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Let's say we have a spin-up electron near x=-10:

$$\psi_1 = |\uparrow\rangle\alpha(x_1)$$

$\alpha$ being the space part: a bump near x=-10. Similarly we have a second spin-down electron near x=+10:

$$\psi_2 = |\downarrow\rangle\beta(x_2)$$

What if we want to consider these noninteracting, unentangled electrons as a system? I understand that it must be antisymmetric, so it seems that we should get:

$$\psi_\textrm{total} = \frac{1}{\sqrt2}( |\uparrow\downarrow\rangle\alpha(x_1)\beta(x_2) - |\downarrow\uparrow\rangle\beta(x_1)\alpha(x_2))$$

Is this correct? Would it be wrong to use only one term according to QM formalism?

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  • $\begingroup$ You may be interested in this question $\endgroup$ Nov 24, 2022 at 10:20

1 Answer 1

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As far as notation is concerned, you have different options. Often, the spatial information is included in the state: $$ |\uparrow\rangle\alpha(x_1) \rightarrow |\alpha(x_1),\uparrow\rangle ~~~ \text{or}~\text{just} ~~~~ |x_1,\uparrow\rangle . $$ If you just write down the state in a formal way then one term is OK, with the tacit understanding that it is anti-symmetrized. However, when you use it in calculations, both terms are necessary. Then $$\psi_\textrm{total} = \frac{1}{\sqrt{2}} \left( |x_1,\uparrow\rangle |x_2,\downarrow\rangle -|x_2,\downarrow\rangle |x_1,\uparrow\rangle \right) . $$ Sometimes, people combine the kets of the two particles, as you did, but I think it is clearer to keep them separate.

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  • $\begingroup$ Thanks for the answer. I'm trying to understand your notation. Do you use $x_1$ to mean "located at x=-10" and the first ket in each term to be for particle 1? (I was using $x_1$ to mean the position of particle 1; if you also intend that, then the two last arrows should be flipped, and the ket ordering is not significant.) $\endgroup$
    – Travis Lee
    Nov 24, 2022 at 9:09

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