# Finding rotation axis of an operator on Bloch sphere

I would like to find the action of an operator $V$ on the Bloch sphere. The matrix representation of $V$ is

$V=\sqrt{\frac{i}{2}}\begin{pmatrix} i &-1 \\ -1 &i\end{pmatrix}$.

$\rightarrow |\det(V)|=1$

This means, it should be either a reflection or a rotation. Since it maps $|0\rangle \to \sqrt{i/2}(i \cdot|0\rangle-|1\rangle):=|a\rangle$ and $|1\rangle \to \sqrt{\frac{i}{2}}(-|0\rangle+i|1\rangle):=|b\rangle$. In order to represent those states on the Bloch sphere, the prefactor of $|0\rangle$ needs to be 1. Therefore

$|a\rangle \propto |+\rangle$ and $|b\rangle \propto |-\rangle$.

My first question is, if I'm allowed to neglet different phase factors, i.e. for $a$ I neglected $i^{3/2}$ and for $b$ I neglected $-1$. Is this correct?

If this was true, then it would correspond to a hadamard gate. Knowing that any unitary gate can be represented by $U=e^{i \frac{\theta}{2} \vec{n}\cdot\vec{\sigma}}$, so it would be possible to solve for $\theta$, but isn't there a quicker and more elegant way to solve this?

• Fine to neglect overall phase factors. Based on the final states, this looks like a rotation of $+\pi$ on the Bloch sphere about the $y$-axis. Jul 14 '16 at 20:06
• ok, thanks for your answer! but how did you find out? This is a pretty simple case, but how do you find out about the axis $\vec{n}$ in a more difficult case? Jul 14 '16 at 20:39
• Exponential of Pauli vector, and of course read section 1.2 above it. Jul 14 '16 at 22:21
• Sorry, I meant a rotation by $\pi/2$. I obtained this just by noting that states can be represented on the Bloch sphere by $cos(\frac{\theta}{2})\mid 0 \rangle + e^{i\phi}sin(\frac{\theta}{2}) \mid 1 \rangle$, and matching terms. Your transformed states are, to within a global phase factor, $\frac{1}{\sqrt{2}}[\mid 0 \rangle \pm i \mid 1 \rangle]$ which works out to $\theta=\pm \pi/2$ and $\phi=\pi/2$. Jul 14 '16 at 22:30

Since the action of a unitary matrix $U$ on the Bloch sphere is defined by

$$\sigma_j\mapsto U\,\sigma_j\,U^\dagger = U\,\sigma_j\,U^{-1}\tag{1}$$

with $\sigma_j$ standing for the Pauli matrices, any scalar phase factor $e^{i\,\phi}$ is quotiented away by the map (which is the big A Adjoint representation of the unitary group). Unitary matrices always act on the Bloch sphere by rotation, never reflexion. This is because $U(2)$, as opposed to $O(3)$, is a topologically connected group and a reflexion matrix, say $\left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$, is simply the $SU(2)$ member $i\,\sigma_z=\left(\begin{array}{cc}i&0\\0&-i\end{array}\right)$ multipied by the phase factor $i$. So the action of the two are the same rotation.

Given 1, the Bloch vector $S =\sum\limits_j s_j\,\sigma_j$ becomes $S^\prime = \sum\limits_j s_j^\prime\,\sigma_j$ where:

$$s_j^\prime = \frac{1}{2}\,\mathrm{tr}(\sigma_j\,\sum\limits_k\,(s_k\,U\,\sigma_k\,U^\dagger))\tag{2}$$

So the matrix of your Bloch sphere rotator is the $3\times 3$ matrix $\left[\frac{1}{2}\,\mathrm{tr}(\sigma_j\,U\,\sigma_k\,U^\dagger))\right]$.

You can find the axis of rotation by finding the eigenvector of the rotation matrix that corresponds to the eigenvalue 1, which is usually the only real eigenvector (unless the rotation is trivial).

