# Quantum mechanics: Myths and facts

1. Nov 9, 2006

### Demystifier

A common understanding of quantum mechanics (QM) among students and practical users is often plagued by a number of "myths", that is, widely accepted claims on which there is not really a general consensus among experts in foundations of QM. These myths include wave-particle duality, time-energy uncertainty relation, fundamental randomness, the absence of measurement-independent reality, locality of QM, nonlocality of QM, the existence of well-defined relativistic QM, the claims that quantum field theory (QFT) solves the problems of relativistic QM or that QFT is a theory of particles, as well as myths on black-hole entropy. The fact is that the existence of various theoretical and interpretational ambiguities underlying these myths does not yet allow us to accept them as proven facts. In
http://arxiv.org/abs/quant-ph/0609163
I review the main arguments and counterarguments lying behind these myths and conclude that QM is still a not-yet-completely-understood theory open to further fundamental research.

2. Nov 9, 2006

### zbyszek

Hi,
I just have seen your e-print on arXive. I need more time to analyze it throughly but have already two remarks. I was particularly interested in time-energy uncertainty relations
of QM, so I jumped there firts to see your views on the subject. Two things:

1. The Pauli's argument on nonexistance of self-adjoint time operator in QM due to
the unboundness of Hamiltonian operators is mathematically unsound.
He didn't pay much attention to the domains of the operators. You can find
the explanation in http://xxx.lanl.gov/abs/quant-ph/9908033 by Galapon.

2. It can be shown explicitly that one can actually determine all energetic properties
of a quantum system with any accuracy in arbitrarly short time. I have worked out
a general example in http://xxx.lanl.gov/abs/quant-ph/0412073 that eliminates several versions of the uncertainty relation.
It is simply shown how to get energies from arbitrarily short autocorrelation function.

Don't get discouraged! The myth tracking is as important as discovering new phenomena.

Cheers!
zbyszek

3. Nov 9, 2006

### Demystifier

Thanks for the remarks.
Note however that my paper is not intended to be an exchaustive review with all relevant references, but rather an introductory pedagogic review.

4. Nov 9, 2006

### zbyszek

I don't demand the paper to be complete. I would be happy if it was correct, but it is not.

On first four pages you perpetuate more myths than you demistify. To be precise
you demistify none.

1. You start well with the observation that QM provides probabilities that have been confirmed in numerous experiments. Then you try to argue that an electron is a wave
rather than a particle. How do you know that QM describes a single electron?
From the introduction one would expect that QM is about statistical ensambles (i.e.
a wave function of ELECTRON corresponds to the ensamble of electrons, independently
prepared and handled) rather than to a single particle. Deduction of properties
of a member of the ensamble from the wave function needs justification. Do you have one?
What you actually do is to ascribe wave- or particle-like properties to an ensamble.

2. In the section with Heisenberg uncertainties you write that "one cannot measure both the particle momentum and the particle position with arbitrary accuracy". Can you back
this statement up with any evidence? Experiment shows that if you prepare N quantum
objects in state |psi> each then measurement of position on N/2 of them and measurement of momentum on the second half will give dispersions related by the
Heisenberg uncertainty. It says nothing about measurement of position and momentum
of a single particle, does it?

3. Along those lines, can anybody predict an outcome of an experiment performed
on a single quantum object like a single electron passing through double slit? Can anybody
tell where the electron will hit a screen with certainty? If the answer is NO then QM does
not describe a single electron but an ensamble of single electrons.

4. You argument on time-energy uncertainty is bad. You say that because time is
not an operator in QM but a mere parameter then pluggin it into Heisenberg uncertainty
makes no sense. Well, it makes sense. A parameter is a special case of an self-adjoint
operator and one can introduce it into the uncertainty relation. The problem is that
the relation becomes 0 >= 0, which in turn does not give any grounds for the uncertainty
relations.

5. You defend non-existence of time operator in QM referring to the wrong argument
by Pauli. In fact one can construct a self-adjoint operator that has dimension of time
and is dual to a bounded from bellow Hamiltonian!

6. The argument against some time-energy uncertainty relation relays on a counterexamples. Bohm and Aharonov provided one counterexample to the
Landau's uncertainty relation and I gave
a class of counterexamples that handle all discrete spectrum quantum systems
and most of the false time-energy uncertainty relations.

7. There are valid time-energy uncertainty relations. They are rigorously derived from
axioms of QM. One of them worked out by Mandelshtam and Tamm in 1945. You will find
a reference in the Bohm and Aharonov's paper.

The myths in QM are still in the textbooks and it is very difficult to get rid of them.
If you care fighting them, do it right.

Cheers!

5. Nov 13, 2006

### Demystifier

Zbyszek, your remarks show that one of the main points of my paper is correct: That there is no consensus among the experts about some fundamental questions on QM. Do you agree at least with that?

