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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.

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  • $\begingroup$ 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. $\endgroup$ Aug 29, 2020 at 19:01

2 Answers 2

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A derivation of this is given in Mark Wilde's book https://arxiv.org/abs/1106.1445 equation 9.173, pages 274-275.

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$\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.

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  • $\begingroup$ Thanks for the answer, but I specifically wanted a reference to something that I could cite. $\endgroup$
    – BHT
    Aug 31, 2020 at 20:45
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    $\begingroup$ @ChronusZed sure, but someone stumbling into the question in the future might be looking for something else $\endgroup$
    – glS
    Aug 31, 2020 at 20:46

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