I could not find any lower bound on the diamond norm for two uniformly random unitaries of dimension D sampled from the haar measure.


2 Answers 2


This answer won't actually give you a bound, but will provide some information that may help you in your search. You may be able to find an answer in the random matrix theory literature if you translate the question into different terms, as I will describe.

First, suppose that $\Phi_0(X) = U_0 X U_0^{\dagger}$ and $\Phi_1(X) = U_1 X U_1^{\dagger}$ are any two unitary channels. In this special case, the diamond norm distance between these two channels can be expressed as follows: $$ \|\Phi_0 - \Phi_1\|_{\diamond} = 2 \sqrt{1 - \delta(U_0^{\dagger} U_1)^2}, $$ where $\delta(A)$ denotes the minimum absolute value taken over the numerical range of a given operator $A$. That is, $$ \delta(A) = \min_{|\psi\rangle} |\langle \psi | A | \psi\rangle|, $$ where the minimum is over all unit vectors $|\psi\rangle$.

Now, if you're interested in $U_0$ and $U_1$ being Haar-random, then you might as well set $U_0 = I$ and $U_1 = U$ for $U$ Haar-random, by the fact that the diamond norm is unitarily invariant. To understand the distribution of the diamond norm distance, you therefore need to understand something about the numerical range of a Haar-random unitary. Because unitary operators are normal, their numerical range is the convex hull of their eigenvalues, so you're essentially trying to understand something about the eigenvalues of a Haar-random unitary. This topic has been studied extensively in random matrix theory.

I do not know enough about this topic to give you precise bounds, but it is clear that as the dimension $D$ grows, the probability that the diamond norm distance equals the maximum possible value 2 (which means the two channels can be discriminated perfectly) approaches 1. Indeed, in order to have diamond norm distance strictly less than 2, all of the eigenvalues of $U$ (or $U_0^{\dagger} U_1$) must fall within an arc on the unit circle having length strictly less than $\pi$, which is extremely unlikely for large $D$ (surely the probability is exponentially small in $D$).


The result you're looking for is effectively Proposition 19 of the paper: Almost all quantum channels are equidistant; which I'm rewriting here for convenience:

Let $U, V \in \mathcal{U}(d)$ be two independent random variables, at least one of them being Haar-distributed. Then, with overwhelming probability as $d \rightarrow \infty$, the quantum channels $\Phi(X)=U X U^{\dagger}$ and $\Psi(X)=V X V^{\dagger}$ become perfectly distinguishable: for $d$ large enough, $$ \mathbb{P}\left[\|\Phi-\Psi\|_{\diamond}=2\right] \geq 1-\exp \left(-\frac{\log 2}{2} d^{2}\right) $$

If you're familiar with typicality results, then this is an example of concentration of measure phenomenon.

Note that this result is general than the one you're looking for since it requires only one of the channels to be Haar. Moreover, it fits nicely with the insights that John Watrous provided above; and the rest of the paper has generalizations to random matrix theory.

  • $\begingroup$ Thank you. The reference has some really nice results. And indeed the unitary invariance property of diamond norm indicates that one needs to choose only one unitary to be haar random while the other can be fixed to show perfect distinguishability with exponentially high probability. Do you know if one can also expect perfect distinguishability when the distinguishability is defined via any other shatten-p norm, instead of diamond norm? $\endgroup$ Jun 3, 2020 at 5:30
  • $\begingroup$ I'd think so, yes. Off the top of my head, some simple bounds on the Schatten 1-norm follow just from the properties relating the 1-norm to the diamond norm (see, for example, Watrous' notes: cs.uwaterloo.ca/~watrous/TQI/TQI.3.pdf; page 171). Moreover, the 1-norm can be bounded using the other p-norms in turn. Would that suffice or are you looking for tighter bounds? $\endgroup$ Jun 8, 2020 at 3:42

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