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

Question

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.

Answer 1

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.

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

Answer 2

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.