
#19
Jan1413, 05:54 AM

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P: 28,804

Zz. 



#20
Jan1513, 05:14 AM

P: 977





#21
Jan1513, 04:05 PM

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P: 10,824

I don't see any classical calculation in my post.




#22
Jan1613, 07:05 AM

P: 977

does your eqn take into account the uncertainty principle.I don't think so.




#23
Jan1613, 09:04 AM

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P: 10,824

The p_{i} take the uncertainty principle into account.
Seriously, I don't understand your problem with my equation. You just repeat "uncertainty principle" every time. It is like I would counter this with "particle mass" all the time. I mean, where is your point? 



#24
Jan1613, 10:05 AM

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P: 28,804

The problem here is that people think that the HUP is a starting point in QM rather than merely a consequence. It is as if one has to forcibly introduce the HUP into something, rather than realizing that it is naturally built in when one adopts the QM formalism.
Zz. 



#25
Jan1713, 12:55 AM

P: 977

Also the formula written by you is classical one,it is called Gibbs entropy S=k_{B}Ʃp_{i} lnp_{i} quoting wiki For a classical system (i.e., a collection of classical particles) with a discrete set of microstates, if E_{i} is the energy of microstate i, and p_{i} is its probability that it occurs during the system's fluctuations, then the entropy of the system is S=k_{B}Ʃp_{i} lnp_{i} 



#26
Jan1713, 01:52 AM

Sci Advisor
P: 3,371

Quantum mechanically, the statistical entropy is defined by von Neumann's formula
##S=k\mathrm{Tr}\rho \ln \rho ## where ##\rho## is the so called density matrix. This expression can be evaluated in a basis which diagonalizes the density matrix. The density matrix only has nonnegative eigenvalues $p_i$ which can be interpreted as occupation probabilities. Hence von Neumann's expression reduces to Gibbs expression. Especially for a pure quantum mechanical state, only one of the p_i is 1, the others being 0. Hence the entropy vanishes for a pure state. Already for a single particle there are pure states with nonvanishing velocity or current. Hence finite current and vanishing entropy are not contradictory, even when QM effects are taken into account. 



#27
Jan1713, 06:43 AM

P: 38





#28
Jan1713, 06:58 AM

P: 38





#29
Jan1713, 07:18 AM

P: 977

in terms of a density matrix this expression is fine and yields the required classical form.It will also yield zero for entropy if pure state is used.But it seems like to build up on classical analogy.It seems to be a matter of definition.One can go with it.But then usual definition of entropy must not apply(randomness) 



#30
Jan1713, 07:55 AM

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P: 10,824

It can be different from zero (with a degenerate groundstate), but it does not have to. The dependencies look like that: [Basics of QM] > uncertainty principle [Basics of QM] > zero entropy at zero temperature (possible) Therefore, it is meaningless to ask how you can "derive", or "include" the uncertainty principle in (QM) statistical mechanics. It is impossible to avoid it! 



#31
Jan1713, 08:30 AM

Sci Advisor
P: 3,371





#32
Jan1713, 09:13 AM

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P: 10,824

That is true. I just keep track of that special case as it is not forbidden.




#33
Jan1813, 01:06 AM

P: 977





#34
Jan1813, 03:55 AM

P: 38





#35
Jan1813, 08:04 AM

P: 38




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