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?

  • 4
    $\begingroup$ 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

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.