Axiomatic derivation of [X,P]?

[Gerenuk]

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
<x|p>=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.

 

[strangerep]

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
<x|p>=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.

It comes from translations that we can perform to move from x="here" to x="there".

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.)

 

[Gerenuk]

It comes from translations that we can perform to move from x="here" to x="there".
[...]

Sounds good! I'll surely get this book.
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?

 

[meopemuk]

The operator of momentum P is *defined* as a generator of space translations. Then the translation by a finite distance a is represented by the exponential function of the generator e^{\frac{i}{\hbar}Pa}. The action of this transformation on the position operator X should result in a shift

e^{-\frac{i}{\hbar}Pa} X e^{\frac{i}{\hbar}Pa} = X-a

From this postulate you can get the desired commutator

[X,P] = i \hbar

Eugene.

 

[espen180]

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

 

[Gerenuk]

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?

 

[espen180]

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

 

[dextercioby]

[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.

 

[strangerep]

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?

Ballentine also covers this. In sect 3.3 he covers the generators of the Galilei group, one of which is the P mentioned above, but at that point its meaning is merely geometrical, and not yet identified with the "momentum" concept from classical dynamics.

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.)

 

[Gerenuk]

[...]
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.)

It's OK. I'll get the book, so no need to reproduce what is well written somewhere else ;)

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 necessary 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?

 

[strangerep]

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 free nonrelativistic case, we just use p = mv = m dq/dt. To get a
velocity operator from Galilean generators, we just use
<br /> V = i[H,Q]<br />
(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.

For the relativistic case I do the same argument with the Poincare group?

In the relativistic case, things are trickier since there's no position
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.

Does Ballentine give all the necessary maths I need to understand full how
that emerges?

It's not a maths textbook, but anyone with reasonable proficiency in calculus
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. :-)

A more philosophical question:
Does all this treatment mean, that everything has to be a particle and not
some sort of field?

No. We're really finding representations of Lie algebras as operators on a
Hilbert space. In some case these representations turn out to be
finite-dimensional, others infinite-dimensional. Some of the latter are field
theories. Fields can have momentum too... :-)

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 constraints, 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?

In the group theoretic development of quantum theory we proceed from the other
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.

 

[Gerenuk]

It's not a maths textbook, but anyone with reasonable proficiency in calculus
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. :-)

Oh, it gets more and more interesting and I suppose it's time for me to study books! I've ordered Ballentine, yet I believe it won't outline all ideas you have presented here?!

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?

 

[strangerep]

Oh, it gets more and more interesting and I suppose it's time for me to study books! I've ordered Ballentine, yet I believe it won't outline all ideas you have presented here?!

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.

Start with Ballentine's "QM - A Modern Development", and see whether his treatment of Lie groups via the example of the Galilei group is sufficient for you to get through the rest of the book.

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

Maybe one more question:
Are dimensions of space in any way predicted by these methods?

If you mean the 3+1 dimensionality of space-time, then no (afaik).

 

[naffin]

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.

 

[Gerenuk]

Start with Ballentine's "QM - A Modern Development", and see whether his treatment of Lie groups via the example of the Galilei group is sufficient for you to get through the rest of the book.

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


If you mean the 3+1 dimensionality of space-time, then no (afaik).

A revivial of the thread... Meanwhile I've had a look at Ballentine and I'm quite pleased with that type of derivation :) However, I'd find it better to see it more general and rigorous derivation (but still understandable for a physicist :) ). Perhaps even one for the Poincare group. Maybe I can grab hold of the other book.

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??

 

[strangerep]

A revivial of the thread... Meanwhile I've had a look at Ballentine and I'm quite pleased with that type of derivation :) However, I'd find it better to see it more general and rigorous derivation (but still understandable for a physicist :) ). Perhaps even one for the Poincare group. Maybe I can grab hold of the other book.

I mentioned Greiner & Muller only because it contains some introductory material on Lie groups in a physics-relevant context. I don't think you'll find the same kind of derivation of $[X,P]$ there, nor much about Poincare.

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??

No. The central element in the quantum Galilei algebra (which later gets identified with mass) is a consequence of the requirement that probabilities be invariant. For some groups it can be disposed of by redefining some generators, but not in the Galilei case.

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 3-space, 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 central-extended 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.

 

[strangerep]

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?

Did you read Ballentine's appendices A & B ?

Complaints about mis-sequencing 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.

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.

At that point, we know at least that Q and P are linearly independent generators. To go from this to irreducibility, one must understand that irreducibility means that no subspace of the space of states is left invariant by these operators. But detailed understanding of this point needs material on wave functions in the subsequent ch4. Herein lies, perhaps, a weakness in this sequence of presentation -- but a more advanced approach based on general quantization of Hamiltonian dynamics would be too difficult at that stage for the intended readership, imho.

 

[Gerenuk]

No. The central element in the quantum Galilei algebra (which later gets identified with mass) is a consequence of the requirement that probabilities be invariant. For some groups it can be disposed of by redefining some generators, but not in the Galilei case.

Sure, but density matrix and wave functions should be equivalent representations (the former being slightly more general). The density matrix has phase invariance included just by its mathematical form. So can you express the whole derivation with density matrices?

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

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.

I've seen that and it seems possible interactions have to be of mathematical "electric-field or magnetic field type". No other are possible?

 

[bhobba]

[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.

Exactly - the derivation from the Galilei group (ie Galilean Invariance) can be found in Chapter 3 of Ballentine. It my preferred method - but logically it is an assumption with the same logical status as assuming it in the first place.

