In my lecture the professor said that the mean value of a physical quantity- since it must be real- must satisfy the following condition: $$\begin{align} \left<F\right>=\left<F\right>^* \end{align} $$

$$\begin{align} \int\psi^*(\hat F\psi)\,\mathrm dr=\int\psi(\hat F\psi)^*\,\mathrm dr.....(\alpha)& \end{align} $$

With this, he's definiting a hermitian operator, as the operator $\hat F$ that makes the previous condition true.

Then he used it in the following passage:

$$\begin{align} \int\psi^*(\hat F-\left<F\right>)^2\psi\,\mathrm dr=\int\psi^*(\hat F-\left<F\right>)(\hat F-\left<F\right>)\psi\,\mathrm dr=\int[(\hat F-\left<F\right>)\psi]^*[(\hat F-\left<F\right>)\psi]\,\mathrm dr& \end{align} $$ I really don't see why the last step is true. The professor said he's only applying the definition of hermitian conjugate given in $(\alpha)$ to the part of the operator(that is to $(\hat F-\left<F\right>)$) but I still don't see it.

This is the only definition of hermitian operator we have seen, so please try to explain it without using matrices or brackets or any other notation, or without seeing the operator as a matrix, I know some about this stuff, but I would like to understand it the way he's doing it.

Any help would be really appreciated.

  • $\begingroup$ It's the identity $(AB)^*= B^* A^*$ if star represents the hermitian conjugate $\endgroup$
    – Triatticus
    Feb 19, 2020 at 23:20
  • $\begingroup$ but this holds for matrices, what if I am only seeing operators as linear combination of derivatives or multiplications and the wave function as a complex scalar function? $\endgroup$
    – J.C.VegaO
    Feb 19, 2020 at 23:27
  • $\begingroup$ indeed you can write matrix elements for the derivative operator and the multiplication-by-$x$ operator.. $\endgroup$ Feb 20, 2020 at 5:24
  • $\begingroup$ The thing is $\psi (x)$ is just shorthand for things like $\langle x | \psi \rangle$ using this and the completeness relation it doesn't matter if you have scalar functions or vectors $\endgroup$
    – Triatticus
    Feb 24, 2020 at 21:36

2 Answers 2


Since $$\begin{align} \int\psi^*(\hat F\psi)\,\mathrm dr=\int\psi(\hat F\psi)^*\,\mathrm dr.....(\alpha)& \end{align} $$ is valid for every $\psi$, replacing $\psi$ for $\psi+\phi$ we obtain $$ \int\psi^*(\hat F\psi)\,\mathrm dr+ \int\phi^*(\hat F\phi)\,\mathrm dr+ \int\phi^*(\hat F\psi)\,\mathrm dr + \int\psi^*(\hat F\phi)\,\mathrm dr$$ $$=\int\psi(\hat F\psi)^*\,\mathrm dr+ \int\phi(\hat F\phi)^*\,\mathrm dr+\int\psi(\hat F\phi)^*\,\mathrm dr +\int\phi(\hat F\psi)^*\,\mathrm dr$$ Using ($\alpha$) again, we find $$ \int\phi^*(\hat F\psi)\,\mathrm dr + \int\psi^*(\hat F\phi)\,\mathrm dr$$ $$=\int\psi(\hat F\phi)^*\,\mathrm dr +\int\phi(\hat F\psi)^*\,\mathrm dr\:.\tag{$\beta$}$$ Now, start again with ($\alpha$) but replacing $\psi$ for $\psi+i\phi$. The same procedure yields $$ -i\int\phi^*(\hat F\psi)\,\mathrm dr + i\int\psi^*(\hat F\phi)\,\mathrm dr$$ $$=-i\int\psi(\hat F\phi)^*\,\mathrm dr +i\int\phi(\hat F\psi)^*\,\mathrm dr\:.$$ Namely $$ -\int\phi^*(\hat F\psi)\,\mathrm dr + \int\psi^*(\hat F\phi)\,\mathrm dr$$ $$=-\int\psi(\hat F\phi)^*\,\mathrm dr +\int\phi(\hat F\psi)^*\,\mathrm dr\:.$$ Summing this result to ($\beta$), we have $$\int\psi^*(\hat F\phi)\,\mathrm dr = \int\phi(\hat F\psi)^*\,\mathrm dr$$ also written $$\int\psi^*(\hat F\phi)\,\mathrm dr = \int(\hat F\psi)^*\phi \,\mathrm dr\:.\tag{$\gamma$}$$ This identity is true for all $\hat F$ satisfying ($\alpha$) and all $\psi,\phi$. Now observe that $\hat F - \langle F\rangle$, viewing $\langle F\rangle$ as a fixed real number, satisfies ($\alpha$) as well, so that it also satisfies ($\gamma$): $$\int\psi^*((\hat F- \langle F\rangle)\phi)\,\mathrm dr = \int((\hat F- \langle F\rangle)\psi)^*\phi \,\mathrm dr\:.$$ To conclude use $\phi= (\hat F- \langle F\rangle)\psi$ and the found identity becomes $$\int\psi^*((\hat F- \langle F\rangle)^2\psi)\,\mathrm dr = \int((\hat F- \langle F\rangle)\psi)^*(\hat F- \langle F\rangle)\psi \,\mathrm dr\:.$$


It is usually assumed that $\psi$ can be expressed as a linear sum $\sum_i a_i\psi_i$ of eigen function $\psi_i$ with eigen values $\lambda_i$. So $F \psi = \sum_i a_i\lambda_i\psi_i$. This converts the problem into complex number algebra.

In effect, this is a rephrasing of - turn the operator into a matrix - but I hope it unpacks it so the idea can be seen in terms of explicit functions and simple differential operators.


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