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When I search for the Pauli basis matrices I find both the following sets but I wonder which one is the right one and why does the first set have an imaginary term which is absent in the second set.

First Set

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Second Set

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    $\begingroup$ What is the source for the second set? $\endgroup$
    – G. Smith
    Jan 13 '21 at 18:42
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    $\begingroup$ Whatever they are, the second set are not Pauli matrices. They are a trace-orthonormal basis set for 2 by 2 matrices, however. $\endgroup$
    – mike stone
    Jan 13 '21 at 18:42
  • $\begingroup$ This is where I found the second set and it is termed as the "Pauli bais": earth.esa.int/documents/653194/656796/… $\endgroup$
    – GISEnthu
    Jan 13 '21 at 19:00
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    $\begingroup$ The first set are what is normally meant by 'the' Pauli matrices. I've not heard them called 'the Pauli basis' really, even though they are a basis of the traceless hermitian matrices. $\endgroup$
    – jacob1729
    Jan 13 '21 at 19:14
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The first set of matrices are what is conventionally called the Pauli matrices. The identity matrix is sometimes included as a Pauli matrix $\sigma_0$. With this included, we have a correspondence between the two sets of matrices:

$S_a= \frac 1 {\sqrt 2} \sigma_0 \\S_b= \frac 1 {\sqrt 2} \sigma_3 \\S_c= \frac 1 {\sqrt 2} \sigma_1 \\S_d= \frac {-i} {\sqrt 2} \sigma_2$

Apart from the common factor of $\frac 1 {\sqrt 2}$, the only other difference is the factor of $-i$ which makes the elements of $S_d$ real. Why is this significant ? The Pauli matrices are both Hermitian i.e. $\sigma_n^\dagger = \sigma_n$ and unitary i.e. $\sigma_n^\dagger = \sigma_n^{-1}$. As a result they are involutory i.e. $\sigma_n^2=I$.

$\sqrt 2 S_d$ is unitary since

$(\sqrt 2 S_d)^\dagger = (\sqrt 2 S_d)^T = (\sqrt 2 S_d)^{-1}$

However $\sqrt 2 S_d$ is not Hermitian or involutory since

$(\sqrt 2 S_d)^\dagger = (\sqrt 2 S_d)^{-1} = -(\sqrt 2 S_d) \\ \Rightarrow (\sqrt 2 S_d)^2 = -I$

Since $(\sqrt 2 S_d)$ is not Hermitian and not involutory it cannot be a Pauli matrix.

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