The generator of Galilean boosts

[Kirjava]

Hey folks,

I'm reading a paper by T.F. Jordan which was referenced in chapter 3 of Ballentine. He's introducing the Galilei group commutation relations, and I'm having trouble with one of them in particular.

He states that in the Heisenberg picture, time dependent position operators can be represented as $$Q(t) = e^{itH}Qe^{-itH}$$. Fine. Then the generators for the transformation $$x \rightarrow x + vt$$ are required to give: $$e^{iv \cdot G}Qe^{-iv \cdot G} = Q(t) + vt$$. I don't see how this could be possible short of requiring that the generators be time dependent (since otherwise we have time independence on the left, and manifest dependence on the right). If so, his next step doesn't seem to follow, because he takes the particular case t=0: $$e^{iv \cdot G}Qe^{-iv \cdot G} = Q$$ to deduce that G commutes with Q. Surely this should only read G(0) commutes with Q. Somehow I think that if this were true he would take the care to mention it. Am I mixed up or what?

tl;dr: Are the generators of Galilean boosts time dependent operators?

Something else that's been on my mind: Why do we need boost transformations in the first place? Can't we model relative velocity by a time dependent spatial translation? i.e. why isn't the unitary transformation $$e^{-iv \cdot G}$$ taken to be the same as $$e^{-ix \cdot Pt}$$?

 

[meopemuk]

Kirjava,

I guess you are talking about paper

Thomas F. Jordan, "Why −i∇ is the momentum"
American Journal of Physics -- December 1975 -- Volume 43, Issue 12, pp. 1089


I think that equation before last on the page 1089 should be written as

$$e^{iv \cdot G}Q(t)e^{-iv \cdot G} = Q(t) + vt$$

Indeed, this becomes clear if you take into account that


$$Q(t) = e^{itH}Qe^{-itH} = Q + Pt/M$$

where P is particle's momentum and M is its mass.

Eugene.

 

[Kirjava]

Thanks for the reply Eugene, you've got the right paper.

Kirjava,
I think that equation before last on the page 1089 should be written as

$$e^{iv \cdot G}Q(t)e^{-iv \cdot G} = Q(t) + vt$$

This does make more sense.

As for your second equation, I'm not really where it fits into the logical development I'm trying to come to grips with. How do you conclude that $$e^{iHt}Qe^{-iHt} = Q(t) + Pt/M$$?

You probably replied before I edited in the afterthought, so do you have any thoughts on why the boost generators are a necessary addition to the theory?

 

[meopemuk]

Thanks for the reply Eugene, you've got the right paper.




This does make more sense.

As for your second equation, I'm not really where it fits into the logical development I'm trying to come to grips with. How do you conclude that $$e^{iHt}Qe^{-iHt} = Q(t) + Pt/M$$?

You probably replied before I edited in the afterthought, so do you have any thoughts on why the boost generators are a necessary addition to the theory?

Keep reading chapter 3 in Ballentine and you'll find the answers yourself. In order to derive the above formula $$e^{iHt}Qe^{-iHt} = Q + Pt/M$$ you need to know the commutator [Q,H], which is the 2nd equation on page 81.

Eugene.

 

[Kirjava]

Yeah thanks I can see where it comes from now, but the only derivation I can think of uses the result derived in problem 3.3, which gives a series expansion for transformed operators. His derivation is obviously far from rigorous, and I don't find it very convincing with my limited mathematical background. Would this kind of result be covered in most books on the mathematical foundations of QM? Can you recommend any?

 

[meopemuk]

The formula in problem 3.3 is one of the most important formulas in all quantum mechanics. You should definitely spend some time to learn how to derive and use this formula. Why don't you like the proof given on page 620 of the book? It is rather convincing, in my opinion.

Ballentine's book is rather unique in the sense that it presents QM in the (very appropriate) language of representations of the Galilei group. Other QM textbooks usually don't follow that path, so it is unlikely that you'll find a more satisfying proof of 3.3 somewhere else. Perhaps there exist entry level mathematical textbooks on Lie algebras and Lie groups. But no specific title comes to mind.

Eugene.

 

[Kirjava]

I spose it's just that I've just never done any serious analysis/calculus with operators involved, so his proof has me wondering about necessary and sufficient conditions for a parametrized operator to be differentiable, what the convergence conditions of such a series expansion would be etc etc. It's a bit like the uneasiness I feel whenever we have to deal with continuous spectra or other sticky matters which are often glossed over in QM texts. Anyway thanks for your guidance so far, I think I'm starting to get the big picture (I wish I'd checked the problem set earlier!).

 

[meopemuk]

I spose it's just that I've just never done any serious analysis/calculus with operators involved, so his proof has me wondering about necessary and sufficient conditions for a parametrized operator to be differentiable, what the convergence conditions of such a series expansion would be etc etc. It's a bit like the uneasiness I feel whenever we have to deal with continuous spectra or other sticky matters which are often glossed over in QM texts. Anyway thanks for your guidance so far, I think I'm starting to get the big picture (I wish I'd checked the problem set earlier!).

There is a school of thought in QM and in quantum field theory that demands rigorous definitions and proofs in all instances. Sometimes these demands are so rigid that only non-interacting theories can satisfy them. One can spend a lot of efforts sorting out unbounded operators and convergence conditions without moving an inch toward understanding of physics. In my persnal opinion, it is OK to relax a bit the mathematical rigor, when doing QM.

Eugene.

 

[Kirjava]

Interesting, I'll keep that in mind.

 

[Kirjava]

It seems rather mysterious that G's commutation relations with P and Q should determine so much don't you think? Not trying to sound suspicious, I'm merely in awe.

They are responsible for fixing G = MQ, which determines P's relation to the velocity operator i[H,Q] (hitherto unsuspected). And as if that weren't enough this information goes on to fix Hamiltonian itself as $$P\cdot P +E_0$$! (which until then was simply the generator of time-evolution!). That Q's eigenkets should be complete in the state space seems to be a crucial ingredient here as well, but wow. Ballentine doesn't seem to make a very big deal about the fact that we were able to deduce so much from so little. I wonder if I missed some aspect of his deduction which would serve to deflate my sense of wonder?

 

[meopemuk]

You are right. This is beautiful stuff. It becomes even more beautiful when you go from the Galilei group to the relativistic Poincare group. I think it is a shame that this wonderful group-based view of physics has not found a more prominent place in modern textbooks. Ballentine's book is a notable exception. Another fine textbook is S. Weinberg "The quantum theory of fields", vol. 1.

Eugene.

 

[strangerep]

There is a school of thought in QM and in quantum field theory that demands rigorous definitions and proofs in all instances. Sometimes these demands are so rigid that only non-interacting theories can satisfy them.

That just means that (some parts of) their starting axioms are not the correct
choices for arriving at a physically-satisfactory model.

One can spend a lot of efforts sorting out unbounded operators and convergence conditions without moving an inch toward understanding of physics. In my personal opinion, it is OK to relax a bit the mathematical rigor, when doing QM.

But surely it's not ok to relax mathematical correctness? (Hmmm, maybe it's
a question of how much incorrectness one is willing to ignore.)

Ballentine doesn't seem to make a very big deal about the fact that we were able to
deduce so much from so little. I wonder if I missed some aspect of his deduction
which would serve to deflate my sense of wonder?

Maybe it's partly because he avoids a lot of the endless nonsense
that tends to accompany older interpretations of QM, and sticks to the basic
statistical interpretation (which I summarize as: physics = statistics + symmetry).