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If the expectation value of operator $A$ in any state is real, then $A$ is Hermitian.

there is an uncompleted proof:

$$ \int(c_1\psi_1+c_2\psi_2)^* A (c_1\psi_1+c_2\psi_2)dx$$ $$=|c_1|^2\int\psi_1^* A \psi_1dx + |c_2|^2\int\psi_2^* A \psi_2dx +c_1^*c_2\int\psi_1^* A \psi_2dx +c_2^*c_1\int\psi_2^* A \psi_1dx $$

My teacher said that because $$c_1^*c_2\int\psi_1^* A \psi_2dx +c_2^*c_1\int\psi_2^* A \psi_1dx$$

is real, then we can conclude that $$\int\psi_1^* A \psi_2dx = (\int\psi_2^* A \psi_1dx)^*$$

but I don't understand this last step, can someone explain it, or give another proof?

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Here is another version of the same proof.

If $\langle \psi | A \psi \rangle \in \mathbb R$ for all $\psi \in \cal H$, then $\langle \psi | A \psi \rangle^* = \langle \psi | A \psi \rangle$ for all $\psi \in \cal H$. Since $\langle \psi | \phi \rangle^* = \langle \phi| \psi\rangle$ we have that $\langle \psi | A \psi \rangle = \langle A\psi | \psi \rangle$ that is $^1$ $\langle \psi | A \psi \rangle = \langle \psi | A^\dagger \psi \rangle$, namely: $$\langle \psi | (A-A^\dagger) \psi \rangle =0 \quad \forall \psi \in {\cal H}\:.\tag{1}$$ Now consider $\psi = \phi + \chi$, where $\phi,\chi \in {\cal H}$ are arbitrary, obtaining: $$\langle\phi + \chi| (A-A^\dagger) (\phi + \chi) \rangle =0 \tag{2}\:.$$ using the fact that $\langle \phi | (A-A^\dagger) \phi \rangle = \langle \chi | (A-A^\dagger) \chi \rangle 0$ in view of (1), and expanding (2) taking the real bilinerity of the scalar product into account we find $$\langle \phi | (A-A^\dagger) \chi \rangle + \langle \chi | (A-A^\dagger) \phi \rangle=0\:.\tag{3}$$ The procedure can be implemented another time, starting again from (1) but now using $\psi = \phi + i\chi$. Exploiting the fact that the scalar product is anti linear in the left argument and linear in the right one, this time we end up with $$i\langle \phi | (A-A^\dagger) \chi \rangle -i \langle \chi | (A-A^\dagger) \phi \rangle=0\:,$$ that is $$\langle \phi | (A-A^\dagger) \chi \rangle - \langle \chi | (A-A^\dagger) \phi \rangle=0\:,$$ Together with (3) it entails in particular that $$\langle \chi | (A-A^\dagger) \phi \rangle =0 \quad \forall \phi, \chi \in {\cal H}\:.$$ We are free to choose $\chi = (A-A^\dagger) \phi$ obtaining $$|| (A-A^\dagger) \phi||^2 = \langle (A-A^\dagger) \phi | (A-A^\dagger) \phi \rangle =0 \quad \forall \phi \in {\cal H}\:,$$ which immediately implies $$ (A-A^\dagger) \phi =0 \quad \forall \phi \in {\cal H}\:.$$ Summing up, we have obtained $$A^\dagger \phi = A\phi\quad \forall \phi \in {\cal H}$$ that is the thesis. If the domain of $A$ is not the whole Hilbert space, the above reasoning holds however in that domain as it is a linear subspace by hypotheses.


(1) I just used the defintion of adjoint, $A^\dagger$, of $A$ that implies both $\langle \psi | A \phi \rangle = \langle A^\dagger \psi | \phi \rangle$ and $\langle A\phi | \psi \rangle = \langle \phi | A^\dagger\psi \rangle$.

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    $\begingroup$ Just for completeness (for the reader): the formula to recover the scalar product $\langle\psi,\phi\rangle$ from just norms (of linear combinations of $\psi$ and $\phi$) is called polarization identity. Also, for unbounded operators the procedure proves symmetry, but not self-adjointness. $\endgroup$ – yuggib Sep 24 '14 at 12:02
  • $\begingroup$ Yes, the procedure equally holds on the domain of $A$, $D(A)$, when $D(A)$ is dense so that $A$ admits adjoint $A^\dagger$. In that case one only obtains that $A^\dagger \supset A$, which is a weaker result (symmetry instead of self-adjointness). If the domain of $A$ is the whole $\cal H$, however the proof works as it stands in my answer for self-adjointness and, in this case, $A$ turns out to me bounded too as a consequence (since $A^\dagger$ is closed by definition and the closed graph theorem holds). $\endgroup$ – Valter Moretti Sep 24 '14 at 12:05

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