Alternatively you should first divide out any phase factor to set your matrix's determinant to 1, so that you're dealing with a member of $SU(2)$. This is not essential for the method above (because the phase is cancelled by the Adjoint representation) but it is here. Now calculate the components of the Pauli matrices in your operator:

$$u_j = -\frac{i}{2}\,\mathrm{tr}(\sigma_j\,U)$$

The vector with the $u_j$ as components is the axis of rotation, where the $j^{th}$ co-ordinate axis is the axis of rotation of the Pauli matrix $\sigma_j$. The magnitude $\sqrt{\sum_j\,u_j^2}$ is the sine of half the rotation angle.

Question from OP

Anyway, why is it possible to find a rotation axis for any unitary matrix?It is possible to decompose a unitary matrix in the product of the three Operators x,y,z , so there shoul be 3 axes? –

Answer: (1) The image of the group $U(2)$ under the Adjoint representation is the group $SO(3)$; then (2) Euler's rotation theorem shows that all members of the latter group are rotations.

I tried to find the rotation axis of the matrix .... This gives me the $e_x$ as rotation axis which is not correct.

This is what I get for your matrix. Applying my second method, the matrix is, modulo phase, equivalent to the unimodular (member of $SU(2)$) matrix $\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\i&1\end{array}\right)$ (multiply everything by $-i$). This, by inspection, is the matrix $\exp\left(i\,\frac{\pi}{4}\,\sigma_x\right) = \cos\left(\frac{\pi}{4}\right)\,\mathrm{id}+ i\,\sin\left(\frac{\pi}{4}\right)\,\sigma_x$. So, depending on how you define your rotation axes, it's a rotation of $2\times \frac{\pi}{4}$, i.e. a right angled turn about the axis corresponding to $\sigma_x$. So, unless I have done something dreadfully wrong, I agree with your calculations and there must be a typo somewhere if the answer is meant to be a rotation about $(1,\,1,\,1)$.

For a matrix equivalent (modulo a phase) to $\frac{1}{\sqrt{2}} \begin{pmatrix} 1&-i\\1 &i\end{pmatrix}$, we multiply by $\exp(-\pi\,i/4)$ to get an $SU(2)$ member. Then my second method entails the following calculations (I checked the matrix was unitary first):

So indeed the axis is along $(1,\,1,\,1)$. Alternatively, we can simply calculate the matrix of the the transformation and work out which eigenvector corresponds to the eigenvalue 1; the Frobenius (trace norm) of the matrix logarithm gives us the angle of rotation. So its a rotation of through an angle of $2\pi/3$ about the axis $(1,1,1)$. These are the relevant calculations:

• Thanks for your answer. I tried to find the rotation axis of the matrix $V=\frac{e^{-i3/4 \pi}}{2} \begin{pmatrix} 1&-i\\1 &i\end{pmatrix}$ with the cross product and the vectors $|0\rangle$ and $|+\rangle$ (I neglected an overall phase) on which $|0\rangle$ is mapped. This gives me the $e_x$ as rotation axis which is not correct. The rotation axis is instead $\begin{pmatrix} 1\\1\\1\end{pmatrix}$. Anyway, why is it possible to find a rotation axis for any unitary matrix?It is possible to decompose a unitary matrix in the product of the three Operators x,y,z , so there shoul be 3 axes? Jul 16 '16 at 7:50
• @user9996 See my updates. I agree with your calculations. So I believe there must be a typo in the original problem statement. Jul 16 '16 at 8:40
• Hey, thanks again. I'm not really sure if you have seen that the Matrix in the comment above is different than the Matrix in the first post. I used another axis in the comment, because the rotation axis of the one in my first post was pretty easy to guess Jul 16 '16 at 17:38
• @user9996 Okay, I see I've failed the intelligence test again :), sorry. See my next updates. Jul 17 '16 at 5:55
• ! Thank you very much for your help. I think I figured it out by now :) Jul 21 '16 at 11:52