Suppose we are given two unitary matrices $U$ and $V$, with the following guarantee,

$$||U - V||_1 \geqslant \delta$$ for some $\delta \geqslant 0$.

We apply an input density state $\rho$ individually to both $U$ and $V$, resulting in the output state $U\rho U^{\dagger}$ and $V\rho V^{\dagger}$ respectively. Then can we say anything about the trace norm of the output states, i.e,

$$||U\rho U^{\dagger} - V\rho V^{\dagger}||_1 \geqslant \hspace{1mm} ?$$

  • $\begingroup$ Why would you measure the distance of unitaries in trace norm?? $\endgroup$ Apr 6, 2020 at 16:00
  • $\begingroup$ Hi Norbert, I have tried different distance measures including any shatten-p norm and also the diamond norm. However as evident now, even if I measure the distance using any norm, its lower bound does not give me any meaningful lower bound on the output states when the unitaries are queried with same input. $\endgroup$ Apr 8, 2020 at 4:14
  • $\begingroup$ Also I was not interested in maximum single shot distinguishability which diamond norm provides. I only wanted to see if I can say anything about distinguishability of output states when my unitaries are far apart using any distance norm. $\endgroup$ Apr 8, 2020 at 4:16
  • $\begingroup$ It is not surprising that lower bounds don't carry over! All such a lower bound could tell you is that there exists a $\rho$ for which the distance of the outputs is lower bounded. -- Regardless, distances of unitaries are in most cases naturally quantified in operator norm. $\endgroup$ Apr 8, 2020 at 9:08

1 Answer 1


For the way round that you've got your inequalities, I don't think there's much that can be said. To see why, let's consider the first expression $$ \|U-V\|_1=\text{Tr}(\sqrt{2I-VU^\dagger-UV^\dagger}). $$ Now, $VU^\dagger$ is a unitary, and hence as a spectral decomposition. Let the eigenvectors be $|\lambda_i\rangle$ with eigenvalues $e^{i\lambda_i}$. $UV^\dagger$ has the same eigenvectors, with eigenvalues $e^{-i\lambda_i}$. Hence, the expression simplifies to $$ \|U-V\|_1=2\sum_i\left|\sin\frac{\lambda_i}{2}\right|\geq\delta. $$ Note that this doesn't tell us anything about an individual $\lambda_i$ (whereas if the inequality were the other way around, we'd know that $\sin\frac{\lambda_i}{2}\leq\delta$ for all $i$). In particular, there could be an $i$ such that $\lambda_i=0$. Let's call this particular vector $|\Lambda\rangle$.

Next, consider the final calculation you want $$ \|U\rho U^\dagger-V\rho V^\dagger \|=\|\rho-U^\dagger V\rho V^\dagger U\|. $$ It should now be clear that if $\rho=|\Lambda\rangle\langle\Lambda |$, this value is 0. Actually, it is for any eigenvector, or mixture of eigenvectors, of $VU^\dagger$. Since 0 is always the minimum value, there's clearly no non-trivial lower bound.

Having started to think about the follow-up comment, there's an even stronger example of why you can't get anything useful. Let $V=e^{i\theta} U$. Obviously, $\|U\rho U^\dagger-V\rho V^\dagger \|=0$, while $\|U-V\|_1=2N\left|\sin\frac{\theta}{2}\right|$ for $N\times N$ matrices.

  • $\begingroup$ Thanks, this makes complete sense. So let me now modify my problem a bit. Instead of showing a non-trivial lower bound for all states (which is of course not possible by the argument you provided), what if I pick a state uniformly randomly from the haar measure states and now I am interested to see if one can make an argument on the existence of a non-trivial lower bound. Like some sort of average-case argument. That is the output state norm would be bigger than something with overwhelming probability. $\endgroup$ Apr 6, 2020 at 9:11
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    $\begingroup$ I think my edit makes clear that even averaging over the input state doesn't help you. $\endgroup$
    – DaftWullie
    Apr 6, 2020 at 9:29
  • $\begingroup$ Thanks a lot. I am convinced now that this method does not give me any meaningful non trivial bound. $\endgroup$ Apr 8, 2020 at 4:17

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