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In many textbooks, it is said that because operators in QM are hermitian, we can write:

$$\int \psi^*\hat{A}\phi\,\mathrm d\tau = \int\phi^*\hat{A}\psi\,\mathrm d\tau$$

An operator $\hat{A}$ is called hermitian iff $\hat{A}^\dagger = \hat{A}$. Using this definition, I tried to prove the above property but I can't. I started from:

$$\langle\psi|\hat{A}|\phi\rangle = {\langle\phi|\hat{A}|\psi\rangle}^\dagger$$

based on the fact that the operator is hermitian. But then I don't know how to continue. Any ideas?

I have found the first relation in the following textbook: Quantum Chemistry and Molecular Interactions, in page 81.

For any two wavefunctions $\psi_i$ and $\psi_j$ $$\int \psi_i^*\hat{H}\psi_j\,\mathrm d\tau = \int\psi_j^*\hat{H}\psi_i\,\mathrm d\tau$$

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    $\begingroup$ Are you very sure about this supposed truth? I've tried for ages to prove it, but can't, and indeed the textbooks I have only state the weaker criterion $$\langle \psi | A | \phi \rangle = \langle \phi | A | \psi \rangle^*,$$ which is fairly easy to show using the property of the inner product $\langle a | b \rangle = \langle b | a \rangle^*$, as well as the Hermiticity of $A$ (I assume that's what you did): $$\langle \psi | A | \phi \rangle = \langle \psi | A\phi \rangle = \langle A\phi | \psi \rangle^* = \langle \phi | A^\dagger | \psi \rangle^* = \langle \phi | A | \psi \rangle^*.$$ $\endgroup$
    – orthocresol ♦
    Nov 19 at 2:37
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    $\begingroup$ Gonna need a citation for "the textbooks". In the integral form it should probably have the complex conjugate too: $$\int \psi A\phi \,\mathrm{d}\tau = \left( \int \phi A\psi \,\mathrm{d}\tau \right)^*.$$ For the adjoint bit, no, that's fine. However, note that the resulting integral is just a number (i.e. $1 \times 1$ matrix), so the complex conjugate $\langle \phi | A | \psi \rangle^*$ is the same thing as the adjoint / complex transpose $\langle \phi | A | \psi \rangle^\dagger$ (and the former is probably more commonly used). $\endgroup$
    – orthocresol ♦
    Nov 19 at 11:07
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    $\begingroup$ @orthocresol is correct. In fact the way I learnt it is that the first relation quoted in his/her comment is the definition of a Hermitian operator (see e.g. sces.phys.utk.edu/~moreo/mm08/hicks.pdf or Atkins and Friedman "Molecular Quantum Mechanics") If textbooks are quoting the relationship in the question ditch them, they are wrong. $\endgroup$
    – Ian Bush
    Nov 19 at 16:36
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    $\begingroup$ Hmm, there is another way of writing the same relation which I've seen a couple of times: $$\langle \psi | A\phi \rangle = \langle A\psi | \phi\rangle,$$ which may make it look as if the complex conjugate is gone, but if you carefully work through that you'll find that it's exactly the same as what I wrote. $\endgroup$
    – orthocresol ♦
    Nov 19 at 16:56
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    $\begingroup$ I have taken a look at the book and the textbook is indeed giving the wrong definition. I would seriously consider using a different book, if it already fails at such a fundamental and basic concept. Physical chemistry books are often not the best sources for basic quantum mechanics. I highly recommend taking a look at physics books that focus on QM. They typically give much clearer definitions and focus on fewer topics with more detail. For example Quantum Mechanics, Vol. 1 by Claude Cohen-Tannoudji, Bernard Diu, Frank Laloe. $\endgroup$
    – Hans Wurst
    Nov 20 at 9:33
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This must be false. As a simple counterexample, if this were true for any Hermitian operator $A$ and any set of basis functions $\phi,\psi$, it must also be true for the identity operator $I$ with $\phi$ a complex function and and $\psi$ a real function. But clearly $$\int\phi^*\psi d\tau\ne\int\psi^*\phi d\tau$$ since only $\phi$ has an imaginary part. As others have stated in the comments, we can only say that $\langle\phi|A|\psi\rangle=\langle\psi|A|\phi\rangle^*$ for Hermitian operators, with the complex conjugation removed if $\phi$ and $\psi$ are known to be real functions and $A$ is a real symmetric operator.

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I dug up this textbook, and I believe that the author is restricting his statement to wavefunctions with no imaginary component, although he does not explicitly state this. He writes on p. 72 that

In our notation, $\Psi$ will often be used to describe the "complete" wavefunction, which may include time-dependent or magnetic terms. Conveniently, we are more often interested only in terms that describe the distribution of the system in space--the spatial wavefunction--for which we will use the lower case $\psi$.

This statement could be interpreted to mean that any use of $\psi$ indicates a real-valued function, and that restriction carries forward to "any two wavefunctions $\psi_i$ and $\psi_j$" on p. 81. Notably, he uses $f(x)$ and $g(x)$ in other cases for generic functions that are not specifically designated as wavefunctions (see for example the definition of linearity of operators which is directly before the quoted definition of Hermitian).

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  • $\begingroup$ Even if the wavefunctions $\psi$ and $\phi$ are purely real, I don't see why that should mean that the integral $\langle \psi | A | \phi \rangle$ is also real for all Hermitian operators $A$, unless I'm missing something(?). The inner product $\langle \psi | \phi \rangle$ will definitely be real, but I'm not sure why any other integral should be. $\endgroup$
    – orthocresol ♦
    Nov 21 at 17:24
  • $\begingroup$ @orthocresol - He also mentions that the expectation value $\left< \psi | \hat{A} |\psi \right>$ is real for any observable, so the set of operators being considered may be restricted to those associated with measurable outcomes. I'm trying to give him the benefit of the doubt that he wasn't completely confused when he wrote the quoted text, but rather just wasn't clear about all the assumptions he was making. But it could be that he just doesn't know what he's talking about. $\endgroup$
    – Andrew
    Nov 21 at 17:40

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