Suppose
Suppose further that where In the first term, let Thus, This is my favourite "proof" of the well-known equation 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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#2
May26-06, 08:53 AM
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P: 20
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#3
May26-06, 09:05 AM
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Mentor
P: 6,235
| 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 | |
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#4
May26-06, 10:31 AM
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Emeritus
PF Gold
P: 8,147
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Dirac Proves 0 =1
Isn't this the one about the domains of the operators?
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#5
May26-06, 12:55 PM
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Mentor
P: 6,235
| Regards, George | |
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#6
May26-06, 01:12 PM
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P: 1,636
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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? | |
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#7
May26-06, 01:37 PM
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Sci Advisor
HW Helper
P: 1,333
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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 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? | |
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#8
May26-06, 02:46 PM
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P: 60
| 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 George, I thought of another 'interpretation' of the 'proof' too: you could prove 0=ih => h=0 => things aren't quantized | |
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#9
May26-06, 03:25 PM
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Mentor
P: 6,235
| 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 | |
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#10
May27-06, 06:29 AM
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Mentor
P: 6,235
| 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. Regards, George | |
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#11
May28-06, 10:10 PM
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Sci Advisor
HW Helper
P: 1,333
| 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 | |
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#12
May30-06, 10:54 PM
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Sci Advisor
P: 1,082
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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 | |
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#13
May30-06, 11:29 PM
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Emeritus
Sci Advisor
PF Gold
P: 16,092
| Avoiding abstraction certainly doesn't prevent one from making mistakes... 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. | |
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#14
May31-06, 12:25 AM
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Emeritus
Sci Advisor
PF Gold
P: 11,155
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Can we go over this again, slowly ? This is something that has bothered me for a little while.
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! 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 | |
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#15
May31-06, 01:44 AM
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P: 1,235
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George Jones,
Physics Monkey, Would that "0=1" contradiction be a proof that no finite-dimentional matrix could satisfy the commutation relation Would it possible to see that easily for two-dimentional matrices? Michel
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