
#1
Aug3113, 03:42 AM

P: 49

in VBT we consider orbitals
but orbitals were given by Schrodinger and his quantum mechanics. so we must also consider Heisenberg in it!! 



#2
Aug3113, 09:47 AM

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P: 10,786

The solutions of the Schrödinger* equation always respect Heisenberg's uncertainty relation.
*that's his name, with dots on the o. 



#3
Aug3113, 10:09 AM

P: 160

I'm not sure I get the point of this post. Please elaborate.




#4
Aug3113, 10:27 AM

P: 219

VBT AND Schrodinger(not sure about spelling)
Schr[itex]\ddot{\text{o}}[/itex]dinger and Heisenberg's mechanics are the same mechanics.
Dirac showed this where in the Schr[itex]\ddot{\text{o}}[/itex]dinger picture, the states evolve in time, and the observables are stationary, while in the Heisenberg picture, the states are stationary, and the observables evolve in time. Every calculation gives you the exact same result in both pictures, though one picture might be easier than the other to work with, depending on the problem at hand. 



#5
Aug3113, 06:49 PM

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P: 10,786

@jfizzix: Ah right, that is another way to interpret the first post.
We'll have to wait for namanjain to make clear what exactly he/she is asking. 



#6
Sep213, 04:19 AM

P: 49

in vbt (valance bond theory of covalent bond)
we consider orbitals which are part of Schrodinger, so must not we respect Heisenberg and say it's idiotic to tell electron cloud's position 



#7
Sep213, 06:59 AM

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P: 10,786

Heisenberg is a person, not a law. I guess you mean the uncertainty principle?
"Electron cloud" means that the electron has no precise position  and it is uncertain enough to satisfy the uncertainty principle. 



#8
Sep313, 09:29 AM

P: 219

The reason Heisenberg's uncertainty relation is obeyed by electrons is because they exhibit waveparticle duality, just as all quantum objects do. The probability distributions that orbitals represent are not those of classical ignorance, but true quantum uncertainty. It's not true that electrons are well defined and we just can't in principle make up an experiment to show us exactly where one is and where it's going (these sorts of interpretations are ruled out by violating Bell inequalities).
The classical perspective of an electron becomes especially hard to maintain with electrons in Porbitals which have a nodal plane where the probability is precisely zero. The electrons somehow can still go from one side to the other of this nodal plane even though they cannot strictly speaking, pass through. When dealing with problems like this it's sometimes best to consider these objects (electrons, photons, etc) as a class of their own, because we just don't experience these sort of phenomena in the everyday macroscopic world. 



#9
Sep413, 10:18 AM

P: 160

Quantum mechanics: Myths and facts http://xxx.lanl.gov/abs/quantph/0609163 Also, the business with the nodes is often overstated. For one, the one electron wave functions don't describe an electron "moving" anywhere, they're a static entity that describe the probability of an electron's position. In addition, the probability at the nodes isn't identically zero in the most accurate (relativistic) treatment. Please see the very neat article: Relativistic quantum chemistry: The electrons and the nodes J. Chem. Educ., 1968, 45 (9), p 558 



#10
Sep413, 01:04 PM

P: 219

Talking about quantum systems in terms of wave particle duality is confusing, I'll admit that.
For me it's a useful way of explaining that quantum objects are in a class of their own. When dealing with position and momentum, we do describe quantum systems with "wave functions", but with other observables like spin and polarization, we have a more general object called a state vector which gives the probability amplitudes for outcomes of a given experiment. That nodal planes don't exist in the full relativistic treatment of the electron is news to me, so thanks for the article (which was indeed a good read). I can imagine it makes sense in that if your probability amplitude is a complex valued function, it doesn't have to pass through zero to go from a positive to a negative value. 


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