
#1
Jan212, 03:59 AM

P: 1,057

How do you derive what quantum mechanical momentum is, from some axioms about reality?
Therefore how do you justify one of the following more or less equivalent statements: [X,P]=ih <xp>=exp(ikx) psi(x,k)=exp(ikx) I've seen argumentations why quantum mechanics is set up with the mathematical framework it has (Hilbert space etc.), but no an explanation for momentum. 



#2
Jan212, 04:13 AM

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P: 1,725

Less flippantly, one can start with the Galilean group of transformations (for the nonrelativistic case), and note that the probabilities we measure in experiments are invariant under these transformations. Ballentine's textbook does a reasonably good job of developing (nonrelativistic) QM from this perspective. (For the relativistic case, start with the Poincare group instead.) 



#3
Jan212, 04:38 AM

P: 1,057

Could you give me a hint, why translation operators and momentum in reality as we know it are related? Translations probably yield the above form, but where does reality turn into translations? And can I derive classical mechanics from translations as well? Now I even wonder... what is the essence of momentum anyway? Is it more than just postulating that it is conserved? 



#4
Jan212, 04:53 AM

P: 1,746

Axiomatic derivation of [X,P]?
The operator of momentum P is *defined* as a generator of space translations. Then the translation by a finite distance [itex]a [/itex] is represented by the exponential function of the generator [itex] e^{\frac{i}{\hbar}Pa} [/itex]. The action of this transformation on the position operator X should result in a shift
[tex] e^{\frac{i}{\hbar}Pa} X e^{\frac{i}{\hbar}Pa} = Xa [/tex] From this postulate you can get the desired commutator [tex] [X,P] = i \hbar [/tex] Eugene. 



#5
Jan212, 05:17 AM

P: 836

This may be interesting:
http://www.hep.upenn.edu/~rreece/doc...esentation.pdf The equivalence you mentioned is arrived at midway. 



#6
Jan212, 05:28 AM

P: 1,057

OK, from these ideas I understand why translation operations give the quantum mechanics form of momentum.
So how are translation generators now consistent with the momentum we learned at school? 



#7
Jan212, 06:16 AM

P: 836

I think that requires some knowledge about Lie Groups. However, I found this short explanation. I haven't looked through it properly, but I looks like it answers your question.
http://www.dfcd.net/articles/momentumop.pdf 



#8
Jan212, 06:17 AM

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HW Helper
P: 11,863

[x,p]=ihbar 1 is the cornerstone of quantum mechanics. It's a postulate, unless one starts off with the Galilei group. So axiomatic derivation would mean to go along Ballentine's arguments and refine them using functional and harmonic analysis.




#9
Jan212, 10:17 PM

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P: 1,725

Then in sect 3.4 he proceeds to identify these generators the dynamical variables from classical mechanics, using the postulate that the dynamics of a free particle are invariant under the full Galilei group. (Sorry, but the entire argument is too long to reproduce here. Maybe Amazon or Google Books will let you read some of those sections of Ballentine.) 



#10
Jan312, 04:09 AM

P: 1,057

So what is actually the definition of the classical momentum? I don't know what it means (yet), but you say momentum is a complete and direct consequence of the Galilei group, which itself is more or less simple translations? For the relativistic case I do the same argument with the Poincare group? Does Ballentine give all the neccessary maths I need to understand full how that emerges? A more philosophical question: Does all this treatment mean, that everything has to be a particle and not some sort of field? Isn't it that when you have invariants, it can mean that your coordinate system is overdetermined? As an example take x, y, z coordinates for a sphere which are overdetermined and contraints, in contrast to euler angles. Does such an idea of other coordinate exists where the Galilei invariance is a natural consequence of the mathematical representation? 



