Commutators and traces

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  • #26
samalkhaiat
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Hopefully Sam will answer but in the meantime this is my understanding.
You have to use analytic continuation, that is, impose dependence on Riemann zeta function, in this particular case for s=1, and the particular series 1+2+3+.... , in other bases you get other convergences. This comes at the cost of losing basis-independence of the trace i.e. incompatibility with linearity that Sam found bizarre above.

I see no question to answer, so I just make the following remarks for added clarification:
1) Divergent series are not “the invention of the devil”. Some divergent series can be summed rigorously. For example, the “usual sum” of [itex]1 - 2 + 3 - 4 + \cdots[/itex] diverges, but we can rigorously show that its Abel sum exists and is equal to [itex]1/4[/itex]. Other divergent series need to be regularized: by introducing suitable cutoff function, one can justify the use of the Euler-Maclaurin formula to them.
2) The Euler sum [itex]\sum_{1}^{\infty} k = - 1 / 12[/itex]:
A) In physics, it has passed the test:
(i) The whole derivation of the Casimir force can be summarized by the replacement [itex]\sum^{\infty} k \to ( - 1 / 12 )[/itex].
(ii) In bosonic string theory, the mass of particles in the [itex]J[/itex] excitation is given by [tex]\alpha^{ ’ } m^{2} = J + \frac{D - 2}{2} \sum_{n = 1}^{\infty} n .[/tex] It is also known that the [itex]J = 1[/itex] excitations are massless. Thus, by using the Euler sum in the relation above, we conclude that bosonic string theory is consistent only in space-time of dimension [itex]D = 26[/itex].
B) In mathematics, “all roads take you to Rome”:
(i) Zeta function regularization leads to [tex]\zeta ( - 1 ) = - \frac{1}{12} .[/tex]
(ii) The Heat Kernel Regularization gives you [tex]\lim_{r \to 0} \left( \frac{e^{r}}{( e^{r} - 1)^{2}} - \frac{1}{r^{2}} \right) = \lim_{r \to 0} \left( - \frac{1}{12} + \mathcal{O} ( r^{2} ) \right) = - 1 / 12 .[/tex]
(iii) Ramanujan’s summation method gives us (in terms of Bernoulli’s numbers) [tex]C = - \frac{B_{1}}{1!} f ( 0 ) - \frac{B_{2}}{2!} f^{ ’ } ( 0 ), [/tex] which for our case, [itex]f(x) = x[/itex], leads to [itex]( - 1 /12)[/itex].
3) There is no such thing as “free lunch” : all regularization methods are either not linear or not stable.
4) The main point of my previous post is to highlight point (3): The trace of infinite-dimensional matrices is neither rigorous nor nice: (i) the very notion of diagonal elements lose its meaning. This is why I wrote “diagonal elements”. (ii) the defining properties of the trace in finite dimension, i.e. linearity and invariance (or cyclicity), can not both be maintained in infinite-dimension. This is what happened when we “traced” the number operator [itex]A^{\dagger} A[/itex]: the trace was invariant (cyclic) but not linear.

Sam
 
  • #27
vanhees71
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Mathematically the manipulation here is on par with all those done for many different calculations on QFT, from Casimir effect to vacuum energy density , to Hawking radiation in curved spacetime, or anything for wich you might use dimensional regularization (that according to wikipedia is equivalent to RZ regularization).
I think the zeta regularization is cleaner than your example and doesn't involve that substraction(has no counterterms) but I'm surely no expert. I know Zeta funtion regularization is routinely used to define traces and determinants of self-adjoint operators.
It is important to make clear that the traces defined this way are not the usual sense traces from linear algebra, they are highly nonlinear as all these regularizations are clearly nonlinear.
Of course, you can use the ##\zeta##-function regularization as well. Then you simply use the definition of the ##\zeta## function
$$\zeta(z)=\sum_{n=1}^{\infty} n^{-z}$$
and then analytically continue it. This immediately gives the ##\zeta##-regularized sum
$$\sum_{n=1}^{\infty} n \simeq \zeta(-1)=-\frac{1}{12}.$$
My point, however is that this cannot be interpreted as a meaningful result for the application to the trace discussed here.

