# It two unitaries are delta apart in trace norm, then what is the trace norm of outputs states when the same input state is applied to two unitaries?

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,

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

• Why would you measure the distance of unitaries in trace norm?? Apr 6, 2020 at 16:00
• 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. Apr 8, 2020 at 4:14
• 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. Apr 8, 2020 at 4:16
• 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. Apr 8, 2020 at 9:08

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.