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Conductivity of metals at 0 kelvin?

by spectrum123
Tags: conductivity, kelvin, metals
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ZapperZ
#19
Jan14-13, 05:54 AM
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Quote Quote by andrien View Post
classical calculation does not apply as I already said.It is pointed out in feynman lectures vol. 1 that uncertainty principle must be invoked for non zero entropy.see some early chapter,it is written there.
What classical calculation? The scattering rates are obtained using QFT!

Zz.
andrien
#20
Jan15-13, 05:14 AM
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Quote Quote by ZapperZ View Post
What classical calculation? The scattering rates are obtained using QFT!
just did not see the second page there(too hurry).I replied to the last post of page 1.Sorry for causing trouble.
mfb
#21
Jan15-13, 04:05 PM
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I don't see any classical calculation in my post.
andrien
#22
Jan16-13, 07:05 AM
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does your eqn take into account the uncertainty principle.I don't think so.
mfb
#23
Jan16-13, 09:04 AM
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The pi 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?
ZapperZ
#24
Jan16-13, 10:05 AM
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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.
andrien
#25
Jan17-13, 12:55 AM
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The problem here is that people think that the HUP is a starting point in QM rather than merely a consequence.
yes,So uncertainty principle must follow from it.I don't see how it is following.Just saying it is already there,does not mean it is really there.
I mean, where is your point?
The point is that you must either provide a reference or something equivalent,if you are saying that uncertianty principle is already taken into account.I believe in what Feynman has said that if uncertainty principle is taken into account then entropy must not be zero at zero temperature.
Also the formula written by you is classical one,it is called Gibbs entropy
S=-kBƩpi lnpi
quoting wiki
For a classical system (i.e., a collection of classical particles) with a discrete set of microstates, if Ei is the energy of microstate i, and pi is its probability that it occurs during the system's fluctuations, then the entropy of the system is
S=-kBƩpi lnpi
DrDu
#26
Jan17-13, 01:52 AM
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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 non-negative 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 non-vanishing velocity or current.
Hence finite current and vanishing entropy are not contradictory, even when QM effects are taken into account.
psmt
#27
Jan17-13, 06:43 AM
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Quote Quote by andrien View Post
The point is that you must either provide a reference or something equivalent,if you are saying that uncertianty principle is already taken into account.
It seems the kind of reference you're after would have to be a general proof of the uncertainty principle, since the 'i's label quantum states, which necessarily satisfy the HUP. I'm sure most QM textbooks will give a derivation. Try Sakurai, Modern Quantum Mechanics second edition for example.
psmt
#28
Jan17-13, 06:58 AM
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Quote Quote by ZapperZ View Post
2. Undergraduate level.
The conductivity is expected to be infinite, i.e. resistivity approaches zero. This is because the predominant source of resistivity (lattice vibrations) diminishes to zero (theoretically) at T=0.
Zz.
It depends how "ideal" our idealised metal is: if the metal is modelled as having a partially filled nearly-free band, and no scattering, then i think you would expect Bloch oscillations giving net zero current, so a small amount of scattering is necessary to get a nonzero conductivity. (I'm basing this on Singleton, Band Theory and Electronic Properties of Solids, chapter 9.1.)
andrien
#29
Jan17-13, 07:18 AM
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It seems the kind of reference you're after would have to be a general proof of the uncertainty principle, since the 'i's label quantum states, which necessarily satisfy the HUP. I'm sure most QM textbooks will give a derivation. Try Sakurai, Modern Quantum Mechanics second edition for example.
just an index does not mean any quantum property.Also I have done sakurai some time ago,it does not say anything useful for it.
Hence von Neumann's expression reduces to Gibbs expression.
wiki says'This upholds the correspondence principle, because in the classical limit, i.e. whenever the classical notion of probability applies, this expression is equivalent to the familiar classical definition of entropy'(gibbs one).
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)
mfb
#30
Jan17-13, 07:55 AM
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Quote Quote by andrien View Post
yes,So uncertainty principle must follow from it.I don't see how it is following.
It is possible to calculate that all physical wavefunctions satisfy the (p,x)-uncertainty principle. This is related to the mathematics of fourier transformations. Alternatively, it is possible to derive it in a pure algebraic way as well. Read some introduction book about quantum mechanics, or see Wikipedia for the general concepts.

I believe in what Feynman has said that if uncertainty principle is taken into account then entropy must not be zero at zero temperature.
Source?
It can be different from zero (with a degenerate ground-state), 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!
DrDu
#31
Jan17-13, 08:30 AM
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Quote Quote by mfb View Post
It can be different from zero (with a degenerate ground-state), but it does not have to.
I don't think there are actual thermodynamical systems whose true ground state is degenerate. Usually at sufficiently low temperatures some very weak interactions break the degeneracy.
mfb
#32
Jan17-13, 09:13 AM
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That is true. I just keep track of that special case as it is not forbidden.
andrien
#33
Jan18-13, 01:06 AM
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It is possible to calculate that all physical wavefunctions satisfy the (p,x)-uncertainty principle. This is related to the mathematics of fourier transformations. Alternatively, it is possible to derive it in a pure algebraic way as well.
every one knows that.
Source?
It been time.I think it was written in vol. 1 of his and early chapters.
psmt
#34
Jan18-13, 03:55 AM
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Quote Quote by andrien View Post
I think it was written in vol. 1 of his and early chapters.
If you could provide a page number, chapter number or something like that that would be helpful, but either way i'm going to check this out when i next visit a suitable library, because i would have expected better of Feynman!
psmt
#35
Jan18-13, 08:04 AM
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Quote Quote by psmt View Post
i would have expected better of Feynman!
He was unambiguously referring to zero point energy, not entropy.


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