Can different density matrices have 100% fidelity with a given pure state?

Question

I am trying to understand fidelity a bit better, to do so consider the bell state: $$|\Psi\rangle=\frac{1}{\sqrt{2}}\left(|01\rangle-|10\rangle\right),$$ the density matrix associated with this state is given by $$\rho=\frac{1}{2}\begin{pmatrix}0 & 0 & 0 & 0\\0 & 1& -1& 0\\0 & -1& 1& 0\\0 & 0 & 0 & 0\end{pmatrix}$$ I have been investigating how a faulty measurement device may effect the measured density matrix of such a state. I find that the reconstructed state is given by

$$\rho=\frac{1}{2}\begin{pmatrix}\epsilon& 0 & -\delta i& 0\\ \delta i& 1-\epsilon& -1& 0\\0 & -1& 1+\epsilon& \delta i\\0 & 0 & -\delta i& -\epsilon\end{pmatrix}$$

This desnity matrix contains very different martix elements to the true input, however the calculated fidelity between the two states is always 100%, regardless of the values of $\epsilon$ and $\delta$. I want to know why density matrices can differe so drastically, yet the state can still be described as 100% fidelity.

Answer 1

If $|\psi\rangle$ is a pure state and $\rho$ is a density matrix, then $F(|\psi\rangle\!\langle\psi|,\rho)=\langle\psi|\rho|\psi\rangle=1$ iff $\rho=|\psi\rangle\!\langle\psi|$. Or more generally, given two density matrices $\rho,\sigma$, you have $F(\rho,\sigma)\equiv\|\sqrt\rho\sqrt\sigma\|_1^2=1$ iff $\rho=\sigma$.

If you drop the assumption of $\rho$ being a density matrix though, this is not necessarily true anymore. For example, keeping positive semidefiniteness but dropping the unit trace, you could have $$\rho \equiv \begin{pmatrix}2&0\\0&0\end{pmatrix}=2|0\rangle\!\langle0|, \qquad |\psi\rangle = \frac1{\sqrt2}(|0\rangle+|1\rangle),$$ and then $\langle\psi|\rho|\psi\rangle=1$ would hold even though $\rho=|\psi\rangle\!\langle\psi|$ doesn't.

As a different example, keeping $\operatorname{tr}(\rho)=1$ but dropping positive semidefiniteness, you could have $$\rho = \begin{pmatrix}2&0\\0&-1\end{pmatrix}, \qquad |\psi\rangle = \sqrt{2/3}|0\rangle+\sqrt{1/3}|1\rangle,$$ and again $\langle\psi|\rho|\psi\rangle=1$ with $\rho\neq |\psi\rangle\!\langle\psi|$.

Answer 2

The question has been answered by the comments and @glS's answer, and this serves as a comment. I was taught before on the following statement:

Suppose $\Omega$ is a Hermitian matrix, $\Omega^2\preceq \mathbb{I}$ and $\langle\psi|\Omega|\psi\rangle = 1$. Then $\Omega$ admits a spectral decomposition $\Omega = |\psi\rangle\langle\psi| + \sum_j a_j|\psi_j^\perp\rangle\langle\psi_j^\perp|$, where $\forall j$ $1\ge |a_j| \ge 0$.

To see this it suffices to show that $\Omega|\psi\rangle = |\psi\rangle$. Since $\langle\psi|\Omega|\psi\rangle = 1$, we have $\Omega|\psi\rangle = |\psi\rangle + c |\psi^\perp\rangle$, where $|\psi^\perp\rangle$ is a state vector orthogonal to $|\psi\rangle$. If $\Omega|\psi\rangle \neq |\psi\rangle$, then $\|\Omega|\psi\rangle\|_2^2 = \langle\psi|\Omega^2|\psi\rangle > 1$, which contradicts with $\Omega^2\preceq \mathbb{I}$.

In OP's case $\Omega$ is replaced with the density matrix $\rho$. As in @glS's answer, the unit trace and positive semidefinite conditions imply that $\forall j$ $a_j = 0$ and $\rho = |\psi\rangle\langle\psi|$. So different density matrices cannot have 100% fidelity with a given pure state.

The proof that $F(\rho,\sigma) = 1$ iff $\rho = \sigma$ for general density matrices is given by the Cauchy-Schwarz inequality: Prove that the quantum state fidelity is upper bounded by 1.