# Relation between trace distance and inner product between pure states

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.

• This is a 2-line derivation, since you can solve it with qubits - the space spanned by the two states. No need to cite anyone. Any basis-independent function of those two states is parameterized by their overlap, and nothing else. – Norbert Schuch Aug 29 '20 at 19:01

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.