# Trace distance between mixed state and pure state vs trace distance between their purifications

Let $$\rho$$ be a mixed state and $$\vert\psi\rangle\langle\psi\vert$$ be a pure state on some Hilbert space $$H_A$$ such that

$$\|\rho - \vert\psi\rangle\langle\psi\vert \|_1 \leq \varepsilon,$$

where $$\|A\|_1 = \text{Tr}\sqrt{A^\dagger A}$$. Does there exist a purification $$\vert\phi\rangle\langle\phi\vert$$ of $$\rho$$ and a (trivial) purification $$\vert\psi\rangle\langle\psi\vert\otimes\vert 0\rangle\langle 0\vert$$ living on Hilbert space $$H_A\otimes H_B$$ such that

$$\|\left(\vert\phi\rangle\langle\phi\vert - \vert\psi\rangle\langle\psi\vert\otimes\vert 0\rangle\langle 0\vert\right) \|_1 \leq \varepsilon? \tag{1}$$

I am aware that the fidelity between purifications can be bounded using Uhlmann's theorem. One can then use the Fuchs van de Graaf inequalities and come back to trace distance but the resulting bound in (1) goes as $$\sqrt{\varepsilon}$$. Can one do better and achieve (1)?

Sanity check: the statement is indeed true when $$\rho$$ is a pure state.
We can start by finding the singular values of the combination of purified systems, which I will write as $$|\phi\rangle$$ and $$|\Psi\rangle$$. Given the Hermitian matrix $$M=|\phi\rangle\langle\phi|-|\Psi\rangle\langle\Psi|$$, we can look for eigenvectors of the form $$\alpha|\phi\rangle+\beta|\Psi\rangle$$ (this is much easier when we have a sum of projectors, not a difference like we have here). Defining $$z=\langle\phi|\Psi\rangle$$, the eigenvalue equation reads \begin{align} M\left(\alpha|\phi\rangle+\beta|\Psi\rangle\right)&=|\phi\rangle\left(\alpha+\beta z\right)+|\Psi\rangle\left(-\beta-\alpha z^*\right)\\ &=\lambda\left(\alpha|\phi\rangle+\beta|\Psi\rangle\right). \end{align} The eigenvalues are given by $$\lambda=\frac{\alpha+\beta z}{\alpha}=-\frac{\beta+\alpha z^*}{\beta},$$ which means that the eigenvectors must obey $$\alpha=\beta\frac{-1\pm\sqrt{1-|z|^2}}{z^*};$$ we thus have the unnormalized eigenvectors and eigenvalues $$|v_\pm\rangle=\left(-1\pm\sqrt{1-|z|^2}\right)|\phi\rangle+z^*|\Psi\rangle,\qquad \lambda_\pm=\mp\sqrt{1-|z|^2}.$$ The entire matrix $$M$$ thus has two degenerate singular values $$\sigma_1=\sigma_2=\sqrt{1-\left|\langle\phi|\Psi\rangle\right|^2}.$$ This means that we can write your purification's trace distance as $$\big|\big||\phi\rangle\langle\phi|-|\Psi\rangle\langle\Psi|\big|\big|_1=2\sqrt{1-\left|\langle\phi|\Psi\rangle\right|^2}.$$ To achieve the smallest value of $$\big|\big||\phi\rangle\langle\phi|-|\Psi\rangle\langle\Psi|\big|\big|_1$$, we require the largest value of $$\left|\langle\phi|\Psi\rangle\right|^2$$.
Now, this expression for the singular values uses the overlap between two purified states, so we can connect it to the fidelity! Labelling all viable purifications of $$\rho$$ by $$|\phi_\rho\rangle$$, we know that $$F(\rho,\psi)=\max_{|\phi_\rho\rangle}\left|\langle\phi_\rho|\Psi\rangle\right|^2,$$ where we have fixed our purification $$|\Psi\rangle$$ of $$|\psi\rangle$$ without loss of generality. So we can collect our results and write $$\min_{|\phi_\rho\rangle}\big|\big||\phi_\rho\rangle\langle\phi_\rho|-|\Psi\rangle\langle\Psi|\big|\big|_1=2\sqrt{1-F(\rho,\psi)}\stackrel{?}{\leq}\epsilon.$$
Our question is now how to relate $$F(\rho,\psi)$$ and $$\epsilon$$. We know that $$1-F(\rho,\psi)\leq {\epsilon}/2$$, so the question is by how much can it be smaller. We can always write this fidelity as $$F(\rho,\psi)=\langle \psi|\rho|\psi\rangle.$$ Choosing some fiducial state $$\rho=p|\psi\rangle\langle\psi|+(1-p)|\psi_\perp\rangle\langle\psi_\perp|$$ with orthonormal $$|\psi\rangle$$ and $$|\psi_\perp\rangle$$, we can calculate (2 times) the trace distance $$\big|\big|\rho-|\psi\rangle\langle\psi|\big|\big|_1=2\left(1-p\right)$$ and the fidelity $$F(\rho,\psi)=p.$$ Since $$2\sqrt{1-p}\geq 2\left(1-p\right),$$ we have a counterexample where $$\min_{|\phi_\rho\rangle}\big|\big||\phi_\rho\rangle\langle\phi_\rho|-|\Psi\rangle\langle\Psi|\big|\big|_1 > \big|\big|\rho-|\psi\rangle\langle\psi|\big|\big|_1,$$ so it does not seem like the desired property always holds for all $$\epsilon$$. In fact, it seems as though this counterexample saturates the $$\sqrt{\epsilon}$$ condition that we have been trying to beat.