#11
Jan412, 01:51 AM

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P: 1,725

velocity operator from Galilean generators, we just use [tex] V = i[H,Q] [/tex] (since d/dt of an operator corresponds to commutation with the Hamiltonian H). But this is a bit of an oversimplification. For less trivial interacting systems, the distinction between "position" and "momentum" gets a bit blurred  if you've done any Hamiltonian dynamics perhaps you've heard of canonical transformations which mix position and momentum, but preserve Hamilton's dynamical equations? But I was only talking about the free case here. operator in the basic algebra. One can be built up, but not for all combinations of mass and spin. E.g., a position operator for the photon remains problematic to this day. But the basic idea is the same: determine the unitary irreducible representations of the Poincare group. That's the "Wignerian" approach. should be able to cope. He covers the basics of Lie group ideas when introducing the Galilei group, but some prior exposure to group theory is never wasted. Since I don't know what your math background is, I can't be more specific. Anyway, you can always ask here on PF if you get stuck. :) Hilbert space. In some case these representations turn out to be finitedimensional, others infinitedimensional. Some of the latter are field theories. Fields can have momentum too... :) direction. One finds a maximal set of commuting operators within the algebra of dynamical variables, which includes the invariants (called Casimirs) and one other. Analysis of the spectrum of these operators determines the necessary dimension of the Hilbert space. 



#12
Jan412, 03:49 AM

P: 1,057

I have a very good understanding of undergrad physics and math and a few more topics I looked up for interest. I don't know Lie groups and I had only one simple course on group theory. Could you recommend books I should read to fully understand the answers you have given? It should be a "physicists" book, by which I mean I care about most steps of derivations, but not about mathematical details (convergence, abstract definitions, etc.) If you give several suggestions, I will pick the most clear for me from the library. Maybe one more question: Are dimensions of space in any way predicted by these methods? 



#13
Jan412, 08:45 PM

Sci Advisor
P: 1,725

If not, or maybe later, you might take a look at Greiner & Muller, "QM  Symmetries". 



#14
Mar1412, 03:27 PM

P: 22

The first thing Ballentine says about a particle with no internal degrees of freedom (pag.80) is that the operators Q,P form an irreducible set.
I think it is a crucial point, but he doesn't say why we are assuming this. What exactly means that Q and P are irreducible? We know only that Q is a position operator and P is the generator of spatial translations. We don't know anything about compatible observables. 



#15
Mar1412, 03:34 PM

P: 1,057

Even better if the derivation was in terms of the density matrix. Could I avoid treating the wavefunction and it's phase invariance this way?? 



#16
Mar1512, 09:19 PM

Sci Advisor
P: 1,725

BTW, if you're now interested in delving a bit deeper into this question, you might consider exactly what's going on a bit later in Ballentine when he deals with the external field case. It involves a modification of the Hamiltonian, and some other adjustments. This leads to a more general perspective that what we're doing is simply finding a quantization of a particular dynamics (meaning that we take classical dynamical variables with their Poisson algebra, and attempt to represent them as operators on a Hilbert space). This yields additional insight into the free case: instead of starting with static Euclidean 3space, plus time, and then invoking free Newtonian dynamics to identify the geometric quantities with dynamical ones, we could just as well start with the dynamical equations of a free particle and try to find the largest group of transformations that preserve these equations. One finds the Galilei group as a subgroup. Of course, this is to be expected since our intuitive picture of 3D space is constructed by extrapolation of the motion of free particles... :) Separately, (and I don't know how much you follow other threads here), but I was recently made aware of how the centralextended Galilei algebra (with ##[X,P] \propto 1##) can be regarded as a consequence of relativistic Poincare invariance  when one takes the nonrelativistic limit of low velocity. See Kaiser: http://arxiv.org/abs/0910.0352 , sect 4.2, esp p95 et seq. I'm starting to prefer Kaiser's explanation over the standard one. 



#17
Mar1512, 09:43 PM

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P: 1,725

Complaints about missequencing on this point have been made before: proper understanding of appendix B relies on having studied at least some of ch 4  which comes later than p80. 



#18
Mar1612, 04:58 AM

P: 1,057

Btw, he proves [X,Px]=[Y,Py]=const only? So, the classical case with zero as a constant is included? 


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