Commutation relations (maths)

[dextercioby]

One of my dilemmas about <standard> quantum mechanics is spelled out in the sequel:

If the position and momentum observables of a single-particle quantum system in 3D are described by the self-adjoint linear operators Q_i and P_i on a seperable Hilbert space $\mathcal{H}$ subject to the commutation relations of Weyl

$$[\exp(iP_{j}u_{j}),\exp(iP_{k}u_{k})]=0$$ (1)

$$[\exp(iQ_{j}v_{j}),\exp(iQ_{k}v_{k})]=0$$ (2)

$$\exp(iP_{j}u_{j})\exp(iQ_{k}v_{k})=\exp(i\hbar \delta_{jk} u_{j}v_{k}) \exp(iQ_{k}v_{k})\exp(iP_{j}u_{j}).$$ (3)

$$u_j, v_k \in \mathbb{R}, i,j,k = 1,2,3.$$

, then they obey the commutation relations of Born and Jordan

$$[P_{j},P_{k}] = 0 (4)$$

$$[Q_{j},Q_{k}] = 0 (5)$$

$$[Q_{j},P_{k}] = i\hbar \delta_{jk} 1 (6)$$

By a generalization of the theorem 6.3, page 340 of Ed. Prugovecki's <Quantum Mechanics in Hilbert Space>, one can prove the statement above: (1-3) imply (4-6). What bothers me is that I have not seen a clear explanation (hence the dilemma) as to why (4-6) DO NOT imply (1-3).

Any ideas/references ?

 

[Bill_K]

I don't know, what's wrong with this? The Baker-Hasudorff Theorem says e^Ae^B = e^{A + B + ½[A,B] + ...} so

e^iP.u e^iQ.v = e^{iP.u + iQ.v + ih/2 u.v}
e^iQ.v e^iP.u = e^{iP.u + iQ.v - ih/2 u.v}

The higher order terms involve double commutators, which vanish. Comparing the two:

e^iP.u e^iQ.v = e^{ih u.v} e^iQ.v e^iP.u

 

[dextercioby]

I see your argument, but it doesn't help, as it uses formal manipulations in case of <objects> in infinite-dimensional spaces. Your derivation makes sense for finite dimesions, though.

 

[strangerep]

By a generalization of the theorem 6.3, page 340 of Ed. Prugovecki's <Quantum Mechanics in Hilbert Space>, one can prove the statement above: (1-3) imply (4-6). What bothers me is that I have not seen a clear explanation (hence the dilemma) as to why (4-6) DO NOT imply (1-3).

Any ideas/references ?

Maybe this is just a boundedness issue? The exponentials are bounded, but
the generators are not?

(BTW, the BCH proofs require norms and limits, etc, on the algebra, iirc. This
gets tricky for unbounded operators.)

 

[dextercioby]

Maybe this is just a boundedness issue? The exponentials are bounded, but
the generators are not?
[...]

Hi, of course it must be a boundedness issue. In section VIII.4 of the first volume of Reed & Simon, the theorem VIII.7 is proven which seems/is to be the reciprocal of Stone's theorem VIII.8.

In my mind this theorem VIII.7 could provide the start in establishing whether (4-6) does or does not imply (1-3). So in light of this, where does it fail in establishing the commutation relations ?

As a side note, before stating and proving this theorem, R&S claim that the exponential of an unbounded self-adjoint operator cannot be defined by a series representation in terms of powers of this operator, because it doesn't strongly converge.

 

[strangerep]

In section VIII.4 of the first volume of Reed & Simon, the theorem VIII.7 is proven which seems/is to be the reciprocal of Stone's theorem VIII.8.

In my mind this theorem VIII.7 could provide the start in establishing whether (4-6) does or does not imply (1-3). So in light of this, where does it fail in establishing the commutation relations ?

Both those theorems deal only with a single generator. To extend it to the 2-generator
case you're interested in, I presume one must establish compatibility of their domains
or work only with an intersection, and then make sure that the limit arguments used
in the proof of Thm VIII.8 (Stone) are still valid. This is unlikely in general.

Also, look at Thm VIII.12 which deals with a generalization to multi-parameter
maps. It ends up only being applicable for commuting generators, afaict.

[Edit: there's quite a lot of other useful discussion in the remainder of ch VIII.
In particular, the corollary after Thm VIII.14 appears to answer precisely the
question in your original post.]

BTW, this is one reason why I prefer to work in RHS, where one's intuition
about analytic functions remains good a lot more often.

See also R&S's definition of, and remarks about, "analytic vectors"
following eq(VIII.9).

 

[dextercioby]

Thank you for the reference. I'll check and revert to you.

EDIT: So indeed, one can find an example of operators which obey (4,6), but not (1,3). This means that essentially Weyl's commutation relations are indeed the fundamental ones and should be presented in an axiomatical approach to QM.

