$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$ If I make a rotation of $\frac{\pi}{4}$ around the x axis, starting from $\ket{0}$, I expect $\alpha = \frac{\sqrt{2 + \sqrt{2}}}{2}$ and $\beta= \frac{\sqrt{2 - \sqrt{2}}}{2}$ when measured with respect to the computational basis states of $\ket{0}$ and $\ket{1}$. I can verify this is correct empirically using a quantum simulator. I want to measure with respect to new basis states $\ket{+}$ and $\ket{-}$, defined in Quantum Computation and Quantum Information as follows $\ket{+} \equiv \frac{\ket{0}+\ket{1}}{\sqrt{2}}$, $\ket{-} \equiv \frac{\ket{0}-\ket{1}}{\sqrt{2}}$, and corresponding to the poles of the x axis of the Bloch sphere.

Looking at the Bloch sphere, I would expect to get $\ket{+}$ half of the time, and $\ket{-}$ the other half, and I have been able to empirically verify this. However, according to the mathematics presented in the aforementioned book, I should be able to express this state $\ket{\psi} = \alpha\ket{0} + \beta\ket{1}$, with the $\alpha$ and $\beta$ previously mentioned, as follows:

$$\ket{\psi} = \alpha\ket{0} + \beta\ket{1} = \alpha\frac{\ket{+} + \ket{-}}{\sqrt{2}} + \beta\frac{\ket{+} - \ket{-}}{\sqrt{2}} = \frac{\alpha + \beta}{\sqrt{2}}\ket{+} + \frac{\alpha -\beta}{\sqrt{2}}\ket{-}$$

This math seems sound to me, however if I attempt to translate to this new basis using my $\alpha$ and $\beta$, I get

$$ \frac{\sqrt{2+\sqrt{2}} + \sqrt{2-\sqrt{2}}}{2\sqrt{2}}\ket{+} + \frac{\sqrt{2+\sqrt{2}} -\sqrt{2-\sqrt{2}}}{2\sqrt{2}}\ket{-} $$

Since $\left(\frac{\sqrt{2+\sqrt{2}} + \sqrt{2-\sqrt{2}}}{2\sqrt{2}}\right)^2 \approx 0.85$ this is not at all what I expect from looking at the Bloch sphere, and does not match what I am able to demonstrate empirically. What am I missing?

For reference, the important bit of my Q# code I've been using to test this is:

Rx(PI()/4.0, qubit);
set state = Measure([PauliX], [qubit]);
  • $\begingroup$ Hi @Techmaster21 and welcome to Quantum Computing SE. What are the obtained empirical results for the last case? $\endgroup$ Apr 19, 2020 at 16:40

1 Answer 1


I think there should be $-i$ in the expression for $\beta$:

\begin{equation} R_x\left(\frac{\pi}{4}\right) |0\rangle= \begin{pmatrix} \cos\left(\frac{\pi}{8}\right) & -i \sin\left(\frac{\pi}{8}\right) \\ -i \sin\left(\frac{\pi}{8}\right) & \cos\left(\frac{\pi}{8}\right) \end{pmatrix} \begin{pmatrix} 1 \\ 0\end{pmatrix} = \\ =\frac{\sqrt{2 + \sqrt{2}}}{2} |0\rangle - i \frac{\sqrt{2 - \sqrt{2}}}{2} |1\rangle \end{equation}

So, we will have:

$$|\psi\rangle = \frac{\sqrt{2 + \sqrt{2}} -i \sqrt{2 - \sqrt{2}}}{2\sqrt{2}} |+\rangle + \frac{\sqrt{2 + \sqrt{2}} + i \sqrt{2 - \sqrt{2}}}{2\sqrt{2}} |-\rangle$$


$$\left|\frac{\sqrt{2 + \sqrt{2}} -i \sqrt{2 - \sqrt{2}}}{2\sqrt{2}}\right|^2 = 0.5 $$

  • 2
    $\begingroup$ Thanks for your answer Davit! You hit on two misunderstandings/mistakes I had made. One is that I forgot that phi is measured from the x axis! I for some reason thought that phi was 0 here, which made me miss an i. Secondly, I also believed that Rx rotated clockwise, but it instead rotates counterclockwise (as it should). These two add together to make my expression for Beta off by a -i. Thank you! $\endgroup$ Apr 19, 2020 at 18:19
  • $\begingroup$ You're welcome @Techmaster21. Also, I want to add that, in this problem, for any $\theta$ (from $R_x(\theta)$), the final probability for $|+\rangle$ will be the same. $\endgroup$ Apr 19, 2020 at 18:32

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