# Prove a Hermitian operator satisfies a property

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?

• 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. Jun 12, 2023 at 14:29
• 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. Jun 12, 2023 at 15:55

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.