# Sufficient conditions for a single-qubit unitary to be the identity

Say I have a unitary $$U = e^{-iHt}$$ where $$H = \alpha X + Z$$.

First, suppose $$U = I$$. Then it rotates a set of initial states to themselves. Say I'm working on a computational basis, then on the Bloch sphere $$U$$ will rotate $$\lvert 0 \rangle$$ and $$\lvert 1 \rangle$$ to $$\lvert 0 \rangle$$ and $$\lvert 1 \rangle$$, respectively.

Now, consider $$\alpha = 0$$ so that $$U = e^{-iZt}$$. For any $$t$$ such that $$U \neq I$$, it physically does the same thing on the Bloch sphere: $$U$$ will rotate $$\lvert 0 \rangle$$ and $$\lvert 1 \rangle$$ to $$\lvert 0 \rangle$$ and $$\lvert 1 \rangle$$, respectively. But definitely, $$U$$ isn't the identity gate.

If someone gives me a description of $$U$$ such that "$$U$$ rotates each basis to itself", then as mentioned above $$U$$ is not necessarily $$I$$. So there should be something more than this description to characterize that $$U = I$$. What are these additional descriptions that I'm missing in terms of the Bloch sphere? It seems like the answer should be related to the relative phase between $$U\lvert 0 \rangle$$ and $$U\lvert 1 \rangle$$, but can we describe this phase on the Bloch sphere? Any insightful answer would be much appreciated.

## 2 Answers

TL;DR: The premise - that $$U$$ which fixes elements of every basis may fail to be the identity - is false.

If a unitary $$U$$ fixes elements of every basis then in particular it fixes every vector and $$U=I$$. In fact, if a single-qubit unitary $$U$$ fixes elements of just two distinct$$^1$$ bases then $$U=I$$. This is a consequence of the fact that two distinct bases supply four points on the Bloch sphere three of which are necessarily distinct and a rotation of the 2-sphere that fixes three distinct points is necessarily the identity. The last statement follows from the fact that a rotation in $$\mathbb{R}^3$$ is either the identity or has exactly one eigenvector, known as the axis of rotation.

The alleged counterexample in the question - a non-trivial $$Z$$ rotation - fails because it fixes the computational basis and no other basis.

$$^1$$ Where we ignore the global phase when comparing basis elements.

• Two distinct bases: do you mean, for example, by $\{\lvert 0 \rangle, \lvert 1 \rangle\}$ and $\{\lvert + \rangle, \lvert - \rangle\}$? Sep 19 at 15:53
• Yes, though they don't even have to be orthonormal. Sep 19 at 15:56
• Thanks! Could you say a few more words about your last point on why the rotation of the 2-sphere that fixes three distinct points is necessarily the identity? Sep 19 at 16:20
• There are a few ways to see this. Spin a sphere and count the points that don't move. Alternatively, write the equation $Rx=x$ where $R$ is a rotation matrix, solve it and count the solutions. Sep 20 at 1:10

Another way of phrasing it is "if $$U$$ maps every computational basis state to itself, such that all the computational basis states have the same global phase, then $$U$$ is identity, up to a global phase". This is because, although it looks like a global phase when acting on the computational basis states, it's actually a relative phase when acting on superpositions, and we need that the relative phase doesn't change in order to implement identity.