3
$\begingroup$

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.

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Two distinct bases: do you mean, for example, by $\{\lvert 0 \rangle, \lvert 1 \rangle\}$ and $\{\lvert + \rangle, \lvert - \rangle\}$? $\endgroup$
    – Hailey Han
    Sep 19 at 15:53
  • $\begingroup$ Yes, though they don't even have to be orthonormal. $\endgroup$ Sep 19 at 15:56
  • $\begingroup$ 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? $\endgroup$
    – Hailey Han
    Sep 19 at 16:20
  • $\begingroup$ 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. $\endgroup$ Sep 20 at 1:10
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.