 

[strangerep]

So indeed, one can find an example of operators which obey (4,6), but not (1,3). This means that essentially Weyl's commutation relations are indeed the fundamental ones and should be presented in an axiomatical approach to QM.

I'm not sure things are as clearcut as that. The example in R&S is a rather pathological representation of the CCRs. But we already knew that the CCRs can have inequivalent representations. Consider the P,Q defined near the bottom of p275:

$$P ~=~ \frac{1}{i} \; \frac{\partial}{\partial x} ~~;~~~~~~ Q ~=~ x ~+~ \frac{1}{i} \; \frac{\partial}{\partial y}$$

which satisfy the CCRs, but whose exponentiated forms do not satisfy the Weyl relations.

The Q above, can be thought of as a "field displacement" transformation like

$$Q ~\rightarrow~ Q' ~=~ Q + c$$

where c commutes with both P and Q. In general, this gives an inequivalent representation of the CCRs.

Similarly, if we differentiate a Lie group element near the identity to obtain a Lie generator, the differentiation silently discards a "constant" -- just like any derivative. So when we try to integrate it, there's an ambiguity. What we're seeing here is a different form of the same ambiguity. In QFT we remove the ambiguity by choosing a vacuum. However, some authors (eg Barton) have argued that keeping our "options" more open may provide a way to evade Haag's thm.

 

[dextercioby]

I see where you're coming from, however the counterexample is enough to ensure the inequivalence. Those "pathological representations of the CCRs" are automatically discarded, once one postulates the Weyl form of the CCRs, since we know that the latter not only do they lead to the Born-Jordan form, but also to the Stone-von Neumann theorem which essentially proclaims the inexistence of "pathological representations".

 

[A. Neumaier]

One of my dilemmas about <standard> quantum mechanics is spelled out in the sequel:

If the position and momentum observables of a single-particle quantum system in 3D are described by the self-adjoint linear operators Q_i and P_i on a seperable Hilbert space $\mathcal{H}$ subject to the commutation relations of Weyl

$$[\exp(iP_{j}u_{j}),\exp(iP_{k}u_{k})]=0$$ (1)

$$[\exp(iQ_{j}v_{j}),\exp(iQ_{k}v_{k})]=0$$ (2)

$$\exp(iP_{j}u_{j})\exp(iQ_{k}v_{k})=\exp(i\hbar \delta_{jk} u_{j}v_{k}) \exp(iQ_{k}v_{k})\exp(iP_{j}u_{j}).$$ (3)

$$u_j, v_k \in \mathbb{R}, i,j,k = 1,2,3.$$

, then they obey the commutation relations of Born and Jordan

$$[P_{j},P_{k}] = 0 (4)$$

$$[Q_{j},Q_{k}] = 0 (5)$$

$$[Q_{j},P_{k}] = i\hbar \delta_{jk} 1 (6)$$

By a generalization of the theorem 6.3, page 340 of Ed. Prugovecki's <Quantum Mechanics in Hilbert Space>, one can prove the statement above: (1-3) imply (4-6). What bothers me is that I have not seen a clear explanation (hence the dilemma) as to why (4-6) DO NOT imply (1-3).

Any ideas/references ?

(1)-(3) imply (4)-(6) iff the Weyl representation is regular (This is needed to conclude existence of P and Q, rather than of their exponentials only.)

(4)-(6) imply 1)-(3) iff P and Q are self-adjoint with a common dense domain. (This is needed to make sufficient sense of the commutators and the exponential.)

 

[A. Neumaier]

As a side note, before stating and proving this theorem, R&S claim that the exponential of an unbounded self-adjoint operator cannot be defined by a series representation in terms of powers of this operator, because it doesn't strongly converge.

But it can be defined by a limit, as in the Hille-Yosida theorem, which provides all that is needed to get the converse.

 

[A. Neumaier]

Those "pathological representations of the CCRs" are automatically discarded, once one postulates the Weyl form of the CCRs, since we know that the latter not only do they lead to the Born-Jordan form, but also to the Stone-von Neumann theorem which essentially proclaims the inexistence of "pathological representations".

No. They aren't. Nonregular representations appear in many physically relevant contexts. There is a very nice paper by Acerbi and Strocchi on that, which I had quoted already a few times here on PF.

 

[dextercioby]

Hi Arnold,

I think you mean the one below, discussed on a thread. It's not freely available (unfortunately for me), but I would be interested if it addresses the issue of these nonregular representations (can you, please, point me to a freely available definition of regular vs nonregular representations of commutation relations, algebras, groups, etc. ? Thanks!) in the context addressed by the quantum mechanics of nonrelativistic systems which use a Galilei space-time, not in the context of quantum field theory in Minkowski space-time.

Acerbi, F. and Morchio, G. and Strocchi, F.,
Infrared singular fields and nonregular representations of canonical commutation relation algebras,
Journal of Mathematical Physics 34 (1993), 899

Regards,

Daniel