# Why does $\langle \psi_1 \psi_1 |H_1 + H_2 |\psi_1 \psi_1 \rangle = 2E_1$ (and not simply $E_1$)? [closed]

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 $$0$$ ?

• @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? – aitfel Jan 14 at 17:31