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I'm trying to understand the relation between the permutation group on all the $2^n$ bitstrings and the Clifford group. My question arises from the fact that the Toffoli gate (which can be thought of as a permutation) cannot be decomposed in terms of Clifford operations. How do we prove it?

glS
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mavzolej
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    Define the 3 qubit pauli group $G = <X_1,X_2,X_3,Z_1,Z_2,Z_3>$ then check that $T G T^\dagger \neq G$ so $T$ is not clifford – unknown Oct 18 '22 at 20:48

1 Answers1

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I guess you're talking about unitaries which preserve computational basis states, i.e. which act as $U|x\rangle = |f(x)\rangle$ where $f:\,\mathbb F_2^n \rightarrow \mathbb F_2^n$ is a reversible Boolean function. These are exactly the permutations of bitstrings.

The only Clifford unitaries preserving computational basis states are those for which $f$ is an affine map. Thus the action is $U_{A,b}|x\rangle = |A x + b\rangle$ for some invertible matrix $A\in\mathrm{Gl}(n,\mathbb F_2)$ and a vector $b\in\mathbb F_2^n$. It is easy to show that the linear part of the transformation is generated by CNOT gates, while the affine shift can be realized with $X$ gates. Now, Toffoli fixes $|0\rangle$, hence if it would be Clifford, then it would have to be realized by a linear map (i.e. one for which the affine shift is $b=0$.).

Finally, the Toffoli gate $T$ is not a linear map on bitstrings: $T(1,1,1) = (1,1,0) \neq (1,0,0) + (0,1,1) = T(1,0,0) + T(0,1,1)$. Thus $T$ is non-Clifford. In fact, the Toffoli gate is universal for classical reversible computation, i.e. every reversible circuit can be decomposed into Toffoli gates (using ancillary bits).

Remark: Of course, you can also check that the Toffoli gate is non-Clifford by letting it act on the generators of the Pauli group. Here, you find that the Toffoli gate maps Paulis to Cliffords, and is thus an element of the third level of the Clifford hierarchy.

Markus Heinrich
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  • Could you please double check the equation? I don't see how it proves that $T$ is not a linear map on bitstrings. – mavzolej Nov 10 '22 at 00:49
  • @mavzolej you're right, the example was wrong. I corrected it. – Markus Heinrich Nov 10 '22 at 07:50
  • Thanks, that really the most straightforward explanation by a counterexample. – mavzolej Nov 11 '22 at 00:12
  • The statement that $f$ is a linear map and the equation $U_A|x\rangle=|Ax\rangle$ imply that $|0\dots 0\rangle$ is necessarily a fixed point of every Clifford that preserves the computational basis states. I think that in general $f$ may be an affine map and its action on the computation basis is given by $U_{A,b}|x\rangle=|Ax+b\rangle$ for some $b\in\mathbb{F}_2^n$. A suitable counterexample is easily constructed from e.g. the Pauli $X$ gate. – Adam Zalcman Feb 17 '23 at 03:28
  • @AdamZalcman Yes, the most general Clifford transformations are affine maps, and the affine shift comes from $X$ gates. They don't play a role here because Toffoli fixes 0. Anyway, the statement is missing that and should be corrected (which I will do in a second). – Markus Heinrich Feb 17 '23 at 13:52
  • Where can I learn more about this affine map result? I'm missing something simple about the notation (if $A_{ij}=b_i=0$, then it seems like $U_{0,0}|x\rangle = |0\rangle$, which is not reversible). – user34722 Mar 27 '23 at 00:59
  • @user34722 the all-zeros matrix is not invertible, so this is not a valid transformation. What do you want to learn? The fact that CNOT circuits correspond to linear invertible maps (i.e. $b=0$) is a simple exercise and used throughout the literature in the context of e.g. Clifford compilation (see also http://arxiv.org/abs/quant-ph/0302002). – Markus Heinrich Mar 28 '23 at 12:52
  • Oh I see, I missed the word 'invertible' and then assumed I must be misunderstanding the notation. Resolved. – user34722 Mar 28 '23 at 17:32