In contrast to that, in the QFT examples (any Feynman diagram at 0 or finite temperature, closed diagrams giving contributions to the thermodynamic potential, vacuum energies for evaluating the Casimir effect etc.) you have clearly defined renormalization conditions to make physical sense. Even if some regularization scheme (as for the sum of all integers the ##\zeta## regularization does) gives a finite result, you still have to make the appropriate subtractions which are well defined by your renormalization scheme, and one can show that this renormalized results within a given scheme are independent of the regularization scheme used. Often you can forget about regularization altogether and just subtract directly the integrands given by your Feynman diagrams and evaluate the then finite integral (BPHZ renormalization).

Now, in the case of the position-momentum commutator there is no such renormalization prescription which makes physics sense out of this trace.
 
  • #28
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Of course, you can use the ##\zeta##-function regularization as well. Then you simply use the definition of the ##\zeta## function
$$\zeta(z)=\sum_{n=1}^{\infty} n^{-z}$$
and then analytically continue it. This immediately gives the ##\zeta##-regularized sum
$$\sum_{n=1}^{\infty} n \simeq \zeta(-1)=-\frac{1}{12}.$$
My point, however is that this cannot be interpreted as a meaningful result for the application to the trace discussed here.

In contrast to that, in the QFT examples (any Feynman diagram at 0 or finite temperature, closed diagrams giving contributions to the thermodynamic potential, vacuum energies for evaluating the Casimir effect etc.) you have clearly defined renormalization conditions to make physical sense. Even if some regularization scheme (as for the sum of all integers the ##\zeta## regularization does) gives a finite result, you still have to make the appropriate subtractions which are well defined by your renormalization scheme, and one can show that this renormalized results within a given scheme are independent of the regularization scheme used. Often you can forget about regularization altogether and just subtract directly the integrands given by your Feynman diagrams and evaluate the then finite integral (BPHZ renormalization).

Now, in the case of the position-momentum commutator there is no such renormalization prescription which makes physics sense out of this trace.

Renormalization prescriptions originally had no physical justification other than getting rid of nonsensical results of the original theory, only three decades later a heuristic justification based on the concept of "effective theories" was put forth. In our case what one "substracts" are the divergent traces if you insist. The independence of results from regularization scheme has already been commented.
We have agreed that there is no finite dimensional representation of the canonical commutation using the trace properties. It is perhaps sensible to try and find a trace that makes sense of the the canonical commutation formula for linear operators in the infinite-dimensional case, for instance if one decides that just the "no-contradiction" of the undefined trace case is not enough formally to justify the commutative relation(the OP for one insistently demanded a formal justification beyond the absence of contradiction).
Apparently the regularized trace also gives a contradictory position-momentum commutator equation, whether this result is physically meaningful or pure nonsense is not for me to decide.
 
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  • #29
vanhees71
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The point of the divergences in perturbative relativistic QFT's is that it is an abuse of mathematics rather than a really physically relevant "bug". The reason for the divergences is to multiply distribution-valued field operators at the same space-time point and take expectation values of such undefined products. If you formulate everything in terms of "smeared" field operators, no such divergence problems exist. This is put forward nicely in the textbook by G. Scharff, Finite QED.

Nevertheless, also with perfectly finite results you have to renormalize, because what you do in terms of perturbation theory is also unphysical. You start with, say, non-interacting electrons, neglecting the electromagnetic field produced by it. Then you add the interaction in a perturbative way, but this also implies the eigenfields of the particles, adding to the energy (and mass) of the described particle-like excitation (usually called a "particle" for convenience). The observed mass is of course that of a full electron, including its em. field around it, and this mass is observed in scattering experiments at a certain energy level. The same holds true for the other parameters appearing in the theory like coupling constants (here the em. fine-structure constant), the wave-function normalization, etc. These free parameters must be defined via a given renormalization scheme. Changing the renormalization scheme doesn't change the physics, leading to the renormalization-group equations.

