Let us consider a unitary $U$ parameterised by $\theta \in \mathbb{R}$, i.e, $U(\theta)$. What are the necessary and sufficient conditions for the output states of this unitary to be smooth? One school of thought may be to consider the following version of the Hölder Inequality, i.e, $$\text{Tr}\left[(\rho - \sigma)U \right] \leq || \rho - \sigma ||_1 || U ||_{\infty}.$$ Where, the interpretation is that states that are close to each other cannot differ significantly in their expectation value. I am unsure if this is indeed true. In particular, does this address cases where $\rho$ and $\sigma$ are orthogonal?

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    $\begingroup$ This should hold, although you may need an appropriate limit and you will need an absolute value on the left-hand side: mathoverflow.net/a/248865 $\endgroup$ Sep 20, 2023 at 13:13
  • $\begingroup$ I don't really see why the statement would help the question posed but maybe I'm misunderstanding the question. If you let $\rho_{\theta} = U(\theta) \rho_0 U(\theta)^\dagger$ are you asking that $\rho_{\theta}$ is a smooth path through the set of density operators? Can you write precisely what it is you want to prove? $\endgroup$
    – Rammus
    Sep 20, 2023 at 17:04
  • $\begingroup$ @Rammus I don't know what OP is trying to do but my guess is to make some inequality for all states $\rho$ and $U\rho U^\dagger$ that non-smooth choices of $U$ would lead to violations of the inequality for the state and its evolved version $\endgroup$ Sep 20, 2023 at 17:13
  • $\begingroup$ @Rammus what I am trying to show that the I/O relationship of quantum circuit that executes the operation $U(\theta)$, is immune to the presence of noise in the input state. Concretely what I'm looking at has to do with how quantum neural networks can avoid overfitting the noise in the input. $\endgroup$ Sep 20, 2023 at 21:37

1 Answer 1


I'm still not sure I fully understand what you're asking but here's my take.

Let $\rho$ be the input to my circuit, at the end of the circuit I receive an output $\rho_U = U \rho U^\dagger$. Now suppose instead, by error, I input $\sigma$ to the circuit, which results in an output $\sigma_U =U \sigma U^\dagger$. I wan't to establish that as long as $\rho$ and $\sigma$ were close then $\rho_U$ and $\sigma_U$ will also be close.

This is true as $$ \|\rho_U - \sigma_U\|_1 = \|U\rho U^\dagger - U \sigma U^\dagger\|_1 = \|U(\rho-\sigma)U^\dagger\|_1 = \|\rho-\sigma\|_1 $$ where the last equality follows from the fact that $\|U X U^\dagger\|_1 = \|X\|_1$ (trace norm is unitarily invariant). So overall if the initial states and close then the end states will also be close.


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