# What is the form of a many-body hamiltonian that are subject to the measurement of the position?

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 $$H'$$?

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

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 $$H'$$.