The only difference between a Dyson-renormalizable and an "effective" field theory is that in the former case you need to define only a finite number of parameters for fixing the renormalization scheme, why you need infinitely many for effective field theories. The latter class of theories is justified by symmetry considerations, giving the only constraints to the infinitely many parameters, leading to some predictive power (e.g., chiral perturbation theory as an effective theory for hadrons, based on the (accidental) symmetries of QCD in the light-quark sector).
 
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  • #30
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I tend to agree that the parallelism with renormalization is a red herring. The OP was about the validity of the position-momentum commutator equation. If we dedecide the trace cannot help us here because it is undefined, the problem remains of how to make sense of the commutator if the product of infinite-dimensional position and momentum is not well defined either(or is it?). I'd like to hear the experts wisdom on this.
 
  • #31
vanhees71
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Why shouldn't the product of operators be well-defined? Of course it is on their domain, which is a proper dense subset of the separable Hilbert space. It's most simple to see in the position representation, where these operators are defined as
$$\hat{x} \psi(x)=x \psi(x), \quad \hat{p} \psi(x)=-\mathrm{i} \mathrm{d}_x \psi(x).$$
These operators are well-defined on, e.g., the Schwartz space of the rapidly falling smooth functions, which is dense on the realization of the separable Hilbert space in terms of square-integrable functions, ##L^2(\mathbb{R},\mathbb{C})##, and they are essentially self-adjoint. So they are properly defined as representants of position and momentum in non-relativistic single-particle quantum theory.
 
  • #32
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Why shouldn't the product of operators be well-defined? Of course it is on their domain, which is a proper dense subset of the separable Hilbert space. It's most simple to see in the position representation, where these operators are defined as
$$\hat{x} \psi(x)=x \psi(x), \quad \hat{p} \psi(x)=-\mathrm{i} \mathrm{d}_x \psi(x).$$
These operators are well-defined on, e.g., the Schwartz space of the rapidly falling smooth functions, which is dense on the realization of the separable Hilbert space in terms of square-integrable functions, ##L^2(\mathbb{R},\mathbb{C})##, and they are essentially self-adjoint. So they are properly defined as representants of position and momentum in non-relativistic single-particle quantum theory.
They are well-defined that way by postulate after the fact that the commutator equation is an axiom of the theory. The whole thread is about trying to specify how the axiom fits in functional analysis before taking it as an axiom, that's the hard part, the textbook answer you give starts from the axiom obviously.
When Born and Jordan firts came to to derive it in 1925 from Heisenberg breakthrough that summer, they did it using semiclassical arguments that were later found not general, and mathematically based on infinite matrices (therefore the derivation was representation-dependent). That was even before the linear operator concept applied to QM appeared.
 
  • #33
vanhees71
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That's why nowadays we prefer an approach to teach quantum theory in the modern way, starting right away with Dirac's formulation.
 
  • #34
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Sure, that's quite practical when encountering QM for the first time, but let's say a fellow physicist researching foundations asked about this.
 
  • #35
aleazk
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Sure, that's quite practical when encountering QM for the first time, but let's say a fellow physicist researching foundations asked about this.

