# Spectral theorem for Pauli matrices

Let $$P$$ be a Pauli matrix. Pauli matrices are normal. So by the spectral theorem $$P$$ can be written as $$P=VDV^{-1}$$ for $$V$$ unitary and $$D$$ diagonal (in other words $$P$$ is unitarily diagonalizable). Can we conclude that $$D$$ must be in the Pauli group? Moreover, can we conclude that $$V$$ must be in the Clifford group?

For certain examples of $$P$$ and certain spectral decompositions this is obviously true. For example $$X=HZH$$ here it is indeed the case that $$Z$$ is a Pauli and $$H$$ is in the Clifford group.

Summary:

The answer by DaftWullie shows that every Pauli matrix is diagonalizable by a Clifford gate $$V$$. This implies (indeed is equivalent to) the fact that the diagonalization of a Pauli matrix is always another Pauli matrix.

Discussion with Bebotron shows that there is no uniqueness to this diagonalizing gate $$V$$. A given Pauli matrix can be diagonalized by both Clifford and non Clifford gates. And multiple different Clifford gates can diagonalize the same Pauli. DaftWullie comments further on the extraordinary non-uniqueness of the diagonalizing gate $$V$$.

Consider a member of the Pauli group on $$N$$ qubits. I'm going to write this (up to a possible $$\pm1,\pm i$$ multiplier) as $$P=X_xZ_z$$ where $$x,z\in\{0,1\}^N$$ denote the places where the $$X$$ and $$Z$$ operators lie. Let $$y$$ be $$x$$ OR $$z$$ applied bitwise (and I really mean OR, not XOR). There is always a unitary that maps this Pauli to $$Z_y$$, which is the diagonal Pauli you wanted. To see this, take each site $$i$$. If $$x_i=0$$, do nothing: the qubit is $$Z^{y_i}$$. If $$x_i=1,z_i=0$$, apply Hadamard to qubit $$i$$. This transforms $$X$$ to $$Z$$ (and is in the Clifford group). If $$x_i=z_i=1$$, the qubit $$i$$ has a $$Y$$. Apply the gate $$H_Y=(Y+Z)/\sqrt{2}$$. In the same way that Hadamard, $$(X+Z)/\sqrt{2}$$ exchanges $$X$$ and $$Z$$, this exchanges $$Y$$ and $$Z$$, as required.

Now all we have to do is verify that $$H_Y$$ is Clifford. That means checking that its action on each of the 3 single-qubit Pauli matrices returns a Pauli matrix (up to phases). It was already constructed to do this on $$Z$$ and $$Y$$. We just have to check $$X$$. But $$H_Y$$ and $$X$$ anticommute, so $$H_YXH_Y=-X$$. Thus, $$H_Y$$ is Clifford.

What I have just proven is that there always exists a unitary transformation that satisfies your conditions. Your question specifies must the unitary be of that form? The simple answer is no. As soon as $$N>1$$, your eigenspace has degeneracy. This means that, in addition to the unitary I have constructed, you can apply any unitary that preserves those two spaces. These can certainly be non-Clifford, and can indeed be universal for quantum computation.

That is correct, that is the basic definition of the Clifford group, a set of unitaries that normalize the Pauli group. To rephrase from the link, the Clifford group is

$$C_n = \{V \in U_{2^n} | VP_nV^{\dagger}=P_n\}$$

where $$P_n$$ is the $$n$$-qubit Pauli group.

Edit: Finally note that if we try using $$T$$ gates,

$$TXT^{\dagger}=\begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&e^{-i\pi/4}\end{pmatrix}\\ =\begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix}\begin{pmatrix}0&e^{-i\pi/4}\\1&0\end{pmatrix}\\=\begin{pmatrix}0&e^{-i\pi/4}\\e^{i\pi/4}&0\end{pmatrix}$$

which is not a Pauli matrix, therefore $$T$$ gate is not Clifford.

• Can you explain more about how this answers my question? Feb 7 at 21:38
• Well, as you mention in your edits, this boils down to is the diagonal representation of the Pauli itself a Pauli? And as you mention, this seems trivial. So given that for $P=VDV^{-1}$, with $P\in P_n$ and $D\in P_n$, by definition, any unitary $V$ that normalizes the Pauli group, as mentioned in my response, is a Clifford. Feb 7 at 23:23
• I guess I mean that we could take the case where $P=Z, D=Z, V=T$ then even though $P=VDV^{-1}$ we still have that $V$ is not in the Clifford group Feb 8 at 0:35
• Hmm, I see your point. But for the case that $P=D$, this will always be true for any unitary $V$. If $P=VPV^{-1}$, then $P=V^{-1}PV$. So you can substitute $P=V(V^{-1}PV)V^{-1}=P$. So perhaps we can say that $V$ is a Clifford if $P=VDV^{-1}$, except for the trivial case where $D=P$, since any unitary $V$ would satisfy that? Feb 8 at 5:31