I read in this article (Apendix III p.8) that for $A\in \mathcal{M}_2$, since the normalized Pauli matrices $\{I,X,Y,Z\}/\sqrt{2}$ form an orthogonal matrix basis. $$A=\frac{Tr(AI)I+Tr(AX)X+Tr(AY)Y+Tr(AZ)Z}{2} $$

I don't understand, where does the Trace coefficient come from ?

  • $\begingroup$ Does this answer your question? Example of Hamiltonian decomposition into Pauli matrices $\endgroup$ Commented Apr 10, 2021 at 20:54
  • $\begingroup$ @user1271772 these don't really seem duplicate to me. This is about general decomposition of $2\times 2$ matrices in the Pauli basis; the other question is less clear, but seems to be about decompositions in the multipartite case. That said, I'm pretty sure there are other questions wrt which this might be marked as duplicate $\endgroup$
    – glS
    Commented Apr 11, 2021 at 8:53
  • $\begingroup$ 2x2 is just a special case of the more general 2^n x 2^n case. $\endgroup$ Commented Apr 11, 2021 at 9:46

1 Answer 1


The Pauli matrices form an orthogonal basis of $\mathcal{M}_2$, this vector space can be endowed with a scalar product called the Hilbert-Schmidt inner product

$$ \langle A,B\rangle=\mathrm{Tr}(A^\dagger B)$$ since the Pauli matrices anticommute, their product is traceless, and since they are Hermitian this implies that they are orthogonal with respect to that scalar product, hence the decomposition noticing they have norm $2$ since their square is the identity.

More explicitly, if $$A=\alpha X +\beta Y +\gamma Z+ \delta I $$


$$\mathrm{Tr}(AX)=\alpha \mathrm{Tr}(X^2)+\beta\mathrm{Tr}(XY)+\gamma\mathrm{Tr}(XZ)+\delta\mathrm{Tr}(X)=2\alpha $$

and same for all the others


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