Can I find the axis of rotation for any single-qubit $U_3$ 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:)

Cryoris · Accepted Answer

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}
begin{aligned}
O &= frac{1}{2} left(text{Tr}(Osigma_0) sigma_0 + text{Tr}(Osigma_x)sigma_x + text{Tr}(Osigma_y)sigma_y + text{Tr}(Osigma_z)sigma_z right) 
 &= frac{1}{2} left(text{Tr}(O) sigma_0 + text{Tr}(Osigma_x)sigma_x + text{Tr}(Osigma_y)sigma_y + text{Tr}(Osigma_z)sigma_z right) 
end{aligned}
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 $b_omega = text{Tr}(Osigma_omega)/2)$ 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)
$$
O = e^{ialpha} R_{vec n}(theta)
$$
where
begin{equation}
begin{aligned}
R_{vec n}(theta) &= e^{i theta/2 (n_x sigma_x + n_y sigma_y + n_zsigma_z)} 
&= cosfrac{theta}{2}sigma_0 - isinfrac{theta}{2}(n_x sigma_x + n_y sigma_y + n_zsigma_z)
end{aligned}
end{equation}
The factor $e^{ialpha}$ 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^{ialpha} R_{vec n}) = e^{2ialpha}det(R_{vec n}(theta)) 
&Leftrightarrow e^{ialpha} = sqrt{det(O)}
end{aligned}
end{equation}
Combining (1) and (2)
Now that we know $e^{ialpha}$  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^{ialpha}cosfrac{theta}{2}text{ and } 
frac{1}{2}text{Tr}(Osigma_omega) &= -isinfrac{theta}{2} text{ for } omega = x, y, z
end{aligned}
end{equation}
You finally obtain
begin{equation}
begin{aligned}
theta = 2cos^{-1}left(e^{-ialpha}frac{text{Tr}(O)}{2}right)
end{aligned}
end{equation}
and
begin{equation}
begin{aligned}
n_omega = frac{e^{-ialpha}text{Tr}(Osigma_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 to 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
$$
vec n = (0, 0, 1).
$$

glS · Answer

A generic $2times2$ (special) unitary matrix decomposes in terms of Pauli matrices as
$$U = a_0 I + i sum_{j=1}^3 a_j sigma_j,$$
for $a_jinmathbb R$ 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 $thetainmathbb R$ and $(n_1,n_2,n_3)inmathbb R^3$ such that $|vec n|=1$.
The eigenvectors of such a matrix are
$$frac{1}{sqrt{2|vec a|(|vec a|mp a_3)}}begin{pmatrix}a_3 mp |vec a|  a_1 + i a_2end{pmatrix},$$
where $|vec a|^2equiv sum_{j=1}^3 a_j^2$.
Given an arbitrary complex vector $(alpha,beta)inmathbb C^2$, you can get the corresponding vector in the Bloch sphere via the mapping
$$begin{pmatrix}alphabetaend{pmatrix}
longleftrightarrow
begin{pmatrix}|alpha|^2-|beta|^22operatorname{Re}(baralphabeta)  2operatorname{Im}(baralphabeta)end{pmatrix}.$$
Putting these facts together you can get the Bloch representation of the eigenvectors of a generic $2times2$ special unitary matrix.

ZR- · Answer

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

Can I find the axis of rotation for any single-qubit $U_3$ gate?

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