機率論公理化 Statistical Mechanics at Fixed Temperature. (Canonical Ensemble).

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC19965/


安德雷·柯爾莫哥洛夫，俄國數學家，主要研究機率論、演算法信息論、拓撲學、直覺主義邏輯、湍流、經典力學和計算複雜性理論，最為人所道的是對機率論公理化所作出的貢獻。他曾說："機率論作為數學學科，可以而且應該從公理開始建設，和幾何、代數的路一樣"。


機率論公理化


http://arxiv.org/pdf/quant-ph/0612153.pdf

Kolmogorov 的數學觀與業績

episte.math.ntu.edu.tw/articles/mc/mc_39_01/

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當我得知蘇聯偉大的數學家，84歲的Andreyii Nikolaevich Kolmogorov 教授於1987年10月20日離開人世時， 我感到像是失去了支柱那樣悲哀與孤寂。 在我還是學生 ...

Canonical ensemble - Wikipedia, the free encyclopedia

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In statistical mechanics, a canonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system which is in thermal ...

‎Applicability of canonical ... - ‎Properties - ‎Example ensembles

Grand canonical ensemble - Wikipedia, the free encyclopedia

en.wikipedia.org/wiki/Grand_canonical_ensemble

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In statistical mechanics, a grand canonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system of particles ...

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The Ensembles

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Chapter 6. The Ensembles. In this chapter we discuss the three ensembles of statistical mechanics, the microcanonical ensemble, the canonical ensemble and ...

4.2 Canonical ensemble

homepage.univie.ac.at/~veselyf2/sp.../node19.html

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University of Vienna

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4.2 Canonical ensemble. We once more put two systems in thermal contact with each other. One of the systems is supposed to have many more degrees of ...

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Canonical Ensemble - UCSD Department of Physics

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University of California, San Diego

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8.044 Lecture Notes. Chapter 6: Statistical Mechanics at Fixed Temperature. (Canonical Ensemble). Lecturer: McGreevy. 6.1 Derivation of the Canonical ...

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Handout 8. Canonical Ensemble - Stanford University

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ME346A Introduction to Statistical Mechanics – Wei Cai – Stanford University – Win 2011. Handout 8. Canonical Ensemble. January 26, 2011. Contents. Outline.

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The Canonical Ensemble - Faculty

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Feb 12, 2001 - We will develop the method of canonical ensembles by considering a ... placed in a heat bath at temperature T. The canonical ensemble is the ...

Canonical ensemble: Definition from Answers.com

www.answers.com › Library › Science › Sci-Tech Dictionary

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canonical ensemble ( knnkl nsmbl ) ( statistical mechanics ) A hypothetical collection of systems of particles used to d.

A classic physics example is that of the ensembles of the particle and field strength observables of light. Each ensemble by itself consists of mutually commutative observables, but the two ensembles do not commute with each other. The quantized light field as a unified entity is mathematically perfectly well defined and entirely explicit, notwithstanding its anthropomorphically unintuitive features, and serves to reconcile simply and effectively the corpuscular and wave aspects of light.

Probability theory is equivalent in all observable aspects to just that special case of a quantum system, modeled as an ensemble of self-adjoint operators together with an expectation value form on the ensemble, in which there is full simultaneous observability. This follows from commutative spectral theory, according to which any ensemble of commuting self-adjoint operators is unitarily equivalent to a corresponding ensemble of multiplications by measurable functions acting on the space of square-integrable functions over a measure space. An “expectation value functional” E on the ensemble satisfying the basic constraints of linearity and positivity (hereafter, simply “state,” for brevity and positivism) then corresponds to integration over the measure space, and conversely.

