Saturday, August 9, 2014

Suppose A is an observable, i.e., a self-adjoint operator, with real eigenvalue a and normalized eigenket |a⟩ .

http://www.physicsforums.com/showthread.php?t=259127


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]=iI,


where I is the identity operator. Compute, with respect to |a , the matrix elements of this equation divided by i :

1ia|[A,B]|a1i(a|AB|aa|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:

1i(aa|B|aaa|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

Super Nade
#2
May26-06, 08:53 AM
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Quote 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]=iI,


where I is the identity operator. Compute, with respect to |a , the matrix elements of this equation divided by i :

1ia|[A,B]|a1i(a|AB|aa|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:


1i(aa|B|aaa|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
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Quote 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
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Dirac Proves 0 =1

Isn't this the one about the domains of the operators?

George Jones
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May26-06, 12:55 PM
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Quote 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
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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?

Physics Monkey
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May26-06, 01:37 PM
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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
x|x=δ(xx)
, so let's have some fun with this formula. Since X and P are canonically conjugate we have that
[X,P]=i
, and we can take matrix elements of both sides. The right side is
x|i|x=iδ(xx)
. The left side is
(xx)(iddxδ(xx))
where I have used
x|P=iddxx|
. 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
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Quote 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 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|(ABBA)|a=a|AB|aa|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
1i(aa)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
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May26-06, 03:25 PM
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Quote 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
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May27-06, 06:29 AM
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Quote 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
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May28-06, 10:10 PM
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Quote 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 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 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
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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

Hurkyl
#13
May30-06, 11:29 PM
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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
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May31-06, 12:25 AM
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Can we go over this again, slowly ? This is something that has bothered me for a little while.
Quote 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
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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
[A,B]=iI
?

Would it possible to see that easily for two-dimentional matrices?

Michel

An observable in QM is to be represented by a self-adjoint/hermitean operator, so that the eigenvalues are real numbers.

I could not find the answer to : are all self-adjoint/hermitean operators actually observables ?

Thank you for any help !
marlon
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Aug9-04, 04:31 PM
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yes !!! Why else would we need them in QM ?

All info is in the observables

regards
marlon
humanino
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Aug9-04, 04:35 PM
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What you are saying is :
if I can exhibit ANY self-adjoint operator, then you can provide an experiment to find its spectrum ?!
Can you elaborate please ?

humanino
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Aug9-04, 04:41 PM
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Self-adjoint/hermitean operator vs observables

Quote Quote by marlon
yes !!! Why else would we need them in QM ?
I read in :
http://www.physicsforums.com/showthread.php?t=9164

that only normal operators are needed :
[A,A]=0
.
And it is even argued that we use self-adjoint operators because they are easy to handle !

Quote Quote by Poska
We chose hermitian obserwables because they are easy. And we have good math. tools to get their spectrum, but there is no specyfic rule it's only rutine.
Poska apparently posted only one message, and received no answer.

I am currently confused
marlon
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Aug9-04, 04:56 PM
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What I am saying is that we need hermitian operators for two reasons. First the eigenvalues are real, which corresponds to fysical results. So yes any such operator produces a spectrum which has to correspond to some kind of real fysical situation that is to be verified in experiments. Second : we need them because they can be diagonalized with the eigenvalus on the diagonal. This makes the math a lot easier. Besides this is what we are always looking for when working with tensors in GTR for example. The more symmetry, the easier the matrix representations of the opertors get. The more zero's the better !!! that is the rule

regards
marlon
humanino
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Aug9-04, 04:56 PM
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I should add that :
http://groups.google.com/groups?sel...0pravda.ucr.edu
helped me to understand that one needs a complete set of commuting operators, spanning the entire space, and that it was realized by normal operators.

Then one can argue that, whatever the basis, one can change it for another one. The position basis for instance. Or the momentum basis. And any result in any basis can be expressed in the preffered basis, thus being observable.

This is rather disappointing to me.
humanino
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Aug9-04, 04:59 PM
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Hey Marlon by the way, I remember we are born almost the same week ! Thank you again for your help ! Please add any comment to my last post !
marlon
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Aug9-04, 05:01 PM
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Quote Quote by humanino
I read in :
http://www.physicsforums.com/showthread.php?t=9164

that only normal operators are needed :
[A,A]=0
.
And it is even argued that we use self-adjoint operators because they are easy to handle !


Poska apparently posted only one message, and received no answer.