Well, this fellow would do something along the following lines: we look for a Hilbert space H equipped with the CCR. We can see this statement as some abstract "equation" for which we seek for solutions. The "solutions" are pairs composed by: i) concrete Hilbert spaces; ii) equipped with a concrete realization of the CCR in terms of operators in this space. The problem will be solved once we find all the possible concrete solutions. The equivalence of solutions is given by unitary equivalence of the pairs. By inspection, we can see that no such solution exists if the Hilbert space is taken to be finite dimensional (because of the trace argument mentioned earlier). On the other hand, the solution offered in post #31 is a valid solution, so at least we know that solutions exist. A much more serious issue is raised by the fact that one can show that at least one of the operators in the (abstract) CCR cannot be bounded. This forces us to take the domain of the operators in consideration, since two pairs can be made inequivalent just by changing the domain of the operators (i.e., the formal expression of the operators is still the same for both, we just change the domain). A possible solution to this problem is to restrict the kind of solutions we look for, i.e., rather than considering some general solution for the CCR, we look for a solution that arises as the infinitesimal generators of the "exponentiated CCR", i.e., the Weyl *-algebra (based on a finite dimensional symplectic vector space, since we are dealing with ordinary QM rather than bosonic QFT). There's actually a physical justification for this restriction, since the Weyl relations are equivalent to the Imprimitivity Condition, which captures the notion of the homogeneity of (physical) space.

So, we have posed the problem in a way that takes into consideration the domain difficulties. What's next? we could keep trying to find solutions by inspection or guessing. But that's useless, since there could be an infinite number of inequivalent solutions!

The best way to attack the problem is to formulate it in such a way so that we can apply some powerful theorem that will give us automatically all of the possible solutions of the problem at once (note that, since this process will give us all the solutions, it already has to contain the previous result we got by inspection, namely, that there's no solution in finite dimensions).

There are two important ways to do this. The first is group theoretical. One notices that asking for a realization of the Weyl relations is equivalent to asking for a (unitary, etc.) representation of certain Lie group (the Heisenberg group, which is the unique connected, simply connected Lie group with the CCR as Lie algebra) and that satistifies certain condition. So, the problem now has been reduced to the mathematical problem of finding all of the possible representations of the Heisenberg group that satisfy certain condition. The Heisenberg group is a non-compact group, so we cannot use the Peter-Weyl theorem here (and that's good, otherwise we would get that the irreducible representations are necessarily finite dimensional, in contradiction with the previous result by inspection). But the Heisenberg group is a semidirect product of a normal, abelian subgroup and a closed subgroup. And there's a very powerful theorem, called the Mackey Machine, that deals with the representation theory of these type of groups. The theorem states that there's a family of irreducible and inequivalent representations (called induced representations) and that all other abstract or generic irreducible representation is equivalent to some representation in this family. Thus, the problem is solved if we can build this family of induced representations. Fortunately, there's an algorithmic recipe for doing this. In the case of the Heisenberg group, the result is the following: there's only one non-trivial induced representation (given by the exponentiation of the representation of the CCR given in post #31, sometimes called the "Schrödinger representation"). So, up to unitary equivalence of solutions, there's essentially only one solution to the problem (as stated in terms of the Weyl relations). This result is called the Stone-von Neumann theorem. Evidently, since this solution is over an infinite dimensional space and all other solution is equivalent to this one, then no solution exists over a finite dimensional space. Also, since the solution is over a separable space and all other solution is equivalent to this one, then no solution exists over a non-separable space.

The other method relies on the algebraic formulation of QM. By putting a norm, the Weyl *-algebra is transformed into a C*-algebra. The GNS construction theorem states that any representation of a C*-algebra that satisfies <f|Raf>=w(a) (where |f> is a cyclic vector, R is the representation of the algebra, and w is an algebraic state), is equivalent to the GNS construction based on the state w. Now, for the case of the finite dimensional Weyl C*-algebra, one can show that for any abstract (i.e., generic) realization R of the algebra on a Hilbert space, there always exists a cyclic vector |f> such that <f|Raf>=w(a), where w is an algebraic state that does not depend on the particular realization, i.e., it's always the same. So, any realization is equivalent to the GNS construction based on this w. It can be shown, of course, that the previous Schrödinger representation satisfies this and in this way both methods simply prove the same result.
 
