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$\psi$ is a state that given $|\psi\rangle=\int d^3x\psi(x)|x\rangle$

How does the wave function change in time? The Hamiltonian can be written as $$H=\int d^3x\frac{1}{2m}\nabla\psi^*\nabla\psi=\int \frac{d^3p}{(2\pi)^3}\frac{p^2}{2m}a_p^{\dagger}a_p$$ So how can I find the $$i\frac{\partial\psi}{\partial t}=-\frac{1}{2m}\nabla^2\psi$$

I need to Schrödinger picture, given ket $|\psi\rangle$ is in Heisenberg picture.

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  • $\begingroup$ What have you tried so far? Once you have the Hamiltonian you can derive the time evolution operator and apply it on the original state in the Schrödinger picture to derive the state at a later time. $\endgroup$ – gented May 28 at 13:52
  • $\begingroup$ I cannot get the second equation because don't have any time-dependent wavefunction to apply Hamiltonien. $\endgroup$ – S. Alkan May 28 at 14:23
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    $\begingroup$ You have to apply the Heisenberg equation of motion $i\hbar\partial\psi/\partial t=[\psi, H]$. $\endgroup$ – Jon May 28 at 14:27
  • $\begingroup$ ? Is your ψ(x) a one particle c-number wavefunction, or an infinite-particle quantum field? You appear to be using similar symbols for both, and in the same breath. Some context might well be salutary. $\endgroup$ – Cosmas Zachos May 28 at 22:12
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The computation is rather straightforward. Consider the commutation rule $$ [\psi^*(x),\psi(x')]=-\delta^3(x-x'). $$ After integration by parts, the Hamiltonian takes the form $$ H=-\int d^3x'\frac{\hbar^2}{2m}\psi^*(x')\Delta_2\psi(x'). $$ Now, promote your Hamiltonian to an operator and you will see that $$ [\psi(x),H]=-\frac{\hbar^2}{2m}\int d^3x'[\psi(x),\psi^*(x')]\Delta_2\psi(x')= \\ -\frac{\hbar^2}{2m}\int d^3x'\delta^3(x-x')\Delta_2\psi(x')=-\frac{\hbar^2}{2m}\Delta_2\psi(x). $$ Then, from Heisenberg equation of motion you get the final result.

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