6. Nov 13, 2006

### zbyszek

Hi Demystifier,

I would say that in MTV era the word 'expert' is heavily overloaded :).

Cheers!

7. Nov 23, 2006

### Crosson

zbyszek, I have a comment on your point 2.

I agree, as part of a sensible physical theory (actual experiments), or a purely mathematical theory (no experiments or measurements, just distributions), the uncertainty relations must be interpreted in terms of ensembles.

But consider the following realist point of view: If one could measure with unlimited accuracy the position and momentum of a single particle at some time, then it would in principle be possible to measure the position and momentum of each one of an ensemble of particles. Taking the contrapositive of the previous implication shows the justification for the the (agreeably naive and ripe for misunderstanding) statment "QM predicts that we cannot know exactly the position and momenta of a particle".

8. Nov 24, 2006

### serienumerica

Give me equations and the solutions ,that is numbers ! if anyone wants to expand or ammend an old theory it must be formulated by means of an equation that improves the old results.

www.geocities.com/serienumerica

9. Nov 24, 2006

### swimmingtoday

<<2. In the section with Heisenberg uncertainties you write that "one cannot measure both the particle momentum and the particle position with arbitrary accuracy". Can you back
this statement up with any evidence? Experiment shows that if you prepare N quantum
objects in state |psi> each then measurement of position on N/2 of them and measurement of momentum on the second half will give dispersions related by the
Heisenberg uncertainty. It says nothing about measurement of position and momentum
of a single particle, does it?>>

The standard theoretical formulation of QM does say something about what occurs to a single particle.

If the particle is measured into a position-eigenstate, then by Fourier Analysis, one sees that the distribution of the spatial frequencies has an infinite spread, and thus the spread in the momenta is infinite--it cannot have a unique momentum.

You of course can try to do an experiment contradicting the accepted theory, but the accepted theory has worked so well in the regime we are discussing that it is doubtful it will be contradicted regarding the sort of things we are discussing.

10. Nov 24, 2006

### zbyszek

Could you clarify to me the first implication?

A) If it means that exact measurement of the position and momentum of a single particle
applyies to each member of an ensemble then your (contrapositive) transposition
is incorrect. Because even if you are able to measure exactly x and p for each
member it does not mean an exact (dispersion free) x and p of an ensemble.

B) If that implyies that exact x and p for an ensemble is possible then this is wrong
from the start.

Here is the correct statement: Lack of dispersion free x and p of an ensemble
does not imply that its members inherit this feature.

Cheers!

11. Nov 24, 2006

### zbyszek

You didn't understand. We are not talking here what will happen with the momentum of
the particle after it has been localized at, say, x0.
We discuss a particle in some state |s>, that is given once and for all. For this state
there is a mathematical theorem called Heisenberg uncertainty principle. The only
measurements we are allowed to perform is on a system in the state |s>.
If one measurement reveals the particle in the state |x0> then we are through with it.

But you are right that if one is interested in state |x0>, or better yet if |s>=|x0>,
then measurement of ensemble of particles prepared in |x0> will give infinite
momentum spread. It doesn't mean however that in a single measurement you will
obtain some momentum with an infinite error bars.

Cheers!

12. Nov 24, 2006

### swimmingtoday

"We discuss a particle in some state |s>, that is given once and for all. For this state
there is a mathematical theorem called Heisenberg uncertainty principle. The only
measurements we are allowed to perform is on a system in the state |s>."

On one hand you call [s> the state of a single particle, and on the other hand you call [s> the state of a system. This cannot be unless there is only one particle in the system.

I also find it curious that you informed me about this thing called the Heisenberg Uncertainty Principle. Did you think I was unaware of it? I actually made reference to it in its natuaral (Fourier) form. Were you not aware that the Uncertainty Principke is a statement of Fourier relations?

You had written <<Heisenberg uncertainty. It says nothing about measurement of position and momentum
of a single particle, does it?>>. That is just wrong. The Heisenberg Uncertainty Principle applies to individual particles, not to ensembles. For example, in its simplest form, you get that the fact that he commutator of the X operator with the p operator is non-zero, implies uncertainty beteween position in momentum. The X operator is the position operator for a SINGLE particle, and likewise for the P operator. What you are doing is not quantum physics, or the Heisneberg Uncertainty Principle. Maybe you are correct and conventional quantum physics is wrong, but I doubt it, and even if you were correct, you should not refer to your beliefs as "The Heisenberg Uncertainty Principle". It isn't.

13. Nov 24, 2006

### gongchangjie

thanks for sharing!

14. Nov 24, 2006

### zbyszek

To be precise measuring a single particle state |s> means performing a measurement
on an ensemble of quantum systems each consisting of one particle prepared
in the same way (labeled by 's').

