Let $|\phi\rangle,|\psi\rangle$ be two state vectors, and let $d=\frac{1}{2}\mathrm{Tr}(\sqrt{(|\phi\rangle\langle\phi|-|\psi\rangle\langle\psi|)^2})$ be their trace distance. Then it will always hold that $d = \sqrt{1-|\langle\phi|\psi\rangle|^2}$.
I'm looking for a reference for this result. A textbook reference would be optimal, but any papers mentioning the result would be fine as well.
A derivation of this is given in Mark Wilde's book https://arxiv.org/abs/1106.1445 equation 9.173, pages 274-275.
$\newcommand{\bra}[1]{\langle #1\rvert}\newcommand{\braket}[2]{\langle #1\rvert #2\rangle}\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\on}[1]{\operatorname{#1}}\newcommand{\ketbra}[2]{\lvert #1\rangle\!\langle #2\rvert} $Define $A\equiv \ketbra\phi\phi - \ketbra\psi\psi$. Being $A$ hermitian, $\on{Tr}(\sqrt{A^2})$ equals the sum of its singular values. Being $A$ spanned by only two vectors, its singular vectors must be of the form $\alpha\ket\phi+\beta\ket\psi$. We then see that $$A(\alpha\ket\phi+\beta\ket\psi)= (\alpha +\beta \braket{\phi}{\psi})\ket\phi - (\alpha \braket{\psi}{\phi} +\beta)\ket\psi, $$ and the expectation value thus reads $$ \langle A\rangle\equiv (\alpha^*\bra\phi + \beta^* \bra\psi)A(\alpha\ket\phi+\beta\ket\psi) = (|\alpha|^2 - |\beta|^2) (1 - |\braket\phi\psi|^2).\tag1 $$ Now remember that the largest singular value is the largest value of $|\langle \Psi|A|\Psi\rangle|$ over all unit vectors $\ket\Psi$. From (1), we easily see that this equals $1-|\braket\phi\psi|^2$.
From $\on{Tr}(A)=0$ we know that the eigenvalues of $A$ are $\pm\sqrt{-\det(A)}$, and thus its singular values are equal. We conclude that $\on{Tr}(\sqrt{A^2})=2(1-|\braket\phi\psi|^2)$.
Of course, the more standard argument passing through the matrix representation of $A$ in a basis $\{\ket\phi,\ket{\phi_\perp}\}$ works equally well.
