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This is "somewhat" related to Why does $\langle \psi_1 \psi_2 | H_1|\psi_1 \psi_2 \rangle= \langle \psi_1 | H_1|\psi_1 \rangle \langle \psi_2 | \psi_2 \rangle$?
but I'm asking about something different. I struggle to understand the solution of an exercise.

I believe this could be helpful to a broader audience because I haven't found any other similar question on the site.


Given is the total hamiltonian $\langle H \rangle = \langle H_1 + H_2 \rangle$ where $H_n$ only acts on particle $n$.
Then we have $\langle \psi_1 \psi_1 |H|\psi_1 \psi_1 \rangle = 2E_1$
where $\psi_1= \sqrt{\frac{2}{L}} \sin(\frac{\pi x}{L})$
(and $2E_1$ is obviously the energy)


My question is: if $H_n$ only acts on particle $n$, then why does $$\langle \psi_1 \psi_1 |H|\psi_1 \psi_1 \rangle = \langle \psi_1 \psi_1 |H_1 + H_2 |\psi_1 \psi_1 \rangle = \langle \psi_1 | \psi_1 \rangle \langle\psi_1 |H_1| \psi_1 \rangle + \langle \psi_1 | \psi_1 \rangle \langle\psi_1 |H_2| \psi_1 \rangle = \langle\psi_1 |H_1| \psi_1 \rangle + \langle\psi_1 |H_2| \psi_1 \rangle$$
equal $2E_1$ and not $E_1 ?$
Shouldn't $\langle \psi_1 |H_2| \psi_1 \rangle$ equal $0$ ?

Thanks for your help !

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  • $\begingroup$ @AaronStevens I'm not asking for help to do my homeworks ! $\endgroup$ – holomorphicfunction Jan 14 at 17:08
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    $\begingroup$ I didn't say you were $\endgroup$ – Aaron Stevens Jan 14 at 17:14
  • $\begingroup$ @holomorphicfunction aren't you abusing the notation too badly here the second set of $\psi$ has to be numbered as 2, not 1 when we are talking of n-th body Hamiltonian? $\endgroup$ – aitfel Jan 14 at 17:31
  • $\begingroup$ @aitfel That’s what disturbs me. It’s written like that in the solutions $\endgroup$ – holomorphicfunction Jan 14 at 19:01