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.


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.

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  • $\begingroup$ Still that tensor product symbol will remain in the third step, where did it vanish? $\endgroup$ – Draco_1125 Oct 8 '17 at 8:54
  • $\begingroup$ @SSP_user5275: because you're dealing with scalar functions, cf edit $\endgroup$ – Christoph Oct 8 '17 at 9:23
  • $\begingroup$ Oh got that. But why in the definition the cross terms are not involved? Is it because we are trying to take the projection of state of particle 2 in the position coordinate of particle 1, which is meaningless? $\endgroup$ – Draco_1125 Oct 8 '17 at 10:22
  • $\begingroup$ @SSP_user5275 It goes back to the definition: $(A\otimes B)(u\otimes v):= (Au)\otimes (Bv)$ with no cross terms. The first slot of $(A\otimes B)$ “sees” the first slot of $(u\otimes v)$, the second slot of $(A\otimes B)$ “sees” the second slot of $(u\otimes v)$ etc by definition of the tensor product. $\endgroup$ – ZeroTheHero Oct 8 '17 at 11:24
  • $\begingroup$ @SSP_user5275: That's correct. The tensor product is used to create arbitrary compound systems (or add degrees of freedom), and operations on one subsystem may be entirely meaningless on a different one. But note that in case of identical particles, only symmetrized states (in case of bosons) or anti-symmetrized states (in case of fermions) appear in nature. $\endgroup$ – Christoph Oct 8 '17 at 11:36

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