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My problem states: Prove that the average force that acts on a bounded particle is zero.

My question is: How do we represent a bounded particle? I believe that a bounded particle represents a localized state, meaning that $p(x\to \pm \infty) = 0$. But I'm not sure if that's the meaning of bounded particle. How is it represented mathematically?

The question will be answered as follows $$i \hbar \frac{d}{dt}\langle \hat p \rangle = \langle\left[\hat p, \hat H\right]\rangle \\ \langle \psi | \frac{d}{dt} \hat p | \psi\rangle = -\langle \psi |V^\prime(\hat x) |\psi\rangle \\ \langle \hat p \rangle = - \int_{-\infty}^\infty V^\prime (x) |\psi(x,t = 0)|^2 \mathrm{d}x$$

I don't know how to apply the concept of bounded particle nor localized state (if localized state is the same as bounded particle). So that I can prove that $\langle F(\hat x)\rangle = - \langle V^\prime (\hat x) \rangle = 0$.

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    $\begingroup$ Have you learned the Hellman Feynman theorem? This allows you to get quantum analogue of force. $\endgroup$
    – Matt Hanson
    Commented Jul 8 at 17:20
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    $\begingroup$ The thing you are calling "Heisenberg's equation" (in your first block-offset equation) is missing a factor of $i$ on the RHS, and worse still, is missing the expectation value on the RHS. (As it currently is written you are equating a number with an operator...) $\endgroup$
    – hft
    Commented Jul 8 at 17:57
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    $\begingroup$ Your third block-offset equation only holds when $\psi$ is an eigenstate of the Hamiltonian. $\endgroup$
    – hft
    Commented Jul 8 at 18:11

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