Timeline for What is the difference between real orbital & complex orbital?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 6, 2020 at 16:20 | comment | added | Antonios Sarikas | But the probability densities are different if we use $p_x$ and $p_y$ instead of the the complex wavefunctions. | |
| May 13, 2019 at 17:53 | comment | added | Antonios Sarikas | So how we "put" electrons on orbitals if they are simultaneously in both states ? | |
| Jun 26, 2015 at 2:10 | vote | accept | CommunityBot | moved from User.Id=36790 by developer User.Id=2911 | |
| Jun 23, 2015 at 14:23 | comment | added | Michael Seifert | "How a wavefunction is related to vector space & bases" is a big topic, and one I can't fully address here. I would encourage you to start out by reading up on bra-ket notation and then posting a new question if there are things that still confuse you. As far as how the electron chooses which eigenfunction to collapse to: that's the measurement problem, and it doesn't have a universally accepted answer (yet). | |
| Jun 22, 2015 at 19:42 | comment | added | user36790 | Sir, thanks a lot. +1. Sir, I am new to the quantum arena. So, if you help me just briefly explaining how wavefunction is related to vector space & bases, I'll be very grateful. Also, a general wavefunction is a linear combination of all eigenfunctions & when we measure, the wavefunction collapses to a definite eigenfunction, right? How does it decide which eigenfunction to choose? I know they are bit irrelevent to the present context, but if you help, I'll be grateful to you:) | |
| Jun 22, 2015 at 19:26 | comment | added | Michael Seifert | Those are also valid states for the electron. It's just that its wave-function won't be real-valued if it does. Moreover, if you're in a situation where there are preferred x, y, or z-axes, these combinations can be more useful in constructing the true orbitals of the electrons (i.e., the molecular orbitals under the influence of all the other atoms as well as the central atom.) | |
| Jun 22, 2015 at 19:18 | comment | added | user36790 | Apart from this, the answer is satisfactory. Sir, cannot the orbital remain in either state where $m$ is definite? I mean to say why not $\psi_{(2,1,1)}$ or $\psi_{(2,1,-1)}$ only? | |
| Jun 22, 2015 at 19:09 | history | edited | user36790 | CC BY-SA 3.0 |
removed typo
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| Jun 22, 2015 at 19:02 | comment | added | user36790 | Isn't there a typo in your answer, sir? $l$ can have values like $$l \in 0,1,2,3,\cdots (n-1)$$ & $l\neq n$. For $n=1$, $l$ can have only one value: $l= 0$. | |
| Jun 22, 2015 at 15:15 | history | answered | Michael Seifert | CC BY-SA 3.0 |