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

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?

• Which online lectures? – Qmechanic Jul 6 '19 at 1:01

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.
• So, can I write $\frac{\mathrm{d} \left \langle Q \right \rangle _\Psi }{\mathrm{d} t} = \lim_{h \rightarrow 0} \frac{1}{h}(\langle \Psi(x,t+h),Q\Psi(x,t+h) \rangle - \langle \Psi(x,t),Q\Psi(x,t) \rangle)$? – IanDsouza Jul 18 '17 at 1:27
• The inner product $\langle\Psi,Q\Psi\rangle$ should be written as $\langle\Psi(t),Q\Psi(t)\rangle$. Remember, the state $|\Psi(t)\rangle$ is not a function of space, only the wavefunction $\Psi(x,t)=\langle x|\Psi(t)\rangle$ is. So your expression is correct if you omit the spatial dependence of your state vectors $|\Psi\rangle$. – Bob Knighton Jul 18 '17 at 1:30