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?


1 Answer 1


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

Singular value decomposition

Recall that if \begin{equation} A=\sum_i s_i|a_i\rangle\langle b_i|\tag1 \end{equation} 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 $|b\rangle$.


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}


Nevertheless, we can prove that \begin{equation} \||\psi\rangle\langle\psi|\sigma\|_1\geqslant\langle\psi|\sigma|\psi\rangle.\tag3 \end{equation}

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 $\sigma|\psi\rangle$.

  • $\begingroup$ Thanks for correcting me! $\endgroup$
    – Tristan Nemoz
    Sep 10, 2023 at 10:37

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