Distance between two states in a C$^*$-algebra $\def\h{\mathcal H}$ $\def\bh{\mathcal B(\h)}$ $\def\cA{\mathcal A}$
Let $\h$ be a Hilbert space, and $\xi,\eta\in\h$ with $\|\xi\|=\|\eta\|=1$. Then
$$
\phi(A)=\langle A\xi,\xi\rangle,\qquad \psi(A)=\langle A\eta,\eta\rangle
$$
are (pure) states on $\bh$. We are looking for a proof that
$$\tag{$*$}
\|\phi-\psi\|=2\sqrt{1-|\langle\xi,\eta\rangle|^2}.
$$
While $(*)$ is well-known,
I couldn't find a proof in the literature available to me. So I'm posting a proof below.
 A: $\def\h{\mathcal H}$ $\def\bh{\mathcal B(\h)}$ $\def\cA{\mathcal A}$ $\def\tr{\operatorname{Tr}}$ $\def\abajo{\\[0.3cm]}$
We want to calculate $\|\phi-\psi\|$. Let $P,Q\in\bh$ be the orthogonal projections onto the span of $\xi,\eta$ respectively. We have
$$
|\phi(A)-\psi(A)|=|\tr(AP)-\tr(AQ)|=|\tr(A(P-Q))|\leq\|A\|\,\|P-Q\|_1. 
$$
We get equality when $A=V^*$, where $P-Q=V|P-Q|$ is the polar decomposition. Thus
$$
\|\phi-\psi\|=\|P-Q\|_1.
$$
Now let us evaluate $\|P-Q\|_1$.  Since the one-norm is unitarily invariant, we may as well calculate $\|E_{11}-Q'\|_1$, where $Q'=W^*QW$ and $W$ is a unitary such that $W^*PW=E_{11}$. We have
$$
(E_{11}-Q')^2=E_{11}+Q'-2\operatorname{Re} E_{11}Q'. 
$$
Since $Q'$ is a rank-one projection,  $Q'=\nu\otimes\nu$ for some $\nu\in\h$ with $\|\nu\|=1$. Also, $E_{11}-Q'$ has rank at most 2, so it has at most two nonzero eigenvalues, say $\alpha,\beta$. The equality $\tr(E_{11}-Q')=0$ forces $\beta=-\alpha$. As $E_{11}-Q'$ is selfadjoint, the eigenvalues $\lambda_1,\lambda_2$ of $|E_{11}-Q'|$ are the square roots of the eigenvalues of $(E_{11}-Q')^2$; that is,  $\lambda_1=|\alpha^2|^{1/2}=|\alpha|$, $\lambda_2=|\beta^2|^{1/2}=|\alpha|=\lambda_1$. Then
\begin{align*}
\|P-Q\|_1
&=\|E_{11}-Q'\|_1
=\lambda_1+\lambda_2\abajo 
&=\sqrt2\,\big(\lambda_1^2+\lambda_2^2\big)^{1/2}=\sqrt2\,\tr\big((E_{11}-Q')^2\big)^{1/2}\abajo
&=\sqrt2\,\tr(E_{11}+Q'-2\operatorname{Re} E_{11}Q')^{1/2}=\sqrt2\,\big(2-2|\nu_1|^2\big)^{1/2}\abajo
&=2\,\sqrt{1-|\langle e_1,\nu\rangle|^2}
=2\,\sqrt{1-|\langle \xi,\eta\rangle|^2}.
\end{align*}
The last equality comes from $e_1=W^*\xi$, and $\nu=W^*\eta$.
Since states extend from any C$^*$-algebra $\cA\subset\bh$ to $\bh$, the argument shows that $(*)$ survives in general as an inequality. Equality will occur when $\cA$ is big enough to contain $V^*$ or at least an approximation of it.
Depending on the reader's familiarity with the trace, the unitary $W$ (and the use of $E_{11}$ and $Q'$) can be avoided, if one notes that
$$
\tr(PQ)=\tr(\xi\xi^*\eta\eta^*)=\tr(\xi^*\eta\eta^*\xi)=|\eta^*\xi|^2=|\langle\xi,\eta\rangle|^2.
$$