Suppose A is an observable, i.e., a self-adjoint operator, with real eigenvalue a and normalized eigenket |a⟩ . In other words, A|a⟩=a|a⟩,⟨a|a⟩=1. Suppose further that A and B are canonically conjugate observables, so [A,B]=iℏI, where I is the identity operator. Compute, with respect to |a⟩ , the matrix elements of this equation divided by iℏ : 1iℏ⟨a|[A,B]|a⟩1iℏ(⟨a|AB|a⟩−⟨a|BA|a⟩)=⟨a|I|a⟩=<a|a>. In the first term, let A act on the bra; in the second, let A act on the ket: 1iℏ(a⟨a|B|a⟩−a⟨a|B|a⟩)=<a|a>. Thus, 0=1. This is my favourite "proof" of the well-known equation 0=1 . What gives? In order not spoil other people's fun, it might be best to put "spoiler" at the top of any post that explains what's happening. Regards, George

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Super Nade

#2 May26-06, 08:53 AM

P: 20

Quote by George Jones Suppose A is an observable, i.e., a self-adjoint operator, with real eigenvalue a and normalized eigenket |a⟩ . In other words, A|a⟩=a|a⟩,⟨a|a⟩=1. Suppose further that A and B[/itex]arecanonicallyconjugateobservables,so[tex][A,B]=iℏI, where I is the identity operator. Compute, with respect to |a⟩ , the matrix elements of this equation divided by iℏ : 1iℏ⟨a|[A,B]|a⟩1iℏ(⟨a|AB|a⟩−⟨a|BA|a⟩)=⟨a|I|a⟩=<a|a>. In the first term, let A act on the bra; in the second, let A act on the ket: 1iℏ(a⟨a|B|a⟩−a⟨a|B|a⟩)=<a|a>. Thus, 0=1. This is my favourite "proof" of the well-known equation 0=1 . What gives? In order not spoil other people's fun, it might be best to put "spoiler" at the top of any post that explains what's happening. Regards, George I don't think you can do that because A and B don't commute?

George Jones

#3 May26-06, 09:05 AM

Mentor P: 6,235	Quote by Super Nade I don't think you can do that because A and B don't commute? That step is OK. One way to see this is to take |b> = A|a> and |c> = B|a>, and then to consider <b|c>. Any is to to look at (AB)^* = B^* A^* = B A, which takes care of the order of the operators. Regards, George

selfAdjoint

#4 May26-06, 10:31 AM

Emeritus PF Gold P: 8,147	Dirac Proves 0 =1 Isn't this the one about the domains of the operators?

George Jones

#5 May26-06, 12:55 PM

Mentor P: 6,235	Quote by selfAdjoint Isn't this the one about the domains of the operators? I don't think the problem is with domains. I think it is possible for the intersection of the domains of A, B, and [A , B] to be dense, and to still have the proof be "true". Regards, George

waht	#6 May26-06, 01:12 PM

P: 1,636	Interesting, but the proof is based on an assumption that A and B are canonically conjugate observables. Therefore 0=1 is constrained to that condition. how does <a|[A,B]|a> = <a|AB|a> - <a|BA|a> btw?

Physics Monkey

#7 May26-06, 01:37 PM

Sci Advisor HW Helper P: 1,333

What a wonderful proof! I have never seen this one before, George. My discussion is below. ***SPOILER*** Think about the real line where we can represent the algebra by the usual quantum mechanical operators X and P. The key is to realize that X and P have no normalizable eigenvectors! The usual "normalization" for position "eigenstates" (lots of scare quotes) is ⟨x|x′⟩=δ(x−x′) , so let's have some fun with this formula. Since X and P are canonically conjugate we have that [X,P]=iℏ , and we can take matrix elements of both sides. The right side is ⟨x|iℏ|x′⟩=iℏδ(x−x′) . The left side is (x−x′)(−iℏddxδ(x−x′)) where I have used ⟨x|P=−iℏddx⟨x| . Thus we appear to have stumbled onto the rather cute identity −xddxδ(x)=δ(x) . Go ahead, try it under an integral, it actually works! I love such silly little formulae between wildly singular objects. A further amusing challenge: It isn't always true that the derivative operator has no eigenstates. Suppose you look at the derivative operator on a finite interval. It turns out that the Neumann indices are (1,1), and thus self adjoint extensions exist which are parameterized by a phase (the boundary condition). One can now find proper eigenfunctions and eigenvalues for a given self adjoint extension of the derivative operator. Are we therefore back to proving that 0 = 1 or what?

