Closeness between two unitaries on the Bloch sphere

Question

The fidelity between two (single-qubit) quantum states can be easily translated into the euclidean distance between the two states on the Bloch sphere (hilbert-schidmit distance). I'm curious if this concept can be extended to unitaries. If we have two single-qubit unitary matrices, denoted as $U_1$ and $U_2$, is there a way to interpret the fidelity (or any other metric) between these matrices with respect to the Bloch sphere?

For example, consider the case of $U_1 = X$ and $U_2 = I$. $U_1$ transforms $\lvert 0 \rangle$ into $\lvert 1 \rangle$ and $\lvert 1 \rangle$ into $\lvert 0 \rangle$ on the Bloch sphere, whereas $U_2$ leaves each basis state unchanged. Perhaps one approach to assessing the distance between these two unitary matrices is to sum the differences in fidelity between the states evolved by each matrix. In other words, $2 - F(U_1 \lvert 0 \rangle, U_2 \lvert 0 \rangle) + F(U_1 \lvert 1 \rangle, U_2 \lvert 1 \rangle)$, where $F$ represents fidelity. In this example, $F = 2$.

So far, it appears feasible to establish a relationship between the metric of two unitaries in terms of the Bloch sphere. However, if we set $U_1 = Z$, then $2 - F(U_1 \lvert 0 \rangle, U_2 \lvert 0 \rangle) + F(U_1 \lvert 1 \rangle, U_2 \lvert 1 \rangle) = 0$, since $Z$ has no effect on either $\lvert 0 \rangle$ or $\lvert 1 \rangle$.

Hence, my question is whether it's ever possible to think about the closeness (the metric can be arbitrary, including HS distance, unitary fideltiy, etc) of the given two unitaries, $U_1$ and $U_2$, in terms of the Bloch sphere?

(This question is inspired by this recent post: Sufficient conditions for a single-qubit unitary to be the identity)

Answer 1

A unitary $U$ corresponds to a rotation on the Bloch sphere. Technically, there is a smooth map from $2 \times 2$ unitaries with determinant 1 (which form the group SU(2)) to $3\times 3$ rotation matrices in 3D, see here. We then can compare those rotation matrices.

Note that a distance like $||U_1 - U_2||$ is not fully descriptive, because $U$ and $e^{i\theta}U$ correspond to the same rotation. So that $||I - (-I)|| = ||2I||>0$, for example.

As a more natural "Bloch wise" distance one can consider $$ d(U_1,U_2)=\sqrt{2-|\text{Tr}(U_1{}^\dagger U_2)}|. $$

One can show that if $d(U_1,U_2)$ is close to $0$, then the corresponding rotations on the Bloch sphere also should be close enough. And vice versa.

Answer 2

There are several ways to quantify the closeness of unitary operators. The best approach will be heavily dependent on context.

The approach you laid out is close to an operator norm, which would look for the largest difference in outputs when iterating over state space. This would translate to: $$\lVert U_1-U_2\rVert_{op}=\inf\lbrace d\ge 0 : \lVert (U_1-U_2)\vert\psi\rangle\rVert \le d\lVert \vert \psi\rangle\lVert \, \forall \psi \in \Psi\rbrace,$$ where $\Psi$ is the set of all possible states, which maps to the Bloch sphere. The norm used inside the braces is your choice of vector norm. I believe this approach is the closest to giving an indication of distance between $U_1$ and $U_2$ in terms of the Bloch sphere.

The Hilbert-Schmidt inner product, $\langle U_1,U_2\rangle_{HS}=\text{Tr}(U_1{}^\dagger U_2)$, also provides a natural measure of closeness, $$d(U_1,U_2)=\sqrt{2-\langle U_1,U_2\rangle_{HS}}.$$ Assuming $\text{Tr}(U_1)=2\cos\alpha$, this gives $d(U_1,U_1)=0$ and $d(U_1,U_1{}^\dagger)=2\sin\alpha$, which are appropriate. Informally, you could think of this approach as transforming a point at the identity of the $\mathbf{U(2)}$ group manifold by $U_2$ and then by the inverse of $U_1$, and measuring how close the resulting point is to the identity.

The last approach I'll mention is the straightforward Euclidean distance between single qubit unitary operators when embedded in $\mathbb{R}^4$. Take $U_1,\,U_2\in \mathbf{SU(2)}$, which does not impact generality. Elements of $\mathbf{SU(2)}$ take the form $$U = \begin{bmatrix} x_0+ix_3 && x_2+ix_1 \\ -x_2+ix_1 && x_0-ix_3 \end{bmatrix},$$ with $\text{det}(U)=\sum\limits_{i=0}^3 x_i{}^2=1.$ The group manifold of $\mathbf{SU(2)}$ is clearly a 3-sphere which embeds trivially in $\mathbb{R}^4$. Embedded in $\mathbb{R}^4$ in this manner, the Euclidean distance between $U_1$ and $U_2$ is simply $\sqrt{\text{det}(U_1-U_2)}$.