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
Second Set
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Sign up to join this communityWhen 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
Second Set
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.