The $n$-fold Pauli operator set is defined as $G_n=\{I,X,Y,Z \}^{\otimes n}$, that is as the set containing all the possible tensor products between $n$ Pauli matrices. It is clear that the Pauli matrices form a basis for the $2\times 2$ complex matrix vector spaces, that is $\mathbb{C}^{2\times 2}$. Apart from it, from the definition of the tensor product, it is known that the $n$-qubit Pauli group will form a basis for the tensor product space $(\mathbb{C}^{2\times 2})^{\otimes n}$.

I am wondering if the such set forms a basis for the complex vector space where the elements of this tensor product space act, that is $\mathbb{C}^{2^n\times 2^n}$. Summarizing, the question would be, is $(\mathbb{C}^{2\times 2})^{\otimes n}=\mathbb{C}^{2^n\times 2^n}$ true?

I have been trying to prove it using arguments about the dimensions of both spaces, but I have not been able to get anything yet.

  • $\begingroup$ The set $ \{I,X,Y,Z \}^{\otimes n} $ that you describe is only $ 1/4 $ of the Pauli group (in fact it is not a group at all since it fails to be closed under multiplication). The set $ \{I,X,Y,Z \}^{\otimes n} $ is mostly famous because it is both a generating set for the Pauli group and, as you point out, an orthogonal basis for the space of $ 2^n \times 2^n $ complex matrices. It is already clear for $ n=1 $ that $ \{ I,X,Y,Z \} $ fails to be a group. For example just the subgroup generated by $ X,Z $ has many extra elements $ <X,Z>=I,X,Z,XZ,XZXZ=-I,-X,-Z, ZX=-XZ $ $\endgroup$ Commented May 13, 2022 at 14:17
  • $\begingroup$ Yeah you are right. I was actually referring to the $n$-fold Pauli operator set. This anyway does not change the question nor the answers. Even if the actual Pauli group was considered, it would also form a basis for such complex space. I edited the question for clarity. $\endgroup$ Commented May 13, 2022 at 21:10
  • $\begingroup$ Ya I agree you are totally right the Pauli set is a nice orthogonal basis. And you are right that the actual Pauli group is a spanning set. But the actual Pauli group would not be a basis since not all the vectors are linearly independent (for example $ X,-X,iX,-iX $ are all linearly dependent) $\endgroup$ Commented May 13, 2022 at 22:08
  • $\begingroup$ Ok, agreed. I was just thinking about generating the complex space, not about the concept of basis. Nice clarification. $\endgroup$ Commented May 14, 2022 at 9:04
  • $\begingroup$ The Pauli set is the best matrix basis that is also (almost) a group...some details... basically the group $ G $ in the linked question is the Pauli group (for $ n $ a prime power the Pauli group is called an extraspecial p group, although for $ p=2 $ it is rather the real Pauli group generated by $ I,X,Z,XZ $ that is extraspecial, i.e. drop the global $ i $ phase) and $ \{ \overline{g_i}: 1 \leq j \leq n^2 \} $ is the Pauli set. When I say this is a group in $ PGL $ that just means its a group if you add some phases like $ \pm1 $. Enjoy! $\endgroup$ Commented May 14, 2022 at 14:18

1 Answer 1


Yes, the set of tensor products of all possible $n$ Pauli operators (including $I$) form an orthogonal basis for the vector space of $2^n \times 2^n$ complex matrices. To see this first we notice that the space has a dimension of $4^n$ and we also have $4^n$ vectors ( the vectors are operators in this case). So we only need to show that they are linearly independent.

We can actually show something stronger. It can be easily seen that the members of the Pauli group are orthogonal under the Hilbert-Schmidt inner product. The H-S inner product of two matrices is defined as $Tr(AB^\dagger)$. We can easily verify from the definition that the Pauli group is a mutually orthogonal set under this inner product. We simply have to use the elementary property $Tr(C \otimes D) = Tr(C)Tr(D)$.

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    $\begingroup$ Thanks for the answer. Does this then imply that by discretization of errors the consideration of the Pauli group as the set of all possible errors, then all the errors are considered too when designing an error correction code? $\endgroup$ Commented Jul 11, 2018 at 8:48
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    $\begingroup$ Yes. In the case of error correction, general errors are decomposed into linear combination of Pauli errors and corrected. A more detailed explanation of how this is done can be found in theory.caltech.edu/people/preskill/ph229/notes/chap7.pdf. $\endgroup$
    – biryani
    Commented Jul 11, 2018 at 9:12
  • $\begingroup$ @biryaniTo prove that the new set you obtained is linearly independent, wouldn't you need to prove that $Tr(AB)=0$ for every $A$ and $B$ of the new set? In that case, I haven't understood how the trace propriety with respect to the tensor product comes into play. When you compute $Tr(A_1 \otimes ... \otimes A_n) = Tr(A_1)...Tr(A_n)$ you are computing the trace of a new element of the set but not the trace of the product of the elements of that set $Tr(AB)$. In the latter case, it's not allowed to write $Tr(AB)=Tr(A)Tr(B)$ and then expand the two traces as you proposed, so I missed a step. $\endgroup$
    – Enrico
    Commented Sep 26, 2020 at 10:08

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