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I was reading some lectures on quantum information theory and it came with the idea of " mutually unbiased bases" that I never heard about. But the definition is very simple: two basis $ B = \{|b_0> , |b_1> \} $ and $ C = \{|c_0> , |c_1> \} $ are said to be mutually unbiased if $ <b_i|c_j>$ is independent of i and j for every i and j. In other words, if the scalar product does not depend on any vector in some of the basis, but is just a number.

I do not know why this definition can be useful at all. In other words: why would I want a set of basis to be mutually unbiased? More specifically, where can it be useful in quantum information theory?

That was my first question. Now, given to basis: $ Z = \{ |0> , |1> \}$ and $X = \{|+> , |-> \}$, with

$ |+> = \dfrac{1}{\sqrt2}|0> + \dfrac{1}{\sqrt2}|1> $ and $ |-> = \dfrac{1}{\sqrt2}|0> - \dfrac{1}{\sqrt2}|1> $

I need to find another basis Y that is mutually unbiased with both X and Z and show that there is no other basis that is mutually exclusive with X, Y and Z. Of course we all know the answer must be $Y = \{\dfrac{1}{\sqrt2}|0> + \dfrac{i}{\sqrt2}|1> , \dfrac{1}{\sqrt2}|0> - \dfrac{i}{\sqrt2}|1> \}$, but I cannot show this to be true.

My first attempt was to write $Y = \{ |a> , |b> \} $ and write it as a combination of the basis Z as $ |a> = \alpha|0> + \beta|1> $ and $ |b> = \omega|0> + \lambda|1> $ and then try to solve for the coefficients $\alpha$, $\beta$ , $\omega$ , and $\lambda$ given the set of equations $ <a|a> = 1$ , $ <b|b> = 1$ and $<a|b> = <b|a> = 0$ but I did not get anywhere. Any help?

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    $\begingroup$ see Wootters, W.K. and Fields, B.D., 1989. Optimal state-determination by mutually unbiased measurements. Annals of Physics, 191(2), pp.363-381. Note that the condition is not on $\langle i\vert j\rangle$ but on the absolute value $\vert \langle i\vert j\rangle \vert$. $\endgroup$ – ZeroTheHero Jun 8 at 19:49
  • $\begingroup$ I translated the title "neutral" from portuguese. Did not knew it was supposed to be "unbiased". Now it must be easier to look for references, thank you. And thanks for the papper. Probably there was a typo in the lecture notes, because there it was saying nothing about the absolute value. $\endgroup$ – Dimitri Jun 8 at 21:22