mikeu

#8 May26-06, 02:46 PM

P: 60

Quote by waht Interesting, but the proof is based on an assumption that A and B are canonically conjugate observables. Therefore 0=1 is constrained to that condition. Except you can always find such A and B, so you can always find 0=1..... :) Quote by waht how does <a|[A,B]|a> = <a|AB|a> - <a|BA|a> btw? It's just the definition of the commutator and linearity of the inner product: ⟨a|[A,B]|a⟩=⟨a|(AB−BA)|a⟩=⟨a|AB|a⟩−⟨a|BA|a⟩ . Physics Monkey, I've got a question about your spoiler below.... *** SPOILER cont. *** I suspected (based on X and P ) that delta distributions would enter into it, since we end up with 1iℏ(a−a)⟨a|B|a⟩=1 so it is clear that ⟨a|B|a⟩ must be ill-defined (i.e. infinite) to get something like " 0⋅∞=1 ." Recovering the definition of the derivative of the delta was neat. What I still don't see though is what the flaw in the proof is in the case of discrete operators...? George, I thought of another 'interpretation' of the 'proof' too: you could prove 0=ih => h=0 => things aren't quantized

George Jones

#9 May26-06, 03:25 PM

Mentor P: 6,235

Quote by Physics Monkey A further amusing challenge: It isn't always true that the derivative operator has no eigenstates. Suppose you look at the derivative operator on a finite interval. It turns out that the Neumann indices are (1,1), and thus self adjoint extensions exist which are parameterized by a phase (the boundary condition). One can now find proper eigenfunctions and eigenvalues for a given self adjoint extension of the derivative operator. Are we therefore back to proving that 0 = 1 or what? So, you want to take A = P and B = X for the Hilbert space of square-integrable functions on the closed interval [0 , 1], say. SPOILER for Physic Monkey's Challenge. It looks like, appropriately, selfAdjoint was right - domains are important. For the operator PX, operating by X on an eigenfunction of P results in a function that is not in the domain of selfadjointness for P, so P cannot be moved left while remaining to be P. Easy direct calculations in this example reveal a lot. As I said in another thread, if A and B satisfy [A , B] = ihbar, then at least one of A and B must be unbounded. In example of functions on the whole real line, both X and P are unbounded, while for functions on [0 ,1], X is bounded and P is unbounded. Regards, George

George Jones

#10 May27-06, 06:29 AM

Mentor P: 6,235	Quote by Physics Monkey What a wonderful proof! I have never seen this one before, George. Time to come clean! I lifted (and addded a liitle elaboration) this example from the Chris Isham's nice little book Lectures on Quantum Theory: Mathematical and Structural Foundations. My discussion is below. Very interesting discussion! Regards, George

Physics Monkey

#11 May28-06, 10:10 PM

Sci Advisor HW Helper P: 1,333

Quote by George Jones It looks like, appropriately, selfAdjoint was right - domains are important. For the operator PX, operating by X on an eigenfunction of P results in a function that is not in the domain of selfadjointness for P, so P cannot be moved left while remaining to be P. Very good, George. The commutator is indeed ill defined on the momentum eigenstates. Quote by George Jones I lifted (and addded a liitle elaboration) this example from the Chris Isham's nice little book Lectures on Quantum Theory: Mathematical and Structural Foundations. Well then, I think I might have to take a look at Isham's book. Quote by George Jones Very interesting discussion! Thanks for the interesting post! P.S. To all you readers out there, I can't resist telling about some nice physical applications of such ideas. It turns out that the self adjoint extensions of the momentum operator on a finite interval describe physically the problem of a particle on a ring with a magnetic field through the ring. This is in turn equivalent to imposing a 'twisted' boundary condition ψ(x+L)=eiαψ(x) on the wavefunction for a particle on a ring with no magnetic field. But there's more! Impurities in a metal can localize electronic states and cause a metal to become an insulator. One way to tell if you have localized states is to look at how sensitive such states are to the boundary conditions of your sample. The above ideas can then be applied, and you can relate the question of localization to the behavior of the system under an applied magnetic field (a problem which can be attacked with perturbation theory). And you thought self adjoint extensions were dull! Shame on you.

