Using tensor products in the bra-ket notation I'm trying to find the expectation value of the operator $\hat W(x_1,x_2)=\hat x_1 \hat x_2$ with respect to the eigenstates of a system composed of two one dimensional quantum harmonic oscillators. The eigenstate of the total system will be $|n_1n_2\rangle=|n_1\rangle \otimes|n_2\rangle $, with $|n_1\rangle$, $|n_2\rangle$ the eigenstates of each individual oscillator, so the expectation value will be
$$\left((|n_1\rangle \otimes|n_2\rangle )^\dagger,\hat W(|n_1\rangle \otimes|n_2\rangle)  \right)$$
Two questions have arisen to me with this:

*

*Is the bra corresponding to a ket formed by a tensor product just the tensor product of the bras, $\left(|n_1 n_2\rangle \right)^\dagger=\left(|n_1\rangle \otimes|n_2\rangle \right)^\dagger=\langle n_1| \otimes \langle n_2| =\langle n_1 n_2| $?

*Are operators corresponding to different Hilbert spaces associative with respect to the tensor product of different states? That is, $ \hat x_1 \hat x_2 (|n_1\rangle \otimes|n_2\rangle)=\hat x_1 |n_1\rangle \otimes\hat x_2 |n_2\rangle?$

*How do the inner products behave with respect to the tensor products? Would it be just $\big(\langle n_1| \otimes \langle n_2|\big) \big(\hat x_1 |n_1\rangle \otimes\hat x_2 |n_2\rangle \big)= \langle n_1|\hat x_1 |n_1\rangle\otimes\ \langle n_2|\hat x_2 |n_2\rangle$?

 A: The answer to your first question is yes, see for example equations $(1.32)-(1.36)$ in these lecture notes.
To answer the second question, consider a bipartite Hilbert space $\mathscr{H} \equiv \mathscr{H}_1 \otimes \mathscr{H}_2$ and let $o_1$ and $o_2$ denote operators on $\mathscr{H}_1$ and $\mathscr{H}_2$, respectively. We then can define the action of $o_1$ and $o_2$ on $\mathscr{H}$ by
\begin{align}
 O_1 &\equiv o_1 \otimes \mathbb{I}_2 \\
 O_2 &\equiv \mathbb{I}_1 \otimes o_2  \quad ,
\end{align}
where $\mathbb{I}_i$ for $i=1,2$ denotes the identity operator on $\mathscr{H}_i$.
Now let $|\varphi_i\rangle \in \mathscr{H}_i$ and $ \mathscr{H} \ni|\varphi\rangle \equiv |\varphi_1\rangle \otimes |\varphi_2\rangle$.
We compute
\begin{align}
O_1 |\varphi\rangle &= o_1 |\varphi_1\rangle \otimes \mathbb{I}_2 |\varphi_2\rangle\\ 
O_2 |\varphi\rangle &= \mathbb{I}_1 |\varphi_1\rangle \otimes  o_2 |\varphi_2\rangle \quad .
\end{align}
Consequently, by applying both operators successively, we obtain:
$$O_1 \, O_2  |\varphi\rangle= O_2\, O_1 |\varphi\rangle = o_1 |\varphi_1\rangle \otimes o_2 |\varphi_2\rangle  \quad . $$
Additionally, for $O\equiv o_1 \otimes o_2$ we have $O^\dagger = o_1^\dagger \otimes o_2^\dagger$.
Regarding the third question, note that for an inner product on $\mathscr{H}$ it holds that
$$(\varphi_1 \otimes \varphi_2 , \phi_1 \otimes \phi_2)_{\mathscr{H}} = (\varphi_1,\phi_1)_{\mathscr{H}_1}\,(\varphi_2,\phi_2)_{\mathscr{H}_2} \quad .$$
Defining $ \phi_i \equiv o_i \varphi_i$ yields an expression for the expectation value of $O_1\,O_2$.
A more detailed explanation is given in the above linked lecture notes, equations $(1.26)-(1.31)$ or also in the Wikipedia link provided in the other answer.
A: Regarding your second question: Yes, and that's actually the defining property of $\hat x_1 \hat x_2$:
Let $V,V',W,W'$ be vector spaces over a field $F$ (for example, $V'$ and $W'$ can be Hilbert spaces with vector subspaces $V$ and $W$). If
$$
A\colon V\to V'
$$
and
$$
B\colon W\to W'
$$
are linear functions, the function
\begin{align}
V\times W&\to V'\otimes W'\\
(v,w)&\mapsto(Av)\otimes(Bw)
\end{align}
is bilinear and by the universal property of the tensor product, it extends to a unique linear function
$$A\otimes B\colon V\otimes W\to V'\otimes W'$$
satisfying
$$(A\otimes B)(v\otimes w)=(Av)\otimes(Bw)$$
for all $(v,w)\in V\times W$.
Warning: If $H_1$ and $H_2$ are Hilbert spaces, the vector space $H_1\otimes H_2$ together with the unique inner product satisfying
$$\langle v_1\otimes v_2|w_1\otimes w_2\rangle=\langle v_1|w_1\rangle\langle v_2|w_2\rangle$$
is not necessarely a new Hilbert space, which is why we usually consider the Hilbert tensor product $H_1\hat{\otimes} H_2$.
