# closeness between two unitaries on the bloch sphere

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)

• I guess, without having done a calculation, that it's worth thinking about the product of operations, such as $U_1^\dagger U_2$. You can visualise this action on the Bloch sphere, asking which state changes the most. This, I think, equates to the 2-norm between the two unitaries Sep 29 at 7:07
• @DaftWullie How does that correspond to the 2-norm? Oct 3 at 5:53
• iirc, $\|U_1-U_2\|=\|I-U_1^\dagger U_2\|$ Oct 3 at 6:19
• Right, I assume you're saying it because $\Vert A \Vert_2 = \max_{\Vert \lvert \psi \rangle \Vert_2} \Vert A \lvert \psi \rangle \Vert_2$ (where $A = U_1 - U_2$). However, my question is, does $\Vert A \lvert \psi \rangle \Vert_2$ directly correspond to the vector on the Bloch sphere? I don't think so right? Oct 3 at 16:11
• Not directly, no. Oct 4 at 6:53

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.

• How one should rigorously prove the last statement you made? If $U_1 = X$ and $U_2 = I$, then $\lvert \text{Tr}(U_1^\dagger U_2) \rvert = \lvert \langle 0 \rvert X \lvert 0 \rangle + \langle 1 \rvert X \lvert 1 \rangle\rvert$ (or any other set of basis). However, this is not quite the same as $\lvert \langle 0 \rvert X \lvert 0 \rangle \rvert^2 + \lvert \langle 1 \rvert X \lvert 1 \rangle \rvert^2$, which corresponds to how each of the two states are close on the Bloch sphere. Sep 29 at 16:39
• If $d(U_1, U_2) \approx 0$ then $|\text{Tr}(U_1^\dagger U_2)| \approx 2$, which can be only if $\lambda_1 \approx \lambda_2$ for two eigenvalues of $U_1^\dagger U_2$. This means $U_1^\dagger U_2 \approx \lambda_1 I$, hence $U_2 \approx \lambda_1 U_1$, so they are the same up to some phase. Sep 29 at 17:37
• If $d(U_1, U_2) >> 0$ then the eigenvalues of $U_1^\dagger U_2$ must be different. One can use the eigenvector basis of $U_1^\dagger U_2$ to better see a relation with your expression. Sep 29 at 17:48

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 $$\sqrt{\text{det}(U_1-U_2)}$$.

• It's better to take modulus of inner product in the definition of $d(U_1,U_2)$. Otherwise $I$ and $-I$ appear distant from each other. BTW, in your first approach you also neglect that we should compare states up to a global phase, not just vectors. Sep 29 at 14:10
• What you're suggesting is to consider the closeness of automorphisms associated with $U_1$ and $U_2$, rather than closeness of unitary operators $U_1$ and $U_2$. It's a reasonable thing to do, but it would take a lot of handwaving to do it in a couple paragraphs without conflating $U(2)$ and $\text{Aut}(U(2))$. Sep 29 at 18:07
• Danylo is right. In fact, all three quantities - operator norm $\|U_1-U_2\|_{op}$, $d(U_1, U_2)=\sqrt{2-\langle U_1,U_2\rangle_{HS}}$ and $\sqrt{\det(U_1-U_2)}$ - fail to account for the global phase. Moreover, the second quantity is not a real number. Sep 30 at 3:51
• I think you are both conflating global phase with the $Z_2$ (cyclic group with two elements) center of $SU(2)$. Let's be clear, the operators in question are elements of $U(2)$ and $U(2)=(SU(2)/Z_2)\otimes U(1)$. The tensored $U(1)$ is global phase, the $Z_2$ center of $SU(2)$ is not. Sep 30 at 14:09
• $SU(2)$ has an $SO(3)$ inner automorphism group with $SO(3)=SU(2)/Z_2$, where $Z_2$ is an outer automorphism of $SU(2)$. The topology of $SU(2)$ is a 3-sphere. The topology of $SO(3)=SU(2)/Z2$ is a 2-ball with antipodal points identified. This identification of antipodes is what you rely on when you identify $I$ and $-I$. But when you do so, you're comparing elements of $SO(3)$, not unitary operators. Sep 30 at 15:06