0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Which online lectures? $\endgroup$
    – Qmechanic
    Jul 6, 2019 at 1:01

1 Answer 1

2
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ So, can I write $\frac{\mathrm{d} \left \langle Q \right \rangle _\Psi }{\mathrm{d} t} = \lim_{h \rightarrow 0} \frac{1}{h}(\langle \Psi(x,t+h),Q\Psi(x,t+h) \rangle - \langle \Psi(x,t),Q\Psi(x,t) \rangle)$? $\endgroup$
    – Matrix23
    Jul 18, 2017 at 1:27
  • 1
    $\begingroup$ 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 $|\Psi\rangle$. $\endgroup$ Jul 18, 2017 at 1:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.