Is the inner product a function of position $x$?

Question

I'm going through a set of online lectures in Quantum Mechanics.

$Q$ is a linear operator with no explicit dependence on time $t$. $\Psi(x,t)$ is the wave function, where $x$ is a space coordinate. The professor is trying to calculate $\frac{\mathrm{d} \left \langle Q \right \rangle _\Psi }{\mathrm{d} t}$, where $\left \langle Q \right \rangle _\Psi$ is the expectation value of $Q$ in the state $\Psi$. I'm not sure if he's assuming the definition of the inner product on this particular vector space, but is $\left \langle Q \right \rangle _\Psi = \langle \Psi,Q\Psi \rangle$ a function of both $x$ and $t$? I think $x$ and $t$ are independent variables since the wave is delocalized in space, right?

Answer 1

The expectation value $\langle Q\rangle_{\Psi}$ is defined as

$$\langle Q\rangle_{\Psi}=\int\mathrm{d}x\,\Psi^{*}(x,t)\,Q\Psi(x,t),$$

so long as $\Psi$ is normalized. Thus, $\langle Q\rangle_{\Psi}$ does not depend on space, since it is defined as a spatial average weighted by your wavefunction. This is just the quantum analogue of how the average value of a coin toss can't depend on what you got on any particular toss.

I hope this helps!