Henning Makholm

426 reputation

bio	website	henning.makholm.net
	location	Copenhagen, Denmark
	age	39
visits	member for	11 months
	seen	Apr 16 at 13:06
stats	profile views	52

	426 reputation	bio	website	henning.makholm.net	visits	member for	11 months
9 badges		location	Copenhagen, Denmark		seen	Apr 16 at 13:06

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Apr 15	awarded	Nice Answer
Apr 11	answered	Are Mathematical Physics and Occam's Razor compatible?
Dec 30	awarded	Nice Question
Sep 13	revised	Magnetic paradox in relativity? comment on "still quite good"
Sep 13	awarded	Teacher
Sep 13	answered	Magnetic paradox in relativity?
Aug 19	comment	Explosion in space @Frank: I suspect Jerry meant "interstellar" rather than "intergalactic". The density if the interstellar medium varies but is on the rough order of one molecule per cm³. The chance that a molecule will hit something as it passes through any particular cubic centimeter is rather small, but there are so very many cubic centimeters to pass through that it eventually will hit something. (Also note that if the interstellar gas is ionized, the ions don't need to hit each other perfectly in order to exchange some momentum).
Aug 18	comment	Understanding the Bose-Fermi dichotomy Or (lightbulb!) is the axiom that the "physical subspace" must not contain two independent states $\psi$, $\phi$, such that $\langle\psi\|A\|\psi\rangle=\langle\phi\|A\|\phi\rangle$ for all observables $A$? If so, that would connect the rule with some intuitive sense of "indistinguishable".
Aug 18	comment	Understanding the Bose-Fermi dichotomy (cont.d ...) the Hamiltonian and all observable operators preserve that "physical" subspace, then the fact that they commute with every permutation of the particles seems to become a deductive consequence of the "only one state" axiom, rather then an assumption that defines indistinguishability. Is that right?
Aug 18	comment	Understanding the Bose-Fermi dichotomy I think I understand what you're saying here (though I have some trouble identifying the right ontological sense of "there is one and only one state" -- what exactly does it mean for a state to "be" here?). But I'm afraid I have lost sight of what is the role of the Hamiltonian in the argument. Is there any left? If we just adopt as an axiom that there's a "physical" subspace of my Hilbert space that contains only one linear combination of {\|ABC>, \|ACB>, \|BAC>, \|BCA>, \|CAB>, \|CBA>} (and presumably also of {\|AAB>, \|ABA>, \|BAA>} and so forth, right?), and that (... cont.d)
Aug 18	awarded	Commentator
Aug 18	comment	Understanding the Bose-Fermi dichotomy Okay, I can follow that. If every permutation commutes with $H$, then the permutation group operates on each eigenspace separately, and must therefore work as a 1D representation on each 1D eigenspaces. Is what you're saying then that there's an empirical (rather than deductive) finding that vectors in multi-dimensional eigenspaces of $H$ never occur in reality?
Aug 18	comment	Understanding the Bose-Fermi dichotomy "Every eigenvector of $H$ is either symmetric or antisymmetric" is what you mean by "the wavefunctions which diagonalize the Hamiltonian have to be in a one dimensional representation of the permutation group", right?
Aug 18	comment	Understanding the Bose-Fermi dichotomy I'm considering a system of three particles $\{1,2,3\}$ which each can occupy one of the three coordinates $\{A,B,C\}$. The Hilbert space has the basis vectors \|AAA>, \|AAB>, \|AAC>, \|ABA>, \|ABB>, et cetera, 27 of 'em in all. If $P$ swaps particles 1 and 3, then P\|ACC>=\|CCA>, P\|BCA>=\|ACB> and so forth. I don't have any particular Hamiltonian in mind -- you get to choose that. My point is as far as I can see, no possible Hamiltonian commuting with $S_3$ has the property that every eigenvector of it is either a symmetric or antisymmetric combination of the basis vectors.
Aug 18	comment	Understanding the Bose-Fermi dichotomy I'm afraid that's mutual; I don't understand your argument. As I said, I don't know what "degeneracy" (which appears to be a key concept in the argument) means, and "second quantization" is something I recognize as jargon but don't know the meaning of either. In any case, you seem to have turned my discrete example into a continuous one (where, accordingly, dimension-counting won't work), and made an assumption on the shape of the Hamiltonian that I don't see where comes from. And again, I don't see where you're pulling the "it must be a one-dimensional representation" requirement from.
Aug 18	comment	Understanding the Bose-Fermi dichotomy @Qmechanic: I probably shouldn't have mentioned the anyon thing (which I don't understand anyway) -- I only meant that as circumstantial evidence that perhaps the argument Feynman had in mind doesn't quite work anyway. What I'm trying to understand is just what's the deal in ordinary undergraduate QM.
Aug 18	revised	Understanding the Bose-Fermi dichotomy remove advanced tags -- I'm not interesting in spin right now, only how statistics work
Aug 18	comment	Understanding the Bose-Fermi dichotomy I know some basic definitions of representation theory, but I fail to see how you get it to enter here. As far as in can see in the 3×3 example there is no possible Hamiltonian that one can write down such that every eigenvector of the Hamiltonian is either symmetric or antisymmetric, because there would be 16 dimensions missing. What is wrong in that argument?
Aug 18	comment	Understanding the Bose-Fermi dichotomy "the fact that the wavefunctions which diagonalize the Hamiltonian have to be in a one dimensional representation of the permutation group." I must be misunderstanding this somehow, because it seems to say that the Hamiltonian in my 3-particles-in-3-states example can only have 1+10 dimensions of eigenspace in total -- but unless I'm really misremembering linear algebra, the sum of the eigenspace dimensions for a diagonalizable matrix must equal the total dimension of the space it operates on. What's going on here? Am I counting the eigenspace dimensions wrong?
Aug 18	comment	Understanding the Bose-Fermi dichotomy No, I don't "probably know from quantum mechanics, symmetries generically lead to degeneracies". As I said, I'm an amateur and probably reading the wrong books in the wrong order, so I don't have a clear understanding of what "degeneracies" and "degenerate" means in this context. Your "This leaves us with" seems to jump over the very point that confuses me. Sure, eventually I'd like to understand spin-statistics, but for the nonce I'm just trying to to figure out the "statistics" part.