Timeline for Two particles system
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 19, 2020 at 3:27 | comment | added | flippiefanus | Well in that case the phase would depend on $x_1$ and $x_2$ and then the derivation may not work quite out the way you show it. | |
| Jul 18, 2020 at 6:56 | comment | added | Andrei | @flippiefanus I've changed the probability to probability density. Thanks for pointing that out. For the second part of your comment, you are wrong. If two complex vectors have the same magnitude (and complex conjugates do), then they differ by a complex phase only. As mentioned in the answer, the phase depends on the real and imaginary parts. | |
| Jul 18, 2020 at 6:54 | history | edited | Andrei | CC BY-SA 4.0 |
added 15 characters in body
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| Jul 18, 2020 at 4:21 | comment | added | flippiefanus | There are some serious errors here: the first equation should not have the integral. If you integrate out $x_1$ then the result cannot depend on $x_1$ any more. Complex conjugation is not represented by a phase factor. It changes the sign of $i$: so $(a+i b)^*=a-i b$. The rest after that cannot be trusted. | |
| Apr 14, 2018 at 23:20 | comment | added | Andrei | It's not always true. It is strictly valid if the Hamiltonian is separable. That means non-interacting particles. | |
| Apr 14, 2018 at 21:49 | comment | added | Akababa | Why do you assume the probabilities are independent so you can multiply them together? | |
| Jul 27, 2016 at 3:05 | comment | added | Andrei | It is related to probability. The probability density for the two wavefunctions is the same, so they only differ by a complex constant of magnitude 1 | |
| Jul 26, 2016 at 15:03 | comment | added | newbie125 | How did you get $\Psi(x_1,x_2)=e^{i\phi}\Psi(x_2,x_1)$ or $\Psi(x_2,x_1)=e^{i\phi}\Psi(x_1,x_2)$ ? Is it related to $\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) $ & $\Psi(x_2,x_1)=\Psi_a(x_2) \Psi_b(x_1) $ and how did you convert between them? | |
| Jul 25, 2016 at 18:15 | comment | added | Andrei | I guessed (I did not prove - but it should be easy) that $|a|=|b|$. You will get this by normalizing the wavefunction. I do not care for now what $a$ and $b$ are. In fact, even after getting $b=\pm a$, $a$ is still known only up to a complex phase | |
| Jul 25, 2016 at 14:48 | comment | added | newbie125 | So $b=Ae^{i\phi}$ while a=A? How did you obtain them? | |
| Jul 25, 2016 at 2:47 | history | edited | Andrei | CC BY-SA 3.0 |
reformatted equations equations
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| Jul 24, 2016 at 20:10 | history | answered | Andrei | CC BY-SA 3.0 |