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 \begin{equation} \label{e6} \sum_{a,b}|\langle a|b\rangle|^{2}=d \end{equation} Using the above two equations we can have the following \begin{equation} \label{e7} \sum_{a,b}\left(\alpha_{ai}^{*}\beta_{bi}-\frac{1}{d}\langle b|a\rangle\right)\langle a|b\rangle=0 \end{equation}

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

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.


1 Answer 1


Equation $$ \sum_{a,b} \left( \alpha_{ai}^* \beta_{bi} - \frac{1}{d} \left<b|a\right> \right) \left<a|b\right> = 0 \tag{1} $$ (which by the way is correct), even if it holds for every $i$, does not imply the following equation: $$ \alpha_{ai}^* \beta_{bi} = \frac{1}{d} \left<b|a\right> \tag{2} $$

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.


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