Tensor product of kets

Question

When writing down the many particle wavefunction, we write: $$ \begin{aligned} \psi_{\alpha_1\alpha_2\dots\alpha_N}(\vec r_1,\dots,\vec r_N)&=(\vec r_1\dots\vec r_N|\alpha_1\dots\alpha_N)\\ &=(\langle\vec r_1|\otimes\langle\vec r_2|\otimes\dots\otimes\langle\vec r_N|)(|\alpha_1\rangle\otimes|\alpha_2\rangle\otimes\dots\otimes|\alpha_N\rangle)\\ &=\phi_{\alpha_1}(\vec r_1)\phi_{\alpha_2}(\vec r_2)\dots\phi_{\alpha_N}(\vec r_N). \end{aligned} $$

My question is how do we get the third step from second. I know that: $$\phi_{\alpha_1}(\vec{r}_{1}) = \langle\vec{r}_{1}|\alpha_{1}\rangle $$ Are we assuming that: $$\phi_{\alpha_1}(\vec{r}_{1}) = \langle\vec{r}_{1}\otimes\alpha_{1}\rangle $$ If so why are there no cross terms like $$\langle\vec{r}_{1}\otimes\alpha_{2}\rangle $$ ?

Someone please clarify.

Answer 1

An operator $A\otimes B$ acts on a vector $u\otimes v$ as

$$ (A\otimes B)(u\otimes v) = (Au)\otimes(Bv) $$

by definition, ie

$$ (\langle \vec r_1|\otimes\langle \vec r_2|)\,(|\alpha_1\rangle\otimes|\alpha_2\rangle) = \langle \vec r_1|\alpha_1\rangle\otimes\langle \vec r_2|\alpha_2\rangle = \langle \vec r_1|\alpha_1\rangle \cdot\langle \vec r_2|\alpha_2\rangle $$

as the tensor product on scalars is just the regular one.