Is it possible to have a trace fidelity of 1 even if two unitary operations are different?

Question

The gate fidelity of two quantum unitary operations is often described using $\frac{1}{2^n}|\text{tr}(U^\dagger V)|$. Is it ever possible that $U^\dagger V \ne I$ however $\frac{1}{2^n}|\text{tr}(U^\dagger V)| = 1$? Specifically, is it possible to produce a unitary matrix $X=U^\dagger V$, such that $\text{tr}(X) = 2^n$ despite all of the elements along the diagonal not being $1$, i.e., $\sum_i^{2^n} (x_{ii}) = 1$ and $x_{ii} \ne 1$, wher $x_{ii}$ are the diagonal elements of $X$?

Answer 1

If $U$ and $V$ are unitaries, then $U^\dagger V$ is a unitary, which means it has eigenvalues $e^{i\theta_i}$. Thus, $$ \text{Tr}(U^\dagger V)=\sum_ie^{i\theta_i}. $$ There are $2^n$ terms in the sum, all of modulus 1. The only way these could possibly have modulus $2^n$ is if they all line up. i.e. $\theta_i=\theta$, independent of $i$. But if a unitary has all the same eigenvalues, it's just $e^{i\theta}I$. Thus, $$ U^\dagger V=e^{i\theta}I, $$ which also means $$ V=e^{i\theta}U. $$

Answer 2

The only unitary matrices $X$ that satisfy these conditions are those that are proportional to the identity. That, $X= U^\dagger V$ satisfies $|\mathrm{Tr}[X]|=2^n$ if and only if there exists $\theta$ such that $V=\mathrm{e}^{\mathrm{i}\theta}U$.

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Let $\left\{\left|u_i\right\rangle\right\}_i$ be the columns of $U$ and $\left\{\left|v_i\right\rangle\right\}_i$ that of $V$. We have: $$\left|\mathrm{Tr}\left[U^\dagger V\right]\right|=\left|\sum_{i=1}^{2^n}\left\langle u_i\middle|v_i\right\rangle\right|\leqslant\sum_{i=1}^{2^n}\left|\left\langle u_i\middle|v_i\right\rangle\right|\leqslant2^n.$$ Thus, in order for the first term to be equal to $2^n$, the second sum must also be equal to $2^n$. Since the summand of the second sum is lower than or equal to $1$, this sum is equal to $2^n$ if and only if all the terms in the sum are equal to $1$. This can only happen if for all $i$ there exists $\theta_i$ such that: $$\left|v_i\right\rangle=\mathrm{e}^{\mathrm{i}\theta_i}\left|u_i\right\rangle.$$ We then have: $$\left|\mathrm{Tr}\left[U^\dagger V\right]\right|=\left|\sum_{i=1}^{2^n}\mathrm{e}^{\mathrm{i}\theta_i}\right|.$$ This is equal to $2^n$ if and only if all the $\theta_i$ are equal (it's an application of the triangle inequality). All in all, we can have equality if and only if there exists $\theta$ such that $V=\mathrm{e}^{\mathrm{i}\theta}U$, that is, if $U$ and $V$ are equal up to a global phase.