So, you are right system consists of one particle, but state |s> is a property of an
ensemble identically prepared systems.

Notice, that "identically prepared" does not mean the same. They will give different
results if measured.

I was merely pointing out that in that theorem besides operators there is also a state.

I am still not aware of that. UP has nothing to do with Fourier relations. It applies to
all selfadjoint operators with approprieate domains.

It may come as a surprise to you, but a SINGLE electron wave function does not
describe an individual electron. It describes an ensemble of identically prepared single electrons. Did you ever hear that someone measured a single electron that was spread
in space, say, like its wave function? Do I have to elaborate?

Cheers!

15. Nov 24, 2006

### swimmingtoday

<<So, you are right system consists of one particle, but state |s> is a property of an
ensemble identically prepared systems. >>

No, it isn't. There is no "ensemble". You are not doing real physics--you have some sort of meta-physics fetish about emsembles.

States of electrons in hydrogen atoms are labeled in the form [n, l, m. ms>. A hydrogen electron in the state [2, 0, 1. +> refers to a SINGLE electron in a SINGLE hydrogen atom with those 4 quantum numbers--there is nothing that has to do with an ensemble. You must have seen the standard quantum-number desription of an electron (a SINGLE eelctron) in a hydrogen atom.

<<So, you are right system consists of one particle, but state |s> is a property of an
ensemble identically prepared systems. >>

No, the state [ 2, 0, 1, +> refers to one electron. There is no ensemble.

<<[The Uncertainty Principle] has nothing to do with Fourier relations.>>

Ever think it possible that there is something you just don't know about? The uncertainty principle is a rule about Fourier decomposition. Consider, for example, the uncertainty relationship between position and momentum. The momentum spectrum is the spectrum of spatial frequencies. If a wave-function is a single spatial frequency, then there is no uncertainty in the momentum. But if the wavefunction can be decomposed into frequency components with numerous frequencies, frequencies with a large sprerad, then there is a lot of uncertainty in the momentum. Now consider the position spread. If a wave-function is highly localized, let's say it is non-vanishing only at x=0, then by using Fourier analysis, one finds that there is a wide distribution of its spatial frequencies. To construct a function completely localized at x=0, Fourier analysis tells you that you need to add up waves of every spatial frequency--it is a simple problem in Fourier analysis.

Fourier analysis tells us that functions that are spatially compact (as would be the case for a wave-function that had little uncertainty in position) have wide frequency spreads (as would a wave-function highly uncertain in momentum); and that functions that are composed of a single spatial frequency, or a small number of different waves of similar frequency (as would be the case for a wave-function with little uncertainty in momentum) are highly unlocalized spatially (as would be the case of a wave-function with large uncertainty in position)

<<It applies to
all selfadjoint operators with approprieate domains.
>>

That's Fourier analysis!!! Self-adjoint (i.e Hermitian) operators have eigenvectors that form an orthonormal basis. Fourier analysis is the mathematicals of orthonormal bases.

<<It may come as a surprise to you, but a SINGLE electron wave function does not
describe an individual electron. >>

Yes it does. (Technically, because of identical particle physics, an electron is part of a joint wave-function of every electron in the Universe, but that is not what he is talking about)

<<Did you ever hear that someone measured a single electron that was spread
in space, say, like its wave function? >>

Only because you do not often hear of electrons being measured in real experiments. For example a standard way for an electron to be well spread in space is for it to be a conduction electron in a metal. But solid state physics does not usually involved Measurement Theory experiments. It usually involves mundane things like heat conduction etc.

<<Do I have to elaborate?>>

I think you do!

16. Nov 24, 2006

### zbyszek

O.K. Lets start it slow.

I didn't say 'often'. I said 'ever'. Did you?
You don't have to stick to the electron, it can be any other single quantum object.
Did you?

17. Nov 25, 2006

### dextercioby

Funny that the author of the article didn't mention as a myth the Stone-von Neumann uniquness theorem and its known implications. Also the myth that Hilbert space mathematics is enough to account for the description of all phenomena.

Daniel.

18. Nov 26, 2006

### reilly

Demystifier -- My experience over the last 40 years or so is first that virtually nobody refers to "so-called" Feynman diagrams -- because of their extraordinary role in the progress made in high energy physics over the past 60 years or so, most of us greatly congratulate Feynman on his extraordinary genius -- so, in most places the correct terminology is just plain, Feynman Diagrams.

Second, I've never encountered a particle physicist who believed that Feynman's diagrams were an actual pictures of what happens in an interacting system. If, for no other reason, than they represent terms in the perturbation series. Being extremely tolerant of convergence worries for QFT perturbation theories, I can state that it's the sum of the Feynman diagrams which really counts, not any particular diagram. But that's not really the point: as Julian Schwinger once said, "Feynman brought QED computations to the masses" Let's not forget that these "masses" were a very bright bunch of folks, and certainly they were sufficiently sophisticated to know that the name or description of something is not the something itself.

