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I am trying to understand concept of swap operator, introduced in the article http://arxiv.org/pdf/1001.2335v2.pdf by means of simple example.
Swap operator is supposed to act on two identical(?) copies of system swapping configurations within some part of system.
Definition from article: $$Swap_{A}(\sum_{\alpha_1,\beta_{1}}C_{\alpha_1,\beta_{1}} |\alpha_{1}\rangle|\beta_{1}\rangle) \otimes (\sum_{\alpha_2,\beta_{2}}D_{\alpha_2,\beta_{2}} |\alpha_{2}\rangle|\beta_{2}\rangle)=$$

$$= \sum_{\alpha_1,\beta_{1}}C_{\alpha_1,\beta_{1}} \sum_{\alpha_2,\beta_{2}}D_{\alpha_2,\beta_{2}}(|\alpha_{2}\rangle|\beta_{1}\rangle) \otimes (|\alpha_{1}\rangle|\beta_{2}\rangle)$$ I am trying to see what does it mean for simple system: $$Swap_{A}(|\uparrow\rangle_{1}|\downarrow\rangle_1 + |\uparrow\rangle_{1}|\downarrow\rangle_1)\otimes ( |\uparrow\rangle_{2}|\downarrow\rangle_2 + |\uparrow\rangle_{2}|\downarrow\rangle_2)=$$ $$(|\uparrow\rangle_{2}|\downarrow\rangle_1 + |\uparrow\rangle_{2}|\downarrow\rangle_1)\otimes ( |\uparrow\rangle_{1}|\downarrow\rangle_2 + |\uparrow\rangle_{1}|\downarrow\rangle_2)$$

My question is why can't I open brackets and get initial state by means of exchanging some ket vectors? In particular why $$|\uparrow\rangle_{2}|\downarrow\rangle_1 |\otimes \uparrow\rangle_{1}|\downarrow\rangle_2 \neq |\uparrow\rangle_{1}|\downarrow\rangle_1 \otimes |\uparrow\rangle_{2}|\downarrow\rangle_2 $$ If I try to imagine these states as two electrons on the lattice, then aforementioned statement looks counterintuitive to me.

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    $\begingroup$ $Swap$ refers to swapping states between identical copies of the same system, not to reordering the system copies themselves. In the simple example you selected the states are identical, so nothing happens on application of $Swap$. And you can't do what you propose in your last eq. because you'd be swapping the copies, not just states. $\endgroup$
    – udrv
    Apr 7, 2016 at 3:30
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    $\begingroup$ Perhaps the definition is easier to read if the state and copy labels are separated: $$Swap_{A}(\sum_{\alpha_1,\beta_{1}}C_{\alpha_1,\beta_{1}} |\alpha_{1}>_A|\beta_{1}>_B) \otimes (\sum_{\alpha_2,\beta_{2}}D_{\alpha_2,\beta_{2}} |\alpha_{2}>_{A'}|\beta_{2}>_{B'})=\\ = \sum_{\alpha_1,\beta_{1}}C_{\alpha_1,\beta_{1}} \sum_{\alpha_2,\beta_{2}}D_{\alpha_2,\beta_{2}}(|\alpha_{2}>_A|\beta_{1}>_B) \otimes (|\alpha_{1}>_{A'}|\beta_{2}>_{B'})$$ $\endgroup$
    – udrv
    Apr 7, 2016 at 3:31
  • $\begingroup$ So am I right to understand that "two identical copies" means that, both copies should be represented by states from the same Hilbert space? At the same time each of them may be in different state? $\endgroup$
    – Yaroslav Shustrov
    Apr 7, 2016 at 15:35
  • $\begingroup$ For instance would this example be correct? $$Swap_{A}(C_{\uparrow \uparrow} |\uparrow>_A |\uparrow>_B +C_{\downarrow \downarrow} |\downarrow>_A |\downarrow>_B) \otimes (D_{\uparrow \downarrow}|\uparrow>_{A'} |\downarrow>_{B'})=$$ $$=C_{\uparrow \uparrow} |\uparrow>_A |\uparrow>_B \otimes (D_{\uparrow \downarrow}|\uparrow>_{A'} |\downarrow>_{B'} + C_{\downarrow \downarrow} |\uparrow>_A |\downarrow>_B \otimes (D_{\uparrow \downarrow}|\downarrow>_{A'} |\downarrow>_{B'}$$ $\endgroup$
    – Yaroslav Shustrov
    Apr 7, 2016 at 15:44
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    $\begingroup$ Yes, physical copies are represented by identical copies of the same Hilbert space, the total Hilbert space is then the direct product of those copies. And the new example works too. $\endgroup$
    – udrv
    Apr 8, 2016 at 1:54

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