For the past few weeks, I have been studying Quantum Chemistry and lately in these lectures something has bugged me:

If I do a sum:

$1 \times 2 \times 3 \times {...} $

This is the same as simply writing the order in reverse:

$ {...} \times 2 \times 1 $

a bit like $ab = ba$

and for wave functions, I've seen that the order of coordinates inside a wave functions seem to imply that no two wave functions are the same:

$$\psi({r_1},{r_2}) \neq \psi({r_2},{r_1})$$

and from there on there is a discussion that there must be a change in sign and so on and so forth.

But I'm struggling to work out why if I change the order of my variables, that this makes any effect.

I'm just confused by why the order is such a fuss, if that makes sense!

Why are the two wave functions different by the order of $r_1$ and $r_2$?

  • 1
    $\begingroup$ You mean if you do the product. The order matters because maybe $\psi(r_1,r_2) = \phi_a(r_1) \times \phi_b(r_2)-\phi_a(r_2) \times \phi_b(r_1)$ (for instance) in which case swapping labels will give you a different function. $\endgroup$ – Buck Thorn Oct 25 '19 at 20:41
  • $\begingroup$ So are you saying that in reference to your equation, $\psi\left(r_{2}, r_{1}\right)=\phi_{a}\left(r_{2}\right) \times \phi_{b}\left(r_{1}\right)-\phi_{a}\left(r_{1}\right) \times \phi_{b}\left(r_{2}\right)$ as the order that I represent the coordinates in $\psi\left(r_{2}, r_{1}\right)$ is important here? $\endgroup$ – vik1245 Oct 25 '19 at 21:04
  • $\begingroup$ Hopefully the example I presented (for which $\psi(r_1,r_2)=-\psi(r_2,r_1)$ shows you why the order of labels matters. For instance, electrons have this property, swapping the electrons inverts the sign of the wavefunction. $\endgroup$ – Buck Thorn Oct 25 '19 at 21:07
  • $\begingroup$ I removed my comment about commutativity since your focus is on wavefunctions, not operators, by the way. $\endgroup$ – Buck Thorn Oct 25 '19 at 21:08
  • $\begingroup$ @BuckThorn thanks! Is there a mathematical term that describes this property that I can look up on the internet too for reference? $\endgroup$ – vik1245 Oct 25 '19 at 21:11

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