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I am fairly new to Quantum Computing and have a question which might be trivial for all of you, but I am really struggling with it.

From Quantum Computation and Quantum Information I have learned, that the probility of an outcome is calculatet by: $$ P_{|\varphi\rangle}(x)=\langle \varphi| M_x^\dagger M_x|\varphi\rangle $$ With $$ M_0=|0\rangle \langle 0|,~~~M_1=|1\rangle \langle 1 | $$ for the computational base. Which means: $$ M_0 = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right),~ M_1 = \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right),~ $$ If I would try to calculate this for lets say $|1\rangle$ then I assume it would be done like this: $$ P_{|\varphi\rangle}(1)=\langle \varphi|M_1^\dagger M_1|\varphi\rangle = \left( \begin{array}{cc} \alpha & \beta \\ \end{array} \right) \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right) \left( \begin{array}{cc} \alpha \\ \beta \end{array} \right) =\beta^2 $$ I hope I got this part right.

Now I am struggling with doing the same in the hadamard base. In this case I would use: $$ M_+ = \frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right),~ M_- = \frac{1}{2} \left( \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right) $$ Which means for example for $M_+^\dagger M_+$ would look like this: $$ M_+^\dagger M_+= \frac{1}{2} \left( \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right) \frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) = \frac{1}{4} \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) $$ This would result in every calculation beeing zero. I have obviously made some mistake(s) and would be very happy if someone could point me at those.

(I have searched the forum before, but didn't found and answer to this question. Might be the case that I have a fundamental misunderstanding of what I am doing.)

Thanks in advance.

Alex

Edit: DaftWullie found the reason! Thank you so much.

This would mean the following calculation would be correct?

$$ P_{|\varphi\rangle}(+)=\langle \varphi|M_+^\dagger M_+|\varphi\rangle = \left( \begin{array}{cc} \alpha & \beta \\ \end{array} \right) \frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{cc} \alpha \\ \beta \end{array} \right) =\frac{\alpha+\beta}{2} \left( \begin{array}{cc} \alpha \\ \beta \end{array} \right) $$

Then I still don`t understand the result.

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  • $\begingroup$ The probability should be a real number between $0$ and $1$, not a vector. $\endgroup$ Sep 5, 2022 at 11:23

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You got the bit right with the standard basis, so that's a good start.

The problem that you're having with the Hadamard basis is that you're doing something weird with $M_+^\dagger M_+$. Since $M_+$ is real and symmetric, $M_+=M_+^\dagger$, so you should have $$ M_+^\dagger M_+=\frac12\left(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right)\frac12\left(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right)=\frac12\left(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right). $$ I'm not sure why you put in $M_-$ rather than $M_+$ in your product.

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  • $\begingroup$ Thank you. I somehow took the complexe conjugated as a minus before the 1. Yeah, that one is clearly on me. (rolleyes). I am going to edit my question for one clearification. $\endgroup$
    – dribble290
    Sep 5, 2022 at 8:40
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    $\begingroup$ And 1 more little correction, $\langle\phi| = [\alpha^{\ast} \ \beta^{\ast}]$ $\endgroup$
    – Ohad
    Sep 5, 2022 at 8:44
  • $\begingroup$ @Ohad You are right. Thank you. $\endgroup$
    – dribble290
    Sep 5, 2022 at 8:56

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