# Is fidelity of mixed $\sigma$ and pure $|\psi\rangle$ equal to $\||\psi\rangle\langle\psi|\sigma\|_1$?

The quantum state fidelity between a pure quantum state $$\rho:= \vert \psi \rangle \langle \psi \vert$$ and a state $$\sigma$$ is \begin{align} F(\rho, \sigma):= {\rm Tr}[\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}]^2 = \langle \psi \vert \sigma\vert \psi \rangle. \end{align} Is $$\langle \psi \vert \sigma\vert \psi \rangle$$ equal to $$\Vert \vert \psi \rangle \langle \psi \vert \sigma \Vert_1$$, where $$\Vert \cdot \Vert_1$$ is the trace norm?

No, in general $$\langle\psi|\sigma|\psi\rangle\ne\||\psi\rangle\langle\psi|\sigma\|_1$$.

## Singular value decomposition

Recall that if $$$$A=\sum_i s_i|a_i\rangle\langle b_i|\tag1$$$$ is the singular value decomposition of operator $$A$$, then $$\|A\|_1=\sum_i s_i$$ where the positive real numbers $$s_i$$ are the singular values of $$A$$. In particular, $$\||a\rangle\langle b|\|_1=1$$ for every two normalized state vectors $$|a\rangle$$ and $$|b\rangle$$.

## Counterexample

Take $$|\psi\rangle=|0\rangle$$ and $$\sigma=|+\rangle\langle+|$$. Then, $$\langle 0|\sigma=\frac{1}{\sqrt2}\langle+|$$, so \begin{align} \||\psi\rangle\langle\psi|\sigma\|_1&=\frac{1}{\sqrt2}\||0\rangle\langle +|\|_1=\frac{1}{\sqrt2}\ne\frac12=\langle\psi|\sigma|\psi\rangle.\tag2 \end{align}

## Inequality

Nevertheless, we can prove that $$$$\||\psi\rangle\langle\psi|\sigma\|_1\geqslant\langle\psi|\sigma|\psi\rangle.\tag3$$$$

If $$\sigma|\psi\rangle=0$$, then the inequality is satisfied. Assume $$\sigma|\psi\rangle\ne 0$$ and let $$s>0$$ to be the L2 norm of $$\sigma|\psi\rangle$$. Also, define $$|\phi\rangle:=\frac{\sigma|\psi\rangle}{s}$$, so that $$\sigma|\psi\rangle=s\cdot|\phi\rangle$$. We have \begin{align} \||\psi\rangle\langle\psi|\sigma\|_1&=s\cdot\||\psi\rangle\langle\phi|\|_1\tag4\\ &=s\tag5\\ &\geqslant s\cdot\langle\psi|\phi\rangle\tag6\\ &=\langle\psi|\sigma|\psi\rangle\tag7. \end{align} This gives some meaning to the two quantities and sheds some light on why they aren't generally equal. Namely, $$\||\psi\rangle\langle\psi|\sigma\|_1$$ is just the L2 norm of $$\sigma|\psi\rangle$$, but the fidelity also depends on the angle between $$|\psi\rangle$$ and $$\sigma|\psi\rangle$$.

• Thanks for correcting me! Sep 10, 2023 at 10:37