And how did this sophistication show? Through that old literary device, the metaphor. In fact, Feynman diagrams are just one of the languages used to describe interacting systems. Turns out that any formal problems due to perhaps less a than fully rigorous approach are quite trivial when weighed against the spectacular successes of the diagrammatic language. My experience tells me that your Feynman diagram myth is just another myth itself.

And, by the way, the whole idea of "wave-particle" duality comes from an ancient time in which physicists were blindsided by nature and impotent language. There is little doubt that electrons are particles. Their behavior, as seen by Davisson & Germer, blew more than a few minds. Now we are relatively comfortable in stating that electrons, protons, etc are particles described by probabilities derived from wavefunctions, solutions of an appropriate Schrodinger eq. But to Bohr and all the heavyweights of his day, Davisson_Germer ought not to have happened, because it appeared to suggest not wave -or -particle, but wave and particle.

Photons are different -- as Newton and Wigner showed, they cannot be localized, and so cannot really be particles. But photoelectric reactions can be safely described as point reactions, at least for a macroscopic take on things. There's a huge literature on photons, the properties and descriptions thereof, lots of which can be found in Mandel and Wolf's, Optical Coherence and Quantum Optics.

Also, the particle number of any interacting theory is obtained by means of the appropriate number operator. All the states in the theory can be expanded in eigenstates of the number operator (Fock space to be exact). In QED and other relativistic field theories, number operators do not commute with the Hamiltonian, so eigenstates of the Hamiltonian are superpositions of number eigenstates. For example, photon coherent states provide a solution to the problem of a classical current interacting with a quantum E&M field (See Mandel and Wolf). For these states the probability distribution (density) is a Poisson distribution, hence the coherent field is composed of states of all possible numbers of photons. Nonetheless, it is sometimes convenient to work with coherent states in Fock Space -- it all depends. But the number operator is as valid as are the operators for angular momentum, or of iso-spin. (Your assertion re Eq. 93 in your paper is, at best, problematic. Usually we use free particle creation and destruction operators, that is we work in ordinary Fock Space. All of this is spelled out in probably thousands of books and papers over the past 70 years or so.)

Re relativistic QM. Your discussion is a bit on the old side. For a brilliant, compelling, and elegant more modern approach to relativistic QM see Vol. I of Steven Weinberg's, The Quantum Theory of Fields. Ever since the work of Jacob and Wick on using helicity - momentum states, and rotation operators to generate "IN" and "OUT" states in particle physics, there's been a move to more use of group theory in developing the basic states of relativistic QM. There's no ambiguity about anything, and there's a uniform way to deal with particle states of any spin. Further, Weinberg will walk you through the development of QFTs. And, of course, the problems with negative energy go away.

Is QFT a theory about particles? Who cares? It's a theory that can come at you with many different looks. QFT is just normal, standard QM specified in a particular language.

Regards,
Reilly Atkinson

19. Nov 26, 2006

### Careful

That a pretty strange thing to say. Take for example a classical interacting field theory, split the interaction in a bound'' and radiative'' part if possible and solve the linear equation for the bound problem. Then, proceed by iteration, now at each order in the iterative process (hence order in the coupling constant) you will encounter corrections due to scattering processes'' involving lower order scattering amplitudes of the bound solutions. Actually, I would even go further : a *well chosen* perturbative expansion allows you understand the physical process at hand and is superior in that respect to a non perturbative treatment.

You do not have to worry so much about the total sum, I have little problem with chopping off the series at some order - there are actually good physical motivations for doing so. In some sense you can take these diagrams very literally, it is just so that the basis states in which they are expressed are not so well chosen.

You just rephrase wave particle duality here. The question would then be how a set of independent particles would form a pattern dictated by a wave equation.

Of course they can be localized, no sensible person is going to call a plane wave a photon, just consider a better orthonormal set of basis states.

Sure, but to swallow all of this you have to believe in a multiparticle wave function being the correct description (as well as in some physical content behind irreducible group representations) and this without elucidating the physical meaning of the one particle wave (note : you moreover assume that multi particle processes can be given a Hilbert space representation). In doing so, one could confuse a dog for a cat.

Cheers,

Careful

Last edited: Nov 26, 2006
20. Nov 26, 2006

### swimmingtoday

<<Photons are different -- as Newton and Wigner showed, they cannot be localized, and so cannot really be particles.>>

Photons can indeed be localized. A photon need not be a sine wave.

And ironically, an electron cannot be localized. At first glance it might seem likeyou just add up a bunch of Fourier components of different spatial frequencies, but one must remember that each frequency has a different ratio of the first spinor component to the second spinor component.