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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?

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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?

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