Is the SWAP gate a Clifford Gate? How would I express it using the Clifford Gate generators?

Question

By my calculations, it looks like the SWAP gate is a Clifford Gate. See the following table:

I follow the same method as in this paper for showing a gate is a Clifford Gate. I got the above table by performing calculations in Qiskit. How would I express the SWAP gate in terms of the generators of the Clifford group? The generators are the $H$ gate, the $S$ gate and the $CNOT$ gate.

Answer 1

It's well known that you can make a swap out of three CNOTs.

For reference, Stim's gate documentation includes H+S+CX decompositions of a lot of Clifford gates including the swap:

Stabilizer Generators:

X_ -> +_X
Z_ -> +_Z
_X -> +X_
_Z -> +Z_

Unitary Matrix:

[+1  ,     ,     ,     ]
[    ,     , +1  ,     ]
[    , +1  ,     ,     ]
[    ,     ,     , +1  ]

Decomposition (into H, S, CX, M, R):

# The following circuit is equivalent (up to global phase) to `SWAP 0 1`
CNOT 0 1
CNOT 1 0
CNOT 0 1

Answer 2

Yes, SWAP is a Clifford gate.

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Your proof is correct. By definition, an $n$-qubit gate $U$ is Clifford if $UPU^\dagger\in G_n$ for all $P\in G_n$ where $G_n$ is the $n$-qubit Pauli group. However, it is easy to see that we only need to check that $UQU^\dagger\in G_n$ for $Q$ that are generators of $G_n$. Moreover, $G_n$ is generated by $\{i, Z_k, X_k\,|\,k=1,\dots,n\}$ where $X_k$ denotes the tensor product of Pauli $X$ on the $k$th qubit and identity applied to all other qubits and similarly for $Z_k$. Since $UiU^\dagger=iI\in G_n$ for all $U$, we only have to check that $UX_kU^\dagger\in G_n$ and $UZ_kU^\dagger\in G_n$ for $k=1,\dots,n$. This is exactly what the table in the question accomplishes.

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It is easy to check that

$$ \text{SWAP} = C_1NOT_2 \circ C_2NOT_1 \circ C_1NOT_2\tag1 $$

where $C_iNOT_j$ denotes the CNOT gate with qubit $i$ as control and $j$ as the target, e.g. by applying both sides of $(1)$ to the computational basis states.

Answer 3

Perhaps might be of interest to provide a full proof that the SWAP operation is equal to the $3$ CNOT gates described in the other answers. A very interesting way to prove this is using the ZX-calculus. We only need the following re-writing rules, known as (1) the bialgebra rule and (2) the Hopf rule.

Equation (1):

Equation (2):

The proof is taken from this reference.

First, translate the circuit description in terms of the calculus,

Now, using the fact that maintaining the connectivity one can change freely the diagram we have the following equations:

Which concludes the proof. Apart from the non-trivial rules we also used fusion of spiders and that the white or grey circles as gates are equivalent representations of the identity.