hilbert space - Why does $langle psi_1 psi_2 | H_1|psi_1 psi_2 rangle= langle psi_1 | H_1|psi_1 rangle langle psi_2 | psi_2 rangle$?

In the solution of an exercise, the following reasoning is used:

$1)$ $\langle \psi_1 \psi_2 | H_1|\psi_1 \psi_2 \rangle + \langle \psi_1 \psi_2 | H_2|\psi_1 \psi_2 \rangle = \langle \psi_1 | H_1|\psi_1 \rangle \langle \psi_2 | \psi_2 \rangle + \langle \psi_1 | \psi_1 \rangle \langle \psi_2 | H_2|\psi_2 \rangle$

$2)$ $\langle \psi_1 \psi_2 | H_1|\psi_2 \psi_1 \rangle + \langle \psi_1 \psi_2 | H_2|\psi_2 \psi_1 \rangle = \langle \psi_1 | H_1|\psi_2 \rangle \langle \psi_2 | \psi_1 \rangle + \langle \psi_1 | \psi_2 \rangle \langle \psi_2 | H_2|\psi_1 \rangle$

$H_1$ and $H_2$ refer to the Hamiltonian operator.

May I ask you what property - and if possible the name of this property so that I can look it up on the internet - they used to go from the left hand side to the right hand side for both line $1)$ and $2)$?

I was unable to find anything on the internet (maybe I used the wrong terms ?), even on https://en.wikipedia.org/wiki/Bra%E2%80%93ket_notation#Properties

Answer

What you've written from your text/exercise is an abuse of notation, but it is standard.

The composite state $|\psi \alpha\rangle$ is the tensor product of the states $|\psi\rangle \in \mathcal H_1$ and $|\alpha\rangle \in \mathcal H_2$, which is sometimes written $|\psi\alpha\rangle \equiv |\psi\rangle \otimes |\alpha\rangle$. A very typical example of such a state is a spatial wavefunction attached to a spin-1/2 state, in which case $\mathcal H_1 = L^2(\mathbb R)$ and $\mathcal H_2 = \mathbb C^2$.

The space that such composite states belong to is the tensor product of the Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, usually written $\mathcal H_1 \otimes \mathcal H_2$. The inner product on such a space can be inherited from the inner products on $\mathcal H_1$ and $\mathcal H_2$. Letting $\psi,\phi \in \mathcal H_1$ and $\alpha,\beta\in\mathcal H_2$, we have

$$\langle \psi\alpha|\phi\beta\rangle_{\mathcal H_1 \otimes \mathcal H_2} = \langle\psi|\phi\rangle_\mathcal{H_1} \cdot \langle\alpha|\beta\rangle_{\mathcal H_2}$$

Given two operators $A$ and $B$ which act on $\mathcal H_1$ and $\mathcal H_2$, we can define a new operator which acts on $\mathcal H_1\otimes \mathcal H_2$ like this:

$$\big(A \otimes B\big)\big(|\psi\rangle \otimes |\alpha\rangle\big) = \big(A|\psi\rangle\big)\otimes\big(B|\alpha\rangle\big)$$

Putting everything together (and dropping the subscripts on the inner products,

$$\langle \psi\alpha| \big(A\otimes B\big) |\phi \beta\rangle = \langle \psi|A|\phi\rangle \cdot \langle \alpha|B |\beta\rangle$$

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With that out of the way, if you have a composite system made by stitching together the Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, then the Hamiltonian for the composite system is

$$H = H_1 \otimes \mathbb I_2 + \mathbb I_1 \otimes H_2$$

where $H_{1,2}$ are the Hamiltonian operators acting on $\mathcal H_1$ and $\mathcal H_2$, and $\mathbb I_{1,2}$ are the identity operators on $\mathcal H_1$ and $\mathcal H_2$. Plugging this in to what I wrote above provides you with the answer you're looking for.

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More specifically, note that

$$\langle \psi_1\psi_2|H_1\otimes\mathbb I | \psi_2\psi_1 \rangle = \langle \psi_1|H_1|\psi_2\rangle \cdot \langle \psi_2|\mathbb I|\psi_1\rangle = \langle \psi_1|H_1|\psi_2\rangle \cdot \langle \psi_2|\psi_1\rangle$$