# Proof of Wigner's theorem in Weinberg’s book

I was following the proof of Wigner's theorem from Weinberg’s book Quantum Theory of Fields, volume 1, pp.91-94 and got stuck in the middle: the proof proceeds as follows for arbitrary state vector:

Now consider an arbitrary state-vector $$\Psi$$ belonging to an arbitrary ray $$\mathscr{R},$$ and expand it in the $$\Psi_{k}:$$ $$\Psi=\sum_{k} C_{k} \Psi_{k}$$ Any state $$\Psi^{\prime}$$ that belongs to the transformed ray $$T \mathscr{R}$$ may similarly be expanded in the complete orthonormal set $$U \Psi_{k}$$ $$\Psi^{\prime}=\sum_{k} C_{k}^{\prime} U \Psi_{k}$$ $$k=1)$$ $$\left|C_{k}\right|^{2}=\left|C_{k}^{\prime}\right|^{2}$$ while the equality of $$\left|\left(\Upsilon_{k}, \Psi\right)\right|^{2}$$ and $$\left|\left(U \Upsilon_{k}, \Psi^{\prime}\right)\right|^{2}$$ tells us that for all $$k \neq 1:$$ $$\left|C_{k}+C_{1}\right|^{2}=\left|C_{k}^{\prime}+C_{1}^{\prime}\right|^{2}$$ The ratio of Eqs. (2.A.9) and (2.A.8) yields the formula $$\operatorname{Re}\left(C_{k} / C_{1}\right)=\operatorname{Re}\left(C_{k}^{\prime} / C_{1}^{\prime}\right)$$ which with Eq. $$(2 . \mathrm{A} .8)$$ also requires $$\operatorname{Im}\left(C_{k} / C_{1}\right)=\pm \operatorname{Im}\left(C_{k}^{\prime} / C_{1}^{\prime}\right)$$ and therefore either $$C_{k} / C_{1}=C_{k}^{\prime} / C_{1}^{\prime}$$ or else $$C_{k} / C_{1}=\left(C_{k}^{\prime} / C_{1}^{\prime}\right)^{*}$$ Furthermore, we can show that the same choice must be made for each $$k$$. (This step in the proof was omitted by Wigner.) To see this, suppose that for some $$k,$$ we have $$C_{k} / C_{1}=C_{k}^{\prime} / C_{1}^{\prime},$$ while for some $$l \neq k,$$ we have instead $$C_{1} / C_{1}=\left(C_{1}^{\prime} / C_{1}^{\prime}\right)^{*} .$$ Suppose also that both ratios are complex, so that these are really different cases. (This incidentally requires that $$k \neq 1$$ and $$l \neq 1,$$ as well as $$k \neq 1 .$$ ) We will show that this is impossible.

Define a state-vector $$\Phi \equiv \frac{1}{\sqrt{3}}\left[\Psi_{1}+\Psi_{k}+\Psi_{l}\right] .\tag{1}$$ since all the ratios of the coefficients in this state-vector are real, we must get the same ratios in any state-vector $$\Phi^{\prime}$$ belonging to the transformed ray: $$\Phi^{\prime}=\frac{\alpha}{\sqrt{3}}\left[U \Psi_{1}+U \Psi_{k}+U \Psi_{l}\right] \tag{2}$$ where $$\alpha$$ is a phase factor with $$|\alpha|=1 .$$ But then the equality of the transition probabilities $$|(\Phi, \Psi)|$$ and $$\left|\left(\Phi^{\prime}, \Psi^{\prime}\right)\right|$$ requires that $$\left|1+\frac{C_{k}^{\prime}}{C_{1}^{\prime}}+\frac{C_{l}^{\prime}}{C_{1}^{\prime}}\right|^{2}=\left|1+\frac{C_{k}}{C_{1}}+\frac{C_{l}}{C_{1}}\right|^{2} \tag{3}$$ and hence $$\left|1+\frac{C_{k}}{C_{1}}+\frac{C_{l}^{*}}{C_{1}^{*}}\right|^{2}=\left|1+\frac{C_{k}}{C_{1}}+\frac{C_{l}}{C_{1}}\right|^{2} \tag{4}$$ This is only possible if $$\operatorname{Re}\left(\frac{C_{k}}{C_{1}} \frac{C_{l}^{*}}{C_{1}^{*}}\right)=\operatorname{Re}\left(\frac{C_{k}}{C_{1}} \frac{C_{l}}{C_{1}}\right)$$ or, in other words, if $$\operatorname{Im}\left(\frac{C_{k}}{C_{1}}\right) \operatorname{Im}\left(\frac{C_{l}}{C_{1}}\right)=0$$ Hence either $$C_{k} / C_{1}$$ or $$C_{l} / C_{1}$$ must be real for any pair $$k, l,$$ in contradiction with our assumptions. We see then that for a given symmetry transformation $$T$$ applied to a given state-vector $$\sum_{k} C_{k} \Psi_{k},$$ we must have either Eq. (2.A.12) for all $$k,$$ or else Eq. ( $$2 .$$ A. 13 ) for all $$k$$.

My question is definition of the state vector $$\phi$$ itself implies the ratios $$C_k/C_1$$ and $$C_l/C_1$$ would be real. But he considered this ratios to be complex to avoid trivial condition. Isn't it contradictory? I mean how one can consider (1) and get (2) as well as (4) from (3) at the same time?

I think i got the mistake i was doing . I was considering $$C_k$$ and $$C_1$$ to be the expansion coefficient of $$\phi$$ which isn't. Am I right?