reilly

#12 May30-06, 10:54 PM

Sci Advisor P: 1,082

First, if "0 = 1" is true then QM completely falls apart, sorta like proof by contradiction, and "0=1" is certainly a contradiction. That tells me that the various proofs must be incorrect, or most physicists have been living like Alice in Wonderland. The problem is that P X | x> is not equal to P|x> x. As in, go to an x position representation in which P = -i d/dx. That is, P X |x> = -i d/dx x |x> = (-i + {-i x d/dx})|x> Delta functions and domaines are not at issue Sometimes abstraction can lead even the best astray. Think about Wick's Thrm, which would not hold if "0 =1" were true, nor would many standard manipulations of creation and destruction operators be legitimate. . (For the abstract truth about momentum operators see Hille and Phillips, Functional Analysis and Semi Groups, Chap XIX, which discusses translation operators (d/dx) in great and highly rigorous detail. The authors demonstrate that there really is not a problem with such operators. Again, if "0=1" then QM is inherently mathematically trustworthy, which seems to me to be a completely absurd idea. Regards, Reilly Atkinson

Hurkyl

#13 May30-06, 11:29 PM

Emeritus Sci Advisor PF Gold P: 16,092

Sometimes abstraction can lead even the best astray. Any notation can lead people astray. But abstraction has the advantage that there are fewer messy details, which means less opportunities to make mistakes, and less possibility for those mistakes to be obscured. Avoiding abstraction certainly doesn't prevent one from making mistakes... P X |x> = -i d/dx x |x> = (-i + {-i x d/dx})|x> such as overworking your variables. The x in d/dx is not the same as the x as in |x>; the former is the coordinate variable of the position representation, and the latter is a constant denoting which position eigenstate we've selected. If I relabel the variables so x is no longer being overworked, we're looking at -i d/dx x |a>. (And don't forget that x |a> = a |a>) You could rewrite George's entire post in the A-representation (so that A = x, and B = -ih d/dx), but that doesn't resolve the paradox: you still wind up with 0 = 1. Think about Wick's Thrm, which would not hold if "0 =1" were true, nor would many standard manipulations of creation and destruction operators be legitimate. . That's not accurate: if 0=1 were true, then everything is true. (And simultaneously false) Again, if "0=1" then QM is inherently mathematically trustworthy, which seems to me to be a completely absurd idea. I'm completely confused by this.

Gokul43201

#14 May31-06, 12:25 AM

Emeritus Sci Advisor PF Gold P: 11,155

Can we go over this again, slowly ? This is something that has bothered me for a little while. Quote by Physics Monkey To all you readers out there, I can't resist telling about some nice physical applications of such ideas. It turns out that the self adjoint extensions of the momentum operator on a finite interval describe physically the problem of a particle on a ring with a magnetic field through the ring. This is in turn equivalent to imposing a 'twisted' boundary condition ψ(x+L)=eiαψ(x) on the wavefunction for a particle on a ring with no magnetic field. Is L the circumference of the ring ? Does this not destroy the single-valuedness of ψ(x) ? Or is that what is being probed ? I think I've drunk too deep from the cup of Periodic BCs, what with all the goodies like flux quantization in SCs and Brillouin zones in crystals that it has thrown up like so many marshmallows! But there's more! Impurities in a metal can localize electronic states and cause a metal to become an insulator. One way to tell if you have localized states is to look at how sensitive such states are to the boundary conditions of your sample. The above ideas can then be applied, and you can relate the question of localization to the behavior of the system under an applied magnetic field (a problem which can be attacked with perturbation theory). And you thought self adjoint extensions were dull! Shame on you. Help me understand this, please. Let's start with a simple case : the Anderson hamiltonian for non-interacting electrons in a cubic lattice. The Hamiltonian consists of your favorite on-site disorder potential and the usual hopping term (nn, say). You then apply the above boundary condition to the single-particle eigenfunction in one or more directions. Ignoring what this means for now, this allows you to Taylor expand the eigenvalues Ei(α) and look at the coefficients of higher order terms in α . The deviations from 0 of these coefficients is what you call the phase sensitivity? If that's true, how exactly is this a "measure" of localization? Is the point to extract a dimensionless number (like T/U) and looking for a scaling law? And if not, what happens next?

lalbatros

#15 May31-06, 01:44 AM

P: 1,235

George Jones, Physics Monkey, Would that "0=1" contradiction be a proof that no finite-dimentional matrix could satisfy the commutation relation [A,B]=iℏI ? Would it possible to see that easily for two-dimentional matrices? Michel