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Weinberg’s proof of Wigner's theorem

Physics Asked on December 18, 2020

I’m working through the proof of Wigner’s theorem in Weinberg’s The Quantum Theory of Fields Volume 1 Chapter 2 Appendix A pp. 91. Consider the following steps of the proof

Let $left{psi_{k} epsilon mathbb{R}_{k}right}_{k=1}^{n}$ be a complete orthonormal basis. Then,
$$
left|leftlanglepsi_{k} mid psi_{l}rightrangleright|=delta_{k l}
$$

Let $psi_{k}^{prime}$ be some arbitrary choice of vector belonging to the transformed ray $mathbb{R}_{k} .$ Then, we have
$$
left|leftlanglepsi_{k}^{prime} mid psi_{l}^{prime}rightrangleright|^{2}=left|leftlanglepsi_{k} mid psi_{l}rightrangleright|^{2}=delta_{k l}tag{1}
$$

But $left|leftlanglepsi_{k}^{prime} mid psi_{k}^{prime}rightrangleright|$ is real and positive. Hence $left|leftlanglepsi_{k}^{prime} mid psi_{k}^{prime}rightrangleright|=1$
Therefore,
$$
langlepsi_{k}^{prime} mid psi_{l}^{prime}rangle=delta_{k l}tag{2}
$$

Can anyone explain how to get (2) from (1)

One Answer

$|langle psi_k|psi_ellrangle |^2$ is a real positive number, let us call it $c^2$. Then $|langle psi_k|psi_ellrangle |=pm c$. But also $|langle psi_k|psi_ellrangle |$ is a real positive number, so we need to take the positive square root, here $delta_{kell}$.

Correct answer by Oбжорoв on December 18, 2020

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