rangle$, up to a phase. Can it map at least one of the eight other computational basis into a nontrivial superposition? Or do no such Clifford gates exist?

The controlled-Hadamard gate maps $|00\rangle$ to $|00\rangle$, but it maps $|10\rangle$ to a superposition $\frac{|10\rangle + |11\rangle}{\sqrt{2}}$. This gate isn't a Clifford gate, but I'm wondering whether a Clifford gate could ever behave somewhat similarly, in that some computational basis states are preserved but others become superpositions.

user196574
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In a stabilizer state, every computational basis state with non-zero amplitude has the same squared magnitude. This can be proved inductively by knowing you can produce any stabilizer state as a graph state: by starting from $|+\rangle^{\otimes n}$ then applying CZ gates then applying single qubit rotations.

The operation you're describing would produce states that didn't have that "all same size if not zero" property, therefore it's not a Clifford operation because Clifford operations can't create non-stabilizer states.

Craig Gidney
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