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A qubit is given in the following form:

$\left|\psi\right\rangle = \cos\left(\dfrac{\theta}{2}\right)\left|0\right\rangle + e^{i\phi}\sin\left(\dfrac{\theta}{2}\right)\left|1\right\rangle$.

Let's us start at $\left|0\right\rangle$ and rotate about the $x$-axis $180^{\circ}$ (we should end up at $\left|1\right\rangle$). Mathematically, it could be shown easily:

Let $\theta = 180^{\circ}$ and $\phi = 0^{\circ}$:

$\left|\psi\right\rangle = \cos\left(\dfrac{180}{2}\right)\left|0\right\rangle + e^{i(0)}\sin\left(\dfrac{180}{2}\right)\left|1\right\rangle\\ \left|\psi\right\rangle = \cos\left(90\right)\left|0\right\rangle + \sin\left(90\right)\left|1\right\rangle\\ \left|\psi\right\rangle = \left|1\right\rangle $

Now, let's use the rotation matrix instead. The matrix is given as: $R_x(\theta) \equiv e^{-i \theta \mathbb{X}/2} = \cos(\theta/2)\mathbb{I} -i\sin(\theta/2)\mathbb{X} = \begin{bmatrix} \cos(\theta/2) & -i\sin(\theta/2) \\ -i\sin(\theta/2) & \cos(\theta/2)\end{bmatrix}$, where $\mathbb{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $\mathbb{X} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.

Using $R_x(\theta)$, we get

$R_x(180) = \begin{bmatrix} \cos(180/2) & -i\sin(180/2) \\ -i\sin(180/2) & \cos(180/2)\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\\ R_x(180) = \begin{bmatrix}0 & -i\\ -i & 0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\\ R_x(180) = \begin{bmatrix}0\\-i\end{bmatrix}\\ R_x(180) = -i\begin{bmatrix}0\\1\end{bmatrix}. $

Of course, I feel that I am missing something. The vector obtained is correct but with a phase shift of $-i$.

Also, I am wondering why it is okay to let $\phi = 0$ (if it is not correct, then what should be the value?).

Lastly, I would like to know why the rotation matrix only have $\theta$ but not $\phi$.

Thank you in advance!

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  • $\begingroup$ You could use a $R_y$ gate instead $\endgroup$
    – Mauricio
    Oct 3 at 17:36
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Be careful with your choice of notation. You're using $(\theta,\phi)$ to describe the input state, and you're using $\theta$ as the angle of rotation. These two are different $\theta$s.

Now $\theta=\pi$ and $\phi=0$ simply because you chose your initial state to be $|0\rangle$. (Actually, $\phi$ could be arbitrary, so you pick it to be 0 for simplicity.)

It perhaps helps to think about a picture of the Bloch sphere. An arbitrary pure state (on the surface of the sphere) requires two parameters to describe it, $(\theta,\phi)$. An arbitrary rotation requires three parameters - an axis (which is two parameters, entirely equivalent to the $(\theta,\phi)$ of the pure state), and an angle of rotation about that axis. Now, in your example, you have selected a fixed axes, $X$, and the $\theta$ you're using describes the angle of rotation about that axis. See, it's really incomparable to the other $\theta$ you're using.

Finally, you are correct that the $R_x$ operation gives you the answer that you want only up to a global phase factor. But global phase factors make no difference, and can be neglected.

Also, the -i you see outside the state vector is part of global phase. Those are not considered phase shifts (only relative phase shifts are). And since $R_x$ does not introduce any phase shift, $\phi$ is not in its rotation matrix.

PS: As @DaftWullie pointed out, axis needs two parameters.

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  • $\begingroup$ I appreciate your feedback! Can you please elaborate further on the difference between the input state and angle of rotation. Also, you mentioned that "An arbitrary rotation requires three parameters - an axis, and an angle of rotation about that axis". What is the third parameter? $\endgroup$ Aug 14 '19 at 7:19

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