# Pauli Basis Matrices

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

• What is the source for the second set? Jan 13 '21 at 18:42
• Whatever they are, the second set are not Pauli matrices. They are a trace-orthonormal basis set for 2 by 2 matrices, however. Jan 13 '21 at 18:42
• This is where I found the second set and it is termed as the "Pauli bais": earth.esa.int/documents/653194/656796/… Jan 13 '21 at 19:00
• 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. Jan 13 '21 at 19:14

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.