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Assuming $A$ as hermitian operator, then $\langle A \phi|\phi \rangle= \langle \phi|A\phi \rangle$ holds. I need to show that this is equivalent to $\langle A\psi|\psi' \rangle = \langle \psi|A\psi' \rangle$ for $\psi \neq \psi'.$ We may set $\phi = \psi + \psi'$ and get $\langle A(\psi + \psi')|\psi+\psi' \rangle = \langle \psi+\psi'|A(\psi + \psi') \rangle.$ Given the linearity of $A,$ we get: $$\langle A(\psi + \psi')|\psi+\psi' \rangle = \langle A\psi|\psi \rangle + \langle A\psi | \psi' \rangle + \langle A \psi'|\psi \rangle + \langle A\psi'|\psi' \rangle.$$ In the first and fourth term we may use the assumption. I do not know how to go from here. Can somebody provide some hint or a solution proposal?

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    $\begingroup$ Search for polarization identity. This allows you to conclude that $A$ is hermitian iff $\langle \phi, A\phi\rangle \in \mathbb R$ for all $\phi \in H$. Put differently, try now to use $\phi=\psi+i\psi^\prime$; combining both results should do the job. $\endgroup$ Jun 12, 2023 at 14:29
  • $\begingroup$ Many thanks. I substituted $\phi = \psi + i\psi'$ in the equation defining a hermitian operator and just like in the case $\phi = \psi + \psi'$ I get terms that I do not see how to assemble together using the polarization identity. $\endgroup$
    – user996159
    Jun 12, 2023 at 15:55

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Following the attempt of OP, we also compute $$\langle \psi+i \psi^\prime,A(\psi+i\psi^\prime)\rangle =\langle \psi,A\psi\rangle +\langle \psi^\prime,A\psi^\prime\rangle + i\left(\langle \psi,A\psi^\prime\rangle - \langle \psi^\prime,A\psi\rangle\right ) \quad . $$

To conclude the argument, we note that in both cases $\langle \phi,A\phi\rangle \in \mathbb R$ for all $\phi \in H$ by assumption. This gives the conditions $$\mathrm{Re} \langle \psi,A\psi^\prime\rangle = \mathrm{Re} \langle \psi^\prime,A\psi\rangle $$

and $$\mathrm{Im} \langle \psi,A\psi^\prime\rangle = - \mathrm{Im} \langle \psi^\prime,A\psi\rangle \quad ,$$

which in turn imply $$\langle \psi,A\psi^\prime\rangle = \langle \psi^\prime, A\psi\rangle^* = \langle A\psi,\psi^\prime\rangle \quad .$$

The equality has to hold for all $\psi,\psi^\prime \in H$, where $H$ is a finite-dimensional complex Hilbert space, which then shows that $A$ is hermitian. The converse direction is trivial, i.e. if $A$ is hermitian, then all expectation values are real.

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