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  • #36
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Well, this fellow would do something along the following lines: we look for a Hilbert space H equipped with the CCR. We can see this statement as some abstract "equation" for which we seek for solutions. The "solutions" are pairs composed by: i) concrete Hilbert spaces; ii) equipped with a concrete realization of the CCR in terms of operators in this space. The problem will be solved once we find all the possible concrete solutions. The equivalence of solutions is given by unitary equivalence of the pairs. By inspection, we can see that no such solution exists if the Hilbert space is taken to be finite dimensional (because of the trace argument mentioned earlier). On the other hand, the solution offered in post #31 is a valid solution, so at least we know that solutions exist. A much more serious issue is raised by the fact that one can show that at least one of the operators in the (abstract) CCR cannot be bounded. This forces us to take the domain of the operators in consideration, since two pairs can be made inequivalent just by changing the domain of the operators (i.e., the formal expression of the operators is still the same for both, we just change the domain). A possible solution to this problem is to restrict the kind of solutions we look for, i.e., rather than considering some general solution for the CCR, we look for a solution that arises as the infinitesimal generators of the "exponentiated CCR", i.e., the Weyl *-algebra (based on a finite dimensional symplectic vector space, since we are dealing with ordinary QM rather than bosonic QFT). There's actually a physical justification for this restriction, since the Weyl relations are equivalent to the Imprimitivity Condition, which captures the notion of the homogeneity of (physical) space.

So, we have posed the problem in a way that takes into consideration the domain difficulties. What's next? we could keep trying to find solutions by inspection or guessing. But that's useless, since there could be an infinite number of inequivalent solutions!

The best way to attack the problem is to formulate it in such a way so that we can apply some powerful theorem that will give us automatically all of the possible solutions of the problem at once (note that, since this process will give us all the solutions, it already has to contain the previous result we got by inspection, namely, that there's no solution in finite dimensions).

There are two important ways to do this. The first is group theoretical. One notices that asking for a realization of the Weyl relations is equivalent to asking for a (unitary, etc.) representation of certain Lie group (the Heisenberg group, which is the unique connected, simply connected Lie group with the CCR as Lie algebra) and that satistifies certain condition. So, the problem now has been reduced to the mathematical problem of finding all of the possible representations of the Heisenberg group that satisfy certain condition. The Heisenberg group is a non-compact group, so we cannot use the Peter-Weyl theorem here (and that's good, otherwise we would get that the irreducible representations are necessarily finite dimensional, in contradiction with the previous result by inspection). But the Heisenberg group is a semidirect product of a normal, abelian subgroup and a closed subgroup. And there's a very powerful theorem, called the Mackey Machine, that deals with the representation theory of these type of groups. The theorem states that there's a family of irreducible and inequivalent representations (called induced representations) and that all other abstract or generic irreducible representation is equivalent to some representation in this family. Thus, the problem is solved if we can build this family of induced representations. Fortunately, there's an algorithmic recipe for doing this. In the case of the Heisenberg group, the result is the following: there's only one non-trivial induced representation (given by the exponentiation of the representation of the CCR given in post #31, sometimes called the "Schrödinger representation"). So, up to unitary equivalence of solutions, there's essentially only one solution to the problem (as stated in terms of the Weyl relations). This result is called the Stone-von Neumann theorem. Evidently, since this solution is over an infinite dimensional space and all other solution is equivalent to this one, then no solution exists over a finite dimensional space. Also, since the solution is over a separable space and all other solution is equivalent to this one, then no solution exists over a non-separable space.