I am currently confused
Attention, though we have QM-operators for which the above commutation-relation does not count. Such operators have no mutual eigenvalues. This is not good, because when you measure the first one (i.e. project it out on some chosen basis) you automatically influence the eigenvalues (and thus the measurement-results) of the second one. When the commutator is zero the operators can be diagonalized at the same time and their quantumnumbers entirely describe the wave-function of a system. E.G. : l (s-p-d-f-... energylevel) and m (magnetic number like spin up or down) quantumnumbers of an atom.
marlon
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Aug9-04, 05:03 PM
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Quote Quote by humanino
Hey Marlon by the way, I remember we are born almost the same week ! Thank you again for your help ! Please add any comment to my last post !

oui je sais, ceux de janvier sont les meilleurs...(je suis belge, voilà la raison que je parle le français)
marlon
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Aug9-04, 05:04 PM
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Quote Quote by humanino
I should add that :
http://groups.google.com/groups?sel...0pravda.ucr.edu
helped me to understand that one needs a complete set of commuting operators, spanning the entire space, and that it was realized by normal operators.

Then one can argue that, whatever the basis, one can change it for another one. The position basis for instance. Or the momentum basis. And any result in any basis can be expressed in the preffered basis, thus being observable.

This is rather disappointing to me.

why ??? you should be happy with that
humanino
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Aug9-04, 05:18 PM
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Well, not quite. My concern is : if one discovers a new operator, not commuting with the usual ones, but having some relevance and exhibiting self-adjointness. One can claim it is an observable, but does it necessarilly have its own physical interpretation ?

This is what happened with the spin ! The angular momentum was not conserved without spin, so one concluded it is an observable and particles do have intrisic angular momentum.

Now we know there is not any hidden variable, so we can safely hope we did not forget anything to describe entirely a particle, and there is no new operator to discover. I'm not quite confident about this last argument.
marlon
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Aug9-04, 05:30 PM
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Quote Quote by humanino
Well, not quite. My concern is : if one discovers a new operator, not commuting with the usual ones, but having some relevance and exhibiting self-adjointness. One can claim it is an observable, but does it necessarilly have its own physical interpretation ?

This is what happened with the spin ! The angular momentum was not conserved without spin, so one concluded it is an observable and particles do have intrisic angular momentum.

Now we know there is not any hidden variable, so we can safely hope we did not forget anything to describe entirely a particle, and there is no new operator to discover. I'm not quite confident about this last argument.
The way spin was introduced is a very nice example of how QM and even more QFT work. WE are always looking for conserved quantities because they correspond to certain symmetries of the math (here i am again with symmetry). The concept with total spin as a conserved quantity of which the operator is a Casimir-operator gives a very accurate description of the properties of atoms and so on...

You can be certain that for all atomic energy-levels we have the exact amount of necessary quantum-numbers that label each level, corresponding to experiments. More energy-levels and thus more operators could only occur when we change the fundamental properties of the elementary particles, but then we would be changing tyhe basis of QFT. Why do that, keeping in mind it gives us the best model up till now when it comes to atomic-phenomena : i.e. the standard model

regards
marlon
humanino
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Aug9-04, 05:38 PM
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I realize I might just as well question the validity of quantum mechanics ! Well, I hope every physicists needs doubts now and then.

What are you working on Marlon ?
marlon
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Aug9-04, 05:45 PM
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Quote Quote by humanino
I realize I might just as well question the validity of quantum mechanics ! Well, I hope every physicists needs doubts now and then.

What are you working on Marlon ?
just finished university in Ghent, master in theoretical fysics ; MY THESIS WAS ON THE QFT AND QUARKCONFINEMENT. In september i start to study nanotechnology and photonics at the university of Brussels. I hear that Paris is very high leveled on these subjects right ???
humanino
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Aug9-04, 05:54 PM
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They hope so, I can't really compare
Well, I went three month at McGill University last year to study photonics (Multi-tunable-wavelenghts pulsed laser, in fiber) They had pretty much the same level as back in France, but more money

MY THESIS WAS ON THE QFT AND QUARKCONFINEMENT
sounds hot topic (to me at least) I understand we could hear each other. But be carefull : I'm an experimentalist. Please talk slowly
humanino
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Aug9-04, 05:57 PM
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Quote Quote by marlon
MY THESIS WAS ON THE QFT AND QUARKCONFINEMENT.
Read the Polyakov's (father) "Gauge fields and strings" yet ?
I tried. :surprise:
marlon
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Aug9-04, 05:59 PM
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Quote Quote by humanino
They hope so, I can't really compare
Well, I went three month at McGill University last year to study photonics (Multi-tunable-wavelenghts pulsed laser, in fiber) They had pretty much the same level as back in France, but more money



sounds hot topic (to me at least) I understand we could hear each other. But be carefull : I'm an experimentalist. Please talk slowly
be happy, normally them experimentalists have a much bigger bank-account. I want to become one of them too, hence the reason for this nanotechnology thing. Oh, yes, i also find it interesting...

marlon the poor theory-guy

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