Timeline for Initial condition of the wave function while solving the Schrodinger equation
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12, 2021 at 13:31 | comment | added | Andrew | @user21820 I added a footnote. | |
| Mar 12, 2021 at 13:31 | history | edited | Andrew | CC BY-SA 4.0 |
added 565 characters in body
|
| Mar 12, 2021 at 9:09 | comment | added | user21820 | Why do you call it "probability amplitude distribution"? It's not a distribution... | |
| Mar 11, 2021 at 19:40 | comment | added | Andrew | @user713320 In fact the phases are physically observable! Now, the overall phase (1 number) is not observable. But the spatially dependent part of the phase is observable. If you transform the wavefuncton from the position basis to the momentum basis, you will find that the probability distribution for momentum depends on the phase of the wavefunction in the position basis. You may find this question relevant: physics.stackexchange.com/questions/80434/… | |
| Mar 11, 2021 at 19:38 | vote | accept | user713320 | ||
| Mar 11, 2021 at 19:34 | comment | added | user713320 | Thank you for your answer. It made the concept clearer to me. Since the phases are not physically observable, how do we know the initial probability amplitude distribution? Perhaps, we start with an observation which imples that the wavefunction collapses to an eigenstate of that observable. Then we use that eigenstate as the initial condition for the Schrodinger equation? | |
| Mar 11, 2021 at 19:18 | history | edited | Andrew | CC BY-SA 4.0 |
added 2 characters in body
|
| Mar 11, 2021 at 19:12 | history | answered | Andrew | CC BY-SA 4.0 |