Timeline for Two-particle system wave function
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2021 at 7:32 | history | edited | smallest quanta | CC BY-SA 4.0 |
Grammar mistake
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| Jul 18, 2020 at 8:04 | comment | added | user264677 | Thanks for your nice explanation. | |
| Jul 18, 2020 at 7:57 | comment | added | smallest quanta | We have actually ignored the phase, as this phase is overall phase of the wavefunction which is meaningless in this situation, because we want to study the properties of two particles and overall phase is just like a car in which the two particles are travelling, and to study the relation between those two particles, we don't need to study about the car. | |
| Jul 18, 2020 at 7:46 | comment | added | user264677 | Thanks for your explanation, since independent is the notion based on $|\psi(x)|^2$, if we just consider the probability density it's ok to ignore phase, but here is the wave function itself, can you elaborate more on "because the particles are identical, so we cannot relate the two indistinguishable entities"? | |
| Jul 18, 2020 at 7:34 | vote | accept | CommunityBot | ||
| Jul 18, 2020 at 7:34 | comment | added | smallest quanta | we did not include phase in $ \Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) e^{i\phi}$, because the particles are identical, so we cannot relate the two indistinguishable entities, so they must be independent | |
| Jul 18, 2020 at 7:30 | comment | added | smallest quanta | since the particles are identical, under swapping operation the particle wavefunction must evolve with a pure phase, otherwise, if the wavefunction changes, the probability density of measuring them will change, and the particles will no longer remain identical | |
| Jul 18, 2020 at 7:30 | comment | added | user264677 | I still bit confused by the first question why can ignore phase in $\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2) e^{i\phi}$ | |
| Jul 18, 2020 at 7:26 | comment | added | user264677 | Why the exchange operator is linear operator so that $\hat{\rho}\hat{\rho} \psi(A,B)=\psi(B,A) = (e^{\iota\theta})^2 \psi(A,B)$? | |
| Jul 18, 2020 at 6:57 | history | answered | smallest quanta | CC BY-SA 4.0 |