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I am reading An Modern Introduction to Quantum Field Theory by Maggiore. On page 28, it says

Using the property of the Pauli matrices $\sigma^2 \sigma^i \sigma^2 = -\sigma^{i*}$ and the explicit form of $\Lambda_{L,R}$, it is easy to show that $$ \sigma^2 \Lambda^*_L \sigma^2 = \Lambda_R. $$ From this it follows that $$ \sigma^2 \psi^*_L \rightarrow \sigma^2(\Lambda_L \psi_L)^∗ = (\sigma^2 \Lambda^*_L \sigma^2)\sigma^2 \psi_L^∗ = \Lambda_R(\sigma^2 \psi_L^∗ ), $$ where we used the fact that $\sigma^2\sigma^2 = 1$.

My questions are:

  • why is $(\Lambda_L \psi_L)^∗ = \Lambda^*_L \psi_L^∗$? I thought the rule was $(AB)^* = A^*B^*$.
  • What's point of taking the complex conjugate of Pauli matrices? It is Hermitian.

The author gave their definition of Pauli matrices in the previous section by: $$ \sigma^1 = \pmatrix{0 & 1 \\ 1 & 0},\quad \sigma^2 = \pmatrix{0 & -i \\ i & 0} ,\quad \sigma^3 = \pmatrix{1 & 0 \\ 0 & -1}. $$

For my second question and the property mentioned in my quote, I assume the Pauli matrices here were anti-hermitian. The author might have redefined it, but I couldn't find it. I also wonder when we should make the Pauli matrices anti-Hermitian, which I assume was obtained by multiplying the matrices by a factor of $i$.

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  • $\begingroup$ Pauli spin matrices are hermitian, but they are not all real. So indeed $\sigma^2 \sigma^i \sigma^2= - (\sigma^i)^*$ $\endgroup$
    – mike stone
    Commented Mar 4 at 13:34
  • $\begingroup$ @mikestone What are Pauli spin matrices? Are they the Pauli matrices in my post? $\endgroup$
    – user174967
    Commented Mar 4 at 13:51
  • $\begingroup$ Yes. They are the same thing. $\sigma^2$ has purely imaginary entries, so $(\sigma^2)^* = -\sigma^2$. $\endgroup$
    – mike stone
    Commented Mar 4 at 13:58
  • $\begingroup$ Are you by any change confusing complex conjugation with Hermitian conjugation? The fact that a matrix $M$ is Hermitian means that it is equal to its Hermitian adjoint $M^\dagger$. In a complex inner product space $V$, the Hermitian adjoint (or Hermitian conjugate) of an operator $M$ is defined to be an operator $M^\dagger$ such that $$\langle u, M v\rangle = \langle M^\dagger u, v\rangle$$ for any two vectors $u,v\in V$. If $V$ is finite dimensional, $M^\dagger$ is the transpose of the complex conjugate, i.e. $M^\dagger = (M^T)^*$, so if $M$ is Hermitian, then $$M^* = M^T.$$ $\endgroup$
    – Albert
    Commented Mar 4 at 18:23

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