How to use operators on tensor products This problem is from physics, but I have trouble understanding the math.
I have some problem understanding how to use tensors. Let's say in Quantum Optics if I have the state in mode $b$ (where I can have two possible modes $a$ and $b$)
$$|1_b\rangle = |0_a\rangle \otimes|1_b\rangle$$
And I want to obtain $\langle1_b|\hat{a}^\dagger \hat{b}|1_b\rangle$ do I proceed as follows?
$$\langle1_b|\hat{a}^\dagger \hat{b}|1_b\rangle =
\langle1_b|\hat{b} \hat{a}^\dagger |1_b\rangle =
\langle 0_a|\langle1_b|\hat{a}^\dagger \hat{b} |0_a\rangle |1_b\rangle =
\sqrt{2}  \langle0_a| \langle2_b| \hat{a}^\dagger |1_a\rangle |0_b \rangle$$
$$= 2 \langle 0_a| \langle 2_b|2_a \rangle | 0_b \rangle$$
And then what? Do I just do 
$$= 2 \langle0_a|2_a\rangle \langle2_b|0_b\rangle = 2 \cdot 0 \cdot 0 = 0$$
 A: As I see it $\hat b$ should rather be $\operatorname{id}_a\otimes \hat b$ (does nothing on $a$ and acts like $\hat b$ on the $b$ side) and similar for $\hat a$ (?)
Let us not think about tensor products of a bounded and an unbounded operator for now and just formally evaluate this expectation value--to see what is going on here I will not use bra-ket notation but rather the "usual" scalar product:
$$
\langle 1_b|\hat a^\dagger \hat b|1_b\rangle=\langle (0_a\otimes 1_b),(\hat a^\dagger\otimes\operatorname{id}_b)(\operatorname{id}_a\otimes \hat b)(0_a\otimes 1_b)\rangle\,.
$$
Now operators on tensor products act as $(A\otimes B)(x\otimes y)=(Ax)\otimes (By)$ so
$$
\langle 0_a\otimes 1_b,(\hat a^\dagger\otimes\operatorname{id}_b)(\operatorname{id}_a\otimes \hat b)(0_a\otimes 1_b)\rangle=\langle 0_a\otimes 1_b,(\hat a^\dagger 0_a\otimes \hat b 1_b)\rangle=\langle 0_a\otimes 1_b,1_a\otimes 0_b\rangle\,.
$$
Finally, the scalar product on the tensor product of two Hilbert spaces $\mathcal H_1,\mathcal H_2$ has the property $\langle a\otimes b,c\otimes d\rangle_{\mathcal H_1\otimes\mathcal H_2}=\langle a,c\rangle_{\mathcal H_1}\langle b,d\rangle_{\mathcal H_2}$. In total
$$
\langle 1_b|\hat a^\dagger \hat b|1_b\rangle=\langle 0_a\otimes 1_b|1_a\otimes 0_b\rangle=\langle 0_a|1_a\rangle\langle 1_b|0_b\rangle=0\cdot 0=0\,.
$$
Of course we could also use the last property right from the start to get
$$
\langle 1_b|\hat a^\dagger \hat b|1_b\rangle=\langle (0_a\otimes 1_b),(\hat a^\dagger\otimes\operatorname{id}_b)(\operatorname{id}_a\otimes \hat b)(0_a\otimes 1_b)\rangle=\langle 0_a|\hat a^\dagger| 0_a\rangle\langle 1_b|\hat b|1_b \rangle=0\,.
$$
