We show that for flip operator $F$, and the family of maximally entangled states \(\operatorname{MAXE}=\{|{\Phi}\rangle||{\Phi}\rangle=\sum|{j_Aj_B}\rangle/\sqrt{d},\{|j\rangle\}\text{ is ONB}\},\) the range of the overlap between $F$ and $|{\Phi}\rangle$ depends on the parity of the dimension. Specifically, \(\{\langle{\Phi|F|\Phi}\rangle||{\Phi}\rangle\in\operatorname{MAXE}\}=\begin{cases}[-1,1]&d=2k\\ [-\frac{d-2}{d},1]&d=2k+1\end{cases}\)

This is odd at the first sight, because it is hard to imagine that this has something to do with the parity of the dimension. However the eigenspace of $F$ may be the reason: when $d$ is even the $+1$ and $-1$ eigenspaces of $F$ are of the same dimension, while in the odd case $+1$ eigenspace is one-dimensional larger. Then maybe with local unitary one cannot send $|{\Phi_+}\rangle$ to the $-1$ eigenspace completely.

By direct computation one can have \(\begin{aligned} \langle{\Phi|F|\Phi}\rangle=&\langle{\Phi_+|U^\dagger\otimes IFU\otimes I|\Phi_+}\rangle\\ =&\langle{\Phi_+|U^\dagger\otimes II\otimes U|\Phi_+}\rangle\\ =&\langle{\Phi_+|U^\dagger U^{\intercal}\otimes I|\Phi_+}\rangle=\operatorname{Tr}[U\overline{U}]/d.\end{aligned}\)

It is clear that $\operatorname{Tr}[U\overline{U}]$ is real, and $U\overline{U}$ again is unitary. So $\operatorname{Tr}[U\overline{U}]\in[-d,d]$. By taking $U=I$ one gets upperbound $d$, so the question is that whether the lowerbound $-d$ can be obtained.

For even $d$, it is known that there exist orthogonal matrix $M_d$ such that $M_d^2=-I$, by taking anti-diagonal matrix with entries $\pm1$. So $\operatorname{Tr}[U\overline{U}]$ can be $-d$ for even $d$. But for odd $d$, we can prove that

For any unitary $U$, the spectral of $U\overline{U}$ is conjugate-closed: if $\lambda$ is an eigenvalue of $U\overline{U}$, then $\overline{\lambda}$ is also an eigenvalue of $U\overline{U}$.

This is because the coefficients of the characterization polynomial of $A$ is in polynomial of $\operatorname{Tr}[A^{j}]$, which is real for any $A=U\overline{U}$. Then the characterization polynomial has real coefficients, so eigenvalues come in pair.

This lemma implies nothing in the even case, but for odd $d$, it immediately implies that one of the eigenvalues of $U\overline{U}$ is $\pm1$. And note that $\det(U\overline{U})=\det(U)\overline{\det(U)}$ is positive, that eigenvalue must be $1$. So $U\overline{U}=1\oplus W$ for some $d-1$ dimensional unitary $W$, which has trace at least $-(d-1)$. So $\operatorname{Tr}[U\overline{U}]=1+\operatorname{Tr}[W]\ge-d+2$. This new bound can be obtained simply by taking $U=1\oplus O$ where $O^2=-I$ is a $d-1$ dimensional orthogonal matrix.

Acknowledgement

We thank Kean Chen and Jurij Volcic for valuable discussion.