Thanks
Bill

 

[naffin]

At that point, we know at least that Q and P are linearly independent generators. To go from this to irreducibility, one must understand that irreducibility means that no subspace of the space of states is left invariant by these operators. But detailed understanding of this point needs material on wave functions in the subsequent ch4. Herein lies, perhaps, a weakness in this sequence of presentation -- but a more advanced approach based on general quantization of Hamiltonian dynamics would be too difficult at that stage for the intended readership, imho.

True, we know also that \left[ Q_{\alpha},P_{\beta} \right] = i \delta_{\alpha \beta}, but we don't know yet what is the group generated by Q (from a physical point of view).
I've read App.4,5 and Ch.4, but without knowing the meaning of the group generated by Q I can't see how to characterize a particle without internal degrees of freedom. How could Ballentine's derivation proceed further? Do we have to make other assumptions before saying that Q,P are irreducible?

Assuming that Q_i generates translations in the ''space velocity'' (a boost at t=0), I would say that a particle without internal d.o.f. isn't charaterized by variables invariant under translations and boosts, so we can choose a space in which invariant operators are trivial (multiples of identity). I'm quite sure that this is equivalent of saying that Q and P are irreducible.

 

[strangerep]

Sure, but density matrix and wave functions should be equivalent representations (the former being slightly more general). The density matrix has phase invariance included just by its mathematical form. [...]

And thus this extra phase factor is part of the invariance group. It must be accounted for, one way or another.

Btw, he proves [X,Px]=[Y,Py]=const only?

Actually, he first proves, by geometric considerations only, that $[G,P] \propto M$ (const), where G generates velocity boosts and M is some constant (not yet identified with mass). Then, in sect 3.4, he invokes invariance of Newtonian free dynamics under this group to introduce a position operator, and then derives $G = MQ$ eventually on p80, where now M is indeed identified with the Newtonian mass.

So, the classical case with zero as a constant is included?

There's more to it than that. One does not have to follow Ballentine's sequence of arguments. You could start with the (classical) dynamics of a free Newtonian particle, i.e., $\ddot x_i = 0$, and ask for the largest group of transformations of the space & time variables that preserve this equation. One obtains the Niederer-Schrodinger group http://en.wikipedia.org/wiki/Schrodinger_group which contains Galilei as a subgroup. Quantization of the free dynamics can then be regarded alternately as constructing a representation of this dynamical group by operators on a Hilbert space. (This approach is in fact more general -- Ballentine is just covering the easier cases by an easier pedagogical progression.)

$\hbar$ should be seen as just a scaling factor. The classical and quantum cases are distinguished, not by $\hbar\to 0$, but by (e.g.,) the ratio of the system action becoming very large compared to $\hbar$. I.e., $\hbar$ is still there, but it's often convenient to take a classical limit by letting $\hbar\to 0$, though that can obscure what's really going on.

It's misleading to think of the commutator as vanishing in the classical case. For the latter, we should pass from commutators to Poisson brackets, and the Poisson bracket between position and momentum is not zero.

I've seen that and it seems possible interactions have to be of mathematical "electric-field or magnetic field type". No other are possible?

Ballentine is using the so-called "instant form of dynamics", where interaction terms are confined to the Hamiltonian and boost generators. This covers a large range of cases, but there also exist other forms of dynamics in which other generators get interaction terms. I'm less familiar with those and can't say much more about them.

 

[strangerep]

[...] I can't see how to characterize a particle without internal degrees of freedom. How could Ballentine's derivation proceed further? [...]

My previous response to Gerenuk contains part of the answer to this. One must start from a more general framework of a dynamical group.

(I'm not sure how to usefully respond to the rest of your post, except to note that in a classical case, it's useful to find a maximal set of independent variables, such as q,p which parameterize the phase space in simple cases.)

 

[eaglelake]

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
<x|p>=exp(ikx)
psi(x,k)=exp(ikx)

Reality, unfortunately, usually means “classical reality”, but quantum events are decidedly not classical. There are no “axioms about reality” in quantum theory. Also, we have to use some math here, so I apologize, but I think it is required.

In quantum mechanics, position and momentum are operators: \hat x = x{\rm{ and }}\hat p_x = - i\hbar \frac{\partial }{{\partial x}}. Using the definition of the commutator, \left[ {\hat A,\hat B} \right] = \hat A\hat B - \hat B\hat A, we have \left[ {\hat x,\hat p_x } \right]\psi (x) = \left[ {x, - i\hbar \frac{\partial }{{\partial x}}} \right]\psi (x) = x\left( { - i\hbar \frac{{\partial \psi (x)}}{{\partial x}}} \right) - \left( { - i\hbar \frac{\partial }{{\partial x}}x\psi (x)} \right) = i\hbar \psi (x), where we have put in the \psi (x) so the operators have “something to operate on”.

\left\langle x \right|\left. p \right\rangle = e^{ikx} is the eigenfunction of the momentum operator corresponding to the eigenvalue p = \hbar k.


Generally, \psi (x) represents the state function (wavefunction) that is determined by a preparation procedure. We use it to make probability calculations. If \psi (x) = e^{ikx}
then the experiment is done in such a way that the wavefunction \psi (x)is the eigenfunction of momentum corresponding to the eigenvalue p = \hbar k. In such a state we always get the same value p = \hbar k when we measure the momentum. I hope this helps.
Best wishes

 

[lugita15]

In quantum mechanics, position and momentum are operators: \hat x = x{\rm{ and }}\hat p_x = - i\hbar \frac{\partial }{{\partial x}}.

Just a nitpick, these aren't the position and momentum operators, which act on the abstract Hilbert space. These are just the position-space representations of those operators, which act on the position-space wave functions. In general, the position-space representation of an operator O is the operator D such that <x|O|ψ>=D<x|ψ>=Dψ(x). We can also go in the opposite direction, starting with an arbitrary bounded linear operator D acting on position-space wave functions, and getting an abstract Hilbert space operator O.