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Suppose there is a $N$ body hamiltonian, suppose $N=2$ for simplicity:

$$ H = - \frac{1}{2} \nabla_1^2 - \frac{1}{2} \nabla_2^2 + V(r_1,r_2) + \frac{1}{|r_1 - r_2|}. $$ If we make a measurement for position (for example that $r_2 = x_2$) and the particles are in the state $\psi$ (which is a stationary state), then the wave function would collapse and becomes $$ \phi(r_1,x_2) = \frac{|\psi(r_1,x_2)|^2}{\int |\psi(r_1,x_2)|^2 d r_1}. $$ But how does the hamiltonian change under this measurement? Is it: $$ H' = - \frac{1}{2} \nabla_1^2+ V(r_1,x_2) + \frac{1}{|r_1 - x_2|} $$ and $\phi$ solves $H'$?

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  • $\begingroup$ Changing the state doesn't change the Hamiltonian. $\endgroup$
    – d_b
    Jul 8, 2020 at 17:04

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If one measures an observable of some system, the wave function collapses to an eigenstate of the Hamiltonian of this system. The Hamiltonian does not change under the measurement. The resulting wave function is an eigenfunction of the Hamiltonian.

Thus, for your question, $\phi$ solves $H$ after the measurement since $\phi$ is an eigenfunction of $H$.

In other words, after the measurement, if a particles is found at some position (let us say $x_{2}$), the wave function collapses to a state in which a particle is at that position, however the state is still a many-body state and solves the Hamiltonian, $H$. The measurement is applied on a system which is described by Hamiltonian $H$, and not $H'$.

Additionally, a quote from wikipedia clearly explains this.

Before collapse, the wave function may be any square-integrable function. This function is expressible as a linear combination of the eigenstates of any observable. Observables represent classical dynamical variables, and when one is measured by a classical observer, the wave function is projected onto a random eigenstate of that observable.

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