The other method relies on the algebraic formulation of QM. By putting a norm, the Weyl *-algebra is transformed into a C*-algebra. The GNS construction theorem states that any representation of a C*-algebra that satisfies <f|Raf>=w(a) (where |f> is a cyclic vector, R is the representation of the algebra, and w is an algebraic state), is equivalent to the GNS construction based on the state w. Now, for the case of the finite dimensional Weyl C*-algebra, one can show that for any abstract (i.e., generic) realization R of the algebra on a Hilbert space, there always exists a cyclic vector |f> such that <f|Raf>=w(a), where w is an algebraic state that does not depend on the particular realization, i.e., it's always the same. So, any realization is equivalent to the GNS construction based on this w. It can be shown, of course, that the previous Schrödinger representation satisfies this and in this way both methods simply prove the same result.
Thanks for the elaborated answer. Could you maybe expand on the counterexamples to the Stone-von Neumann theorem, and how to avoid issues with the domain of the unbounded operators an exponentiated form of the canonical commutation relations(the Wey relations you mention) have to be used that are not rigorously the same but close enough. My understanding is that this representation of the Heisenberg group is isomorphic to ##L^2(R)## but only as modules.
 
  • #37
aleazk
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Thanks for the elaborated answer. Could you maybe expand on the counterexamples to the Stone-von Neumann theorem, and how to avoid issues with the domain of the unbounded operators an exponentiated form of the canonical commutation relations(the Wey relations you mention) have to be used that are not rigorously the same but close enough. My understanding is that this representation of the Heisenberg group is isomorphic to ##L^2(R)## but only as modules.

I don't have much time to write a detailed answer now. I refer you to the following references:

Quantum Field Theory, a tourist guide for mathematicians - G.B.Folland (pages 43 to 46 for an exposition on how to interpret the problem in terms of the Heisenberg group/Weyl relations; a counterexample is mentioned)

Foundations of Quantum Mechanics - J.Jauch (pages 197, 198 for the relation between the Weyl relations and the Imprimitivity Condition/quantum localization in homogeneous physical space)
 
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  • #38
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I don't have much time to write a detailed answer now. I refer you to the following references:

Quantum Field Theory, a tourist guide for mathematicians - G.B.Folland (pages 43 to 46 for an exposition on how to interpret the problem in terms of the Heisenberg group/Weyl relations; a counterexample is mentioned)

Foundations of Quantum Mechanics - J.Jauch (pages 197, 198 for the relation between the Weyl relations and the Imprimitivity Condition/quantum localization in homogeneous physical space)
Thank you for the great references.
The key point seems to be that in order to avoid counterexamples to the Stone-von Neuman theorem one must use the CanonicalCommutationRelations in the form of the Weyl relations, but the CCR are only equivalent to the Weyl relations in a basis-dependent way due to the unavoidable fact that the position and momentum operators in the CCR cannot be both bounded and that determines their non-uniqueness (unlike the Weyl relations case that is indeed unique up to unitary equivalence) and their inescapable(even in abstract terms) dependence on position or momentum representation.

The absence of a basis-independent unique unitary equivalence between the canonical and Weyl relations triggers questions about the causes of the absence of interaction picture in QFT(Haag's theorem), in QFT there is no position operator and precisely difficulties with the position representation in quantum relativistic mechanics i.e. the strange behaviour of Newton-Wigner localization prescription is a motivation for having to introduce quantum field theory, is this not existence of position operator in the way it exists in QM what prompts the problem with the interaction picture in QFT?, that is however salvaged in QM by the obliged preferred basis?
 
  • #39
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I've just recalled this was already explained by strangerep in a recent thread:

What's happening is that any unitary irreducible representation ("unirrep") of the Lorentz group is necessarily infinite-dimensional. In such cases, the usual Stone von-Neumann uniqueness theorem (concerning unitary equivalence of the unirreps of the canonical commutation relations) does not hold (unlike the situation in nonrel QM).

It is the failure of the S-vN thm in the inf-dim case which gives rise to the ambiguity of representation of the canonical commutation relations in QFT.

I just stressed that in the NRQM situation uniqueness is basis-dependent due to unboundedness of the position and momentum operators, but this gets too far from the OP and deserves a new thread.
 
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