# 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

$$\newcommand{\bra}{\langle #1\rvert}\newcommand{\braket}{\langle #1\rvert #2\rangle}\newcommand{\ket}{\lvert #1\rangle}\newcommand{\on}{\operatorname{#1}}\newcommand{\ketbra}{\lvert #1\rangle\!\langle #2\rvert}$$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.