Expectation of an Observable

Question

If we have the complex scalar $\langle f|\Omega|g\rangle$ as in equation (2.73) of these course notes (where $\Omega$ is Hermitian), and want to evaluate it in the position basis, I would proceed as follows:

$\langle f|\Omega|g\rangle =\iint dx \ dy\langle f|y\rangle \langle y|\Omega|x\rangle \langle x|g\rangle =\iint dx \ dy \ (f(y))^{*}g(x)\langle y|\Omega|x\rangle \quad \quad \mathbf{(1)} $

However, said notes claim that

$\langle f|\Omega|g\rangle=\int dx \ (\Omega f(x))^{*}\ g(x)=\int dx \ (f(x))^{*}\Omega \ g(x) \quad \quad \mathbf{(2)}$

which I don't understand. I'm okay with the first equality in $\mathbf{(2)}$, but have an issue with the second one. $(\Omega f(x))^{*}$ is really $\langle \Omega f | x\rangle=(\langle x|\Omega|f\rangle)^{\dagger}$, so I don't see how the $\Omega$ can just be plucked out.

Context:

I want to evaluate the expectation $\langle \hat{O}\rangle_{\psi}$ of observable $\hat{O}$ using the position basis. By my usual approach in $\mathbf{(1)}$, $\langle \psi| \hat{O} | \psi \rangle=\iint dx\ dy \ (\psi(y))^{*}\psi(x)\langle y|\hat{O}|x\rangle$ but if this can be whittled down to $\langle \psi| \hat{O} | \psi \rangle=\int dx \ \|\psi(x)\|^{2}\langle x|\hat{O}|x\rangle$ then I'd like to know if that is indeed possible! I was trying to follow part 2 of Valerio's answer here.

Answer 1

First, let's be familiar with the notation given on your source.

Consider a Hermitian operator $\Omega$ so that $$\Omega=\Omega^\dagger$$

Look at the matrix element of $\Omega$ in some basis $$\langle u|\Omega|v\rangle=\langle u|\Omega v\rangle$$ or $$\langle u|\Omega|v\rangle=\langle u|\Omega^{\dagger}|v\rangle=\langle \Omega u| v\rangle=\langle u|\Omega v\rangle$$

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Note the following $$\Omega|v\rangle=|\Omega v\rangle$$ and

$$\langle u|\Omega^\dagger=\langle \Omega u|$$

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Now for functions $f(x)$ and $g(x)\in \mathcal{C}$ $$\langle f|g\rangle=\int(f(x))^*g(x)dx$$ where * means complex conjugate. Also for Hermitian operator, we have proved above: $$\langle \Omega f| g\rangle=\langle f|\Omega g\rangle$$ The quantity of interest $$\langle f|\Omega|g \rangle=\langle \Omega f| g\rangle=\langle f|\Omega g\rangle$$

In position basis we can directly write $$\langle \Omega f| g\rangle=\int(\Omega f(x))^*g(x)dx$$ and $$\langle f|\Omega g\rangle=\int(f(x))^*(\Omega g(x))dx$$ $$\int(\Omega f(x))^*g(x)dx=\int(f(x))^*(\Omega g(x))dx$$

Note that $\Omega $ is still a abstract operator because we don't know its matrix element $\langle x|\Omega|x'\rangle$ in position basis. But this doesn't change the equality of those two.

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You can do them another way:

$$\langle f|\Omega| g\rangle=\int \langle f|x\rangle\langle x|\Omega|g\rangle dx=\int (f(x))^* (\Omega g(x))dx$$ or $$\langle f|\Omega| g\rangle=\int \langle f|\Omega|x\rangle\langle x|g\rangle dx=\int (\Omega f(x))^* g(x) dx $$ which are equal.