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

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.

• That's not a density matrix in the last equation. I presume that $-\delta i$ in the top row is in the wrong position, which is probably a typo — but the real issue is that the matrix isn't positive semidefinite. Commented May 9 at 14:26
• Ah right that makes sense! This kind of error is due to measuring two different states w two different detectors, so it shouldnt produce a true density matrix. My question is then: Is the condition that $\mathcal{F}(\hat{\sigma}\hat{\rho})=1$ if and only if $\hat{\sigma}=\hat{\rho}$ only true for $(\hat{\sigma},\hat{\rho})$ that are well defined density matrices (i.e unit trace, positive semidefinite and Hermitian)? Commented May 10 at 9:51
• (And yes that was a typo) Commented May 10 at 9:52
• You would first have to define $\mathcal{F}$ for indefinite matrices (positive and negative eigenvalues), since you're taking the square root of a matrix... Commented May 10 at 13:25
• And yes, the fidelity between two density matrices is 1 if and only if they are equal. Commented May 10 at 17:45

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 $$\rho\neq |\psi\rangle\!\langle\psi|$$.

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.