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In this article: http://home.lu.lv/~sd20008/papers/essays/Clifford%20group%20[paper].pdf , the author calculates the order of the Clifford group and there is one step I'm not convinced.

Consider $n$ qubits and $2n$ operators, $X_i$ and $Z_i$ ($i=1,2,\cdots,n$). All $X$'s and $Z$'s commute, except $X_i$ and $Z_i$ that anti-commute. The author says if we could find another set of operators from $\pm P_n$ denoted as $A_i$ and $B_i$ ($i=1,2,\cdots,n$) which satisfy the same commutative relations with $X$'s and $Z$'s, that is, $A_i$ and $B_i$ anticommute, while all other pairs commute, then there exists a unitary operator $U$ such that $$ UX_iU^\dagger=A_i,\quad UZ_i U^\dagger=B_i $$ $P_n=\{I,X,Y,Z\}^{\otimes n}$. I want to know how to prove the existence of such a $U$.

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All your $B_i$ commute, so that means you can find a set of simultaneous eigenvectors $|\psi_x\rangle$, and can write out their eigenvalues as a vector, e.g. $x=(+1,+1,-1,+1,-1...)$. Next find the basis state $|y\rangle$ whose vector of eigenvalues with the $Z_1$ matches. Now you can define a unitary $$ U=\sum_x|y\rangle\langle \phi|. $$ By construction, this must transform $B_i\rightarrow Z_i$.

Then you have to ask yourself what is $UA_iU^\dagger$. Given commutation/anti-commutation properties are preserved under unitaries, it must be that it anti-commutes with $Z_i$ and commutes with all other $Z_j$. This means that $$ UA_iU^\dagger=\cos\theta_i X_i+\sin\theta_i Y_i. $$ (If you know that you've implemented $U$ with a Clifford, then $X$ and $Y$ are your only options). Now, however, you can find a Z rotation such that $$ e^{i\phi_iZ}(\cos\theta_i X_i+\sin\theta_i Y_i)e^{-i\phi_iZ}=X_i $$ (IIRC, $\phi_i=\pm\theta_i/2$, but you'd want to check). So, replace $$ U\rightarrow \left(\bigotimes_ie^{i\phi_iZ}\right)U. $$ The $Z$ rotations don't change the $Z_i$ operators, so we've constructed it to give the desired output.

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  • $\begingroup$ Can you explain in more detail why the support of $UA_iU^\dagger$ is only on the $i$-th qubit? Why it can't have non-trivial action on other qubits? Thank you very much! $\endgroup$ Commented Sep 13 at 1:28
  • $\begingroup$ Sorry, I missed a step! It can in principle have some Z operators on other qubits. The trick then is to use controlled-phase gates, controlled off the one qubit that has $X/Y$ on it for the given stabilizer to get rid of all the others Zs. $\endgroup$ Commented Sep 15 at 6:19
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If you use that $P_n/U(1)$ is a vector space from the first section, then this result is just rephrasing the change of basis theorem from linear algebra. Not only you can explicitly find the $U$ but you are guaranteed uniqueness (up to phase since we've quotient it out).

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  • $\begingroup$ I don't think so. According to what you say, all basis change are associated with a unitary. But unitary transformation doesn't change commutative relations. $\endgroup$ Commented Sep 13 at 1:38
  • $\begingroup$ That's why the A_i and B_i need to have some the same commutation relations as the X_i and Z_i. $\endgroup$ Commented Sep 13 at 12:10
  • $\begingroup$ @EthanDavies You misunderstood me. I mean when viewing $P_n$ as a vector space, not all the basis change can be generated by a unitary. I just want to know how to prove if a basis change preserves the commutative relations, then there is a unitary generating it. $\endgroup$ Commented Sep 13 at 12:29
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    $\begingroup$ @cos you need to use the fact that the Clifford group maps to the symplectic group over that vector space under this identification. Symplectic maps are exactly those which preserve the commutation relations between Paulis. Hence, any symplectic basis change comes from a Clifford unitary. Such a Clifford unitary $C$ could still give the wrong phases, as you modded them out. However, conjugating the $A_i$ and $B_i$ with some Pauli operator $P$ changes only the phases. In fact, we can find a unique $P$, up to a global phase, such $CP$ gives the correct transformation. $\endgroup$ Commented Sep 15 at 11:56
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    $\begingroup$ Am not sure what a unitary would be over $F_2$, but as @MarkusHeinrich sais, you need to follow the isomorphism backwards to re-describe the change of basis matrix in terms of elements of the $2^n x 2^n$ complex matrices. You are guaranteed that this U is in the normaliser of $P_n$ within $U(2^n)$ that satisfies the required relation. In many contexts eg cs.umd.edu/class/spring2024/cmsc858G/QECCbook-2024-ch1-15.pdf this is exactly the definition of the Clifford group. Note that if $N$ normalises subgroup $H$ of $G$ then $N$ is also a subgroup of $G$. $\endgroup$ Commented Sep 15 at 13:25

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