Can I find the axis of rotation for any single-qubit gate?

Question

Suppose I have an arbitrary qiskit $U_3$ gate: $U_3(\theta,\phi,\lambda)$. Is there a way I can find which axis the gate is rotating around? In other words, given any real numbers $\theta,\phi,\lambda$, can I find the vector $\hat n = (n_x,n_y,n_z)$ that the gate corresponds to, so that I can plot the axis of rotation on the Bloch sphere? I'm thinking about the y-z decomposition, but I'm still unable to find out the elements of $\hat n$. How can I figure that out? Thanks a lot for the help:)

Answer 1

You can derive an expression for the rotation axis by combining (1) the decomposition of your unitary $O$ in the Pauli basis, and (2) by the representation of the general rotation operator $R_{\vec n}(\theta)$ in terms of Paulis.

Step (1)

Let $\{\sigma_0, \sigma_x, \sigma_y, \sigma_z \}$ describe the identity matrix ($\sigma_0 = \mathbb 1$) and Pauli matrices, which together form a basis of $U(2)$, which is the space of single-qubit gates. With the trace as dot-product you can write any matrix $O \in U(2)$ in the basis of Pauli matrices as \begin{equation} O = \frac{1}{2} \left(\text{Tr}(O) \sigma_0 + \text{Tr}(O\sigma_x)\sigma_x + \text{Tr}(O\sigma_y)\sigma_y + \text{Tr}(O\sigma_z)\sigma_z \right) \end{equation} This is looks like your typical decomposition of a vector into a basis $$ O = \sum_{\omega \in \{0, x, y, z\}} b_\omega \sigma_\omega, $$ where the basis coefficients are $b_\omega = \text{Tr}(O\sigma_\omega)/2$ and the basis "vector" (or here matrices) are $\sigma_\omega$.

Step (2)

We know that any single-qubit gate can be written as (see e.g. Nielsen & Chuang) $$ O = e^{i\alpha} R_{\vec n}(\theta) $$ where \begin{equation} \begin{aligned} R_{\vec n}(\theta) &= e^{i \theta/2 (n_x \sigma_x + n_y \sigma_y + n_z\sigma_z)} \\ &= \cos\frac{\theta}{2}\sigma_0 - i\sin\frac{\theta}{2}(n_x \sigma_x + n_y \sigma_y + n_z\sigma_z) \end{aligned} \end{equation}

The factor $e^{i\alpha}$ is fixed by the determinant of $O$. Since $R$ is a rotation the determinant is $1$, but $O$ might not have a determinant of $1$. So we know that \begin{equation} \begin{aligned} &\text{det}(O) = \text{det}(e^{i\alpha} R_{\vec n}) = e^{2i\alpha}\det(R_{\vec n}(\theta)) \\ &\Leftrightarrow e^{i\alpha} = \sqrt{\det(O)} \end{aligned} \end{equation}

Combining (1) and (2)

Now that we know $e^{i\alpha}$ we can match the entries of the Bloch vector $\vec n = (n_x, n_y, n_z)$ with the basis coefficients above! I'll leave the math now out because this post is long enough but if you match \begin{equation} \begin{aligned} \frac{1}{2}\text{Tr}(O) &= e^{i\alpha}\cos\frac{\theta}{2}\text{ and } \\ \frac{1}{2}\text{Tr}(O\sigma_\omega) &= -i\sin\frac{\theta}{2} \text{ for } \omega = x, y, z \end{aligned} \end{equation}

You finally obtain \begin{equation} \begin{aligned} \theta = 2\cos^{-1}\left(e^{-i\alpha}\frac{\text{Tr}(O)}{2}\right) \end{aligned} \end{equation} and \begin{equation} \begin{aligned} n_\omega = \frac{e^{-i\alpha}\text{Tr}(O\sigma_\omega)}{-2i \sin(\theta/2)} \end{aligned} \end{equation} which you could possibly simplify further by plugging in the expression for $\theta$, but, as many textbooks say, I'll leave that exercise for the motivated reader. :)

If you do this for e.g. $$ S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} $$ you obtain $$ \alpha = \frac{\pi}{4}, \theta = \frac{\pi}{2} $$ and $$ \vec n = (0, 0, 1). $$

Answer 2

1. A generic $2\times2$ (special) unitary matrix decomposes in terms of Pauli matrices as $$U = a_0 I + i \sum_{j=1}^3 a_j \sigma_j,\tag U$$ for $a_j\in\mathbb R$ such that $\sum_{j=0}^3 a_j^2=1$. One way to write this condition is to parametrise the coefficients as $$a_0 = \cos(\theta), \qquad a_j = \sin(\theta) n_j$$ for any $\theta\in\mathbb R$ and $(n_1,n_2,n_3)\in\mathbb R^3$ such that $\|\vec n\|=1$. To be more explicit, this corresponds to the matrix $$U = \begin{pmatrix} a_0 + ia_3 & i(a_1-ia_2) \\ i(a_1+ia_2) & a_0 - ia_3\\ \end{pmatrix} = \begin{pmatrix} c+i sn_3 & is(n_1-in_2) \\ is(n_1+in_2) & c-is n_3 \end{pmatrix},$$ where $c\equiv \cos(\theta), s\equiv \sin(\theta)$. The eigenvectors of such a matrix are $$\frac{1}{\sqrt{2\|\vec a\|(\|\vec a\|\mp a_3)}}\begin{pmatrix}a_3 \mp \|\vec a\| \\ a_1 + i a_2\end{pmatrix},\tag1$$ where $\|\vec a\|^2\equiv \sum_{j=1}^3 a_j^2$.

2. Given an arbitrary complex vector $(\alpha,\beta)\in\mathbb C^2$, you can get the corresponding vector in the Bloch sphere via the mapping (more precisely, this is a bijection $\mathbb{CP}^1\to S^2$): $$\begin{pmatrix}\alpha\\\beta\end{pmatrix} \longleftrightarrow \begin{pmatrix}2\operatorname{Re}(\bar\alpha\beta) \\ 2\operatorname{Im}(\bar\alpha\beta) \\ |\alpha|^2-|\beta|^2\end{pmatrix}.\tag2$$

Putting these facts together you can get the Bloch representation of the eigenvectors of a generic $2\times2$ special unitary matrix. More explicitly, using (1) and (2) together, we see that the Bloch representations of the two eigenvectors of the $U$ defined in $(\mathrm U)$ are precisely the (normalised versions of the) vectors $\pm(a_1,a_2,a_3)$, as one might have expected from the beginning.

Answer 3

Thanks all for reading and answering the question, just a correction for the mapping: $$ \begin{pmatrix}\alpha\\\beta\end{pmatrix} \longleftrightarrow \begin{pmatrix}2\operatorname{Re}(\bar\alpha\beta) \\ 2\operatorname{Im}(\bar\alpha\beta)\\|\alpha|^2-|\beta|^2\end{pmatrix}. $$ This could be derived from the spherical-coordinate representation of $\hat n$ (note that $n_z=\cos\theta$) and the single-qubit representation:)