# What is wrong in the following proof? [closed]

Consider an arbitrary $$d\times d$$ pure product state $$|a'\rangle|b'\rangle$$. Note that, as $$|a'\rangle$$ and $$|b'\rangle$$ are pure states one can consider each of them to be a part of complete basis $$\{|a\rangle\}$$ and $$\{|b\rangle\}$$ corresponding to first and second party respectively.

Now, consider the measurement basis for both parties to be $$\{|i\rangle\}$$, where each member of the measurement basis can be written as follows \begin{align} \label{e1} |i\rangle & = \sum_{a}\alpha_{ai}|a\rangle\\ \label{e2} & = \sum_{b}\beta_{bi}|b\rangle \end{align} The normalization condition is given by \begin{align} \sum_{a,b}\alpha_{ai}^{*}\beta_{bi}\langle a|b\rangle & = 1 \end{align}

Note that, as $$\{|a\rangle\}$$ and $$\{|b\rangle\}$$ forms complete basis we can write the following $$$$\label{e6} \sum_{a,b}|\langle a|b\rangle|^{2}=d$$$$ Using the above two equations we can have the following $$$$\label{e7} \sum_{a,b}\left(\alpha_{ai}^{*}\beta_{bi}-\frac{1}{d}\langle b|a\rangle\right)\langle a|b\rangle=0$$$$

As the above equation is valid for all the members of the measurement basis $$\{|i\rangle\}$$ we can have the following relationship for all $$i$$ $$$$\label{e8} \alpha_{ai}^{*}\beta_{bi}=\frac{1}{d}\langle b|a\rangle$$$$

I know that the last equation is wrong as I can find counterexamples for $$d=2$$ when $$|a\rangle=|+z\rangle$$ $$|b\rangle=|+x\rangle$$ and $$|i\rangle=\cos\theta|+z\rangle+\sin\theta|-z\rangle$$. But cannot figure out the fault in the rationale of the above proof.

Equation $$\sum_{a,b} \left( \alpha_{ai}^* \beta_{bi} - \frac{1}{d} \left \right) \left = 0 \tag{1}$$ (which by the way is correct), even if it holds for every $$i$$, does not imply the following equation: $$\alpha_{ai}^* \beta_{bi} = \frac{1}{d} \left \tag{2}$$
Indeed, you can easily check that in your counterexample eq. $$(1)$$ is right, but eq. $$(2)$$ is not. I encourage you to carry out the explicit calculation for this specific case. It will convince you that this implication is wrong.