Equivalence between Weyl relations and CCR

[TrickyDicky]

Due to the fact that the operators in the canonical commutation relations(CCR) cannot be both bounded, in order to prove the Stone-von Neuman theorem one must resort to the Weyl relations.
Now the Weyl relations imply the CCR, but the opposite is not true, the CCR don't imply the Weyl relations without additional not physically justified assumptions. So they are not strictly equivalent, there is a homomorphism from the Weyl relations to the CCR, but it's not bijective, there is no isomorphism between the representations if I'm not missing something here. One must fix a basis to achieve the isomorphism but then the self-adjoint operators in the CCR are not rigorously defined in Hilbert space, aren't they?

 

[dextercioby]

There's no strict equivalence, so the 'isomorphism' you mention doesn't exist: see page 275 of Reed & Simon's 1st volume of their "Methods of Modern Mathematical Physics". The "Stone-von Neumann's theorem" valid for the CCRs is called "Dixmier's theorem". Theorem 4.6.1. of Putnam's "Commutation properties of Hilbert space operators and related topics", or the original article by Dixmier in "Comp. Math.", Vol.13, 1956-1958.

 

[TrickyDicky]

There's no strict equivalence, so the 'isomorphism' you mention doesn't exist: see page 275 of Reed & Simon's 1st volume of their "Methods of Modern Mathematical Physics". The "Stone-von Neumann's theorem" valid for the CCRs is called "Dixmier's theorem". Theorem 4.6.1. of Putnam's "Commutation properties of Hilbert space operators and related topics", or the original article by Dixmier in "Comp. Math.", Vol.13, 1956-1958.

Yes. Those are the references I used for the OP(together with pages 204-5 from Jauch's "Foundations of quantum mechanics").
Now, to me these well known(to the experts) facts obviously lead to certain conclusions and basically explain away mathematically the origin of the "mysteries" in QM(all ultmately related to the "measurement problem) that people (not only science popularizers) keep harping on about and addressing with all kinds of esoterical nonsense, and "interpretations".
If scholars are aware of this fundamental flaw in the QM dynamics formalism why is it not more often mentioned and used to end the tiring debates about interpretations in QM?

 

[dextercioby]

There's no ”fundamental flaw” in the QM dynamics. It's all well governed by the Schrödinger equation.

 

[TrickyDicky]

It's all well governed by the Schrödinger equation.

Sure. Subject to the Schrodinger representation. By von Neuman theorem every irreducible representation of the Weyl relations is unitarily equivalent to the Schrodinger representation, but as discussed above the inverse is not true, not every representation of the CCR is unitarily equivalent to the Weyl relations. The operators that satisfy the CCR are never definable on the entire Hilbert space, even if they are essentially self-adjoint on their own domains, and unitarily equivalence cannot be guaranteed between representations, i.e. in the presence of time dependence(unitary time evolution).

 

[TrickyDicky]

So to be more specific I'm trying to derive some consequence from the lack of isomorphism between the Weyl representation and the canonical representation of the commutation relations. It seems to me that if we use the Schrodinger representation and following the wikipedia page about it we consider unitary transformations in it as examples of active transformations in a fixed basis and we wish to change to the Heisenberg picture(where transformations are considered passive changes of basis), due to the lack of isomorphism between representations their equivalence is only valid as long as a basis change with respect to explicit time-dependency is not performed. This dynamical change of basis would require a complete isomorphism between the Weyl representation and the canonical one to guarantee that transformations are unitary, wouldn't it? I mean if the canonical commutation relations don't imply the Weyl relations and only these are unique irreducible unitary representations(only to them applies the Stone-von Neuman theorem), there cannot be a unitary change of basis in the presence of explicit time-dependence(dynamical case) and no change of representation.
Am I missing something here? Can there actually be a unitary transformation between reperesentations that are not strictly unitarily equivalent (the ones using the CCR)?

 

[TrickyDicky]

I've been searching PF for discussions on this issue and there are not very many. I found this:

The Heisenberg pictures and Schroedinger pictures are equivalent at the non-rigourous level

which seems to imply they are not equivalent at the rigorous level but it doesn't go into specifics or consequences of this.
Could someone confirm at least if it is correct that in the presence of explicit time dependence of the hamiltonian there is no strict equivalence between the Heisenberg and Schrodinger pictures, this being connected to the fact that the unitary operator in this case has as solution an asymptotic series(Dyson series)?

 

[dextercioby]

The Heisenberg and Schrödinger pictures are equivalent at rigorous level. Proven by Hunziker and restated by de la Madrid in RHS language here; http://arxiv.org/abs/quant-ph/0509074v1

 

[TrickyDicky]

The Heisenberg and Schrödinger pictures are equivalent at rigorous level. Proven by Hunziker and restated by de la Madrid in RHS language here; http://arxiv.org/abs/quant-ph/0509074v1

In page 1 of the paper you link the author writes:"For simplicity, we restrict ourselves to pure states and to observables that do not depend explicitly on time." And several times along the paper they comment the need (in order to avoid the well known problems of domain unbounded operators have) of the:"invariance of ##\Phi## under the time evolution group". Doesn't this answer my question in #7 affirmatively?

 

[aleazk]

Both the Heisenberg picture and the Schrödinger picture are rigorously equivalent when they are defined rigorously.

First, we define time evolution as a projective representation ##\mathbb{R}\ni\tau\longmapsto\gamma_{\tau}##, where the maps ##\rho\longmapsto\gamma_{\tau}(\rho)## are Kadison symmetries (##\rho## is a positive trace class operator of trace 1; Kadison symmetries preserve the convexity of the space of states; when reduced to pure states, it's simply a Wigner symmetry). By Kadison's Theorem and using the fact that the group here is ##\mathbb{R}##, we get that there's a unitary one parameter group ##\mathbb{R}\ni\tau\longmapsto U_{\tau}## such that ##\rho_{_{\tau}}=\gamma_{\tau}(\rho)=U_{\tau}\rho U_{\tau}^{-1}##. This gives the Schrödinger evolution for states (all states, not only pure).

Now, taking the symmetries ##\gamma_{\tau}##, characterized by the unitary one parameter group ##U_{\tau}##, we can define the dual action, ##\gamma_{\tau}^{*}##, of the symmetry, which acts on the lattice of orthogonal projectors and preserves its structure: ##\gamma_{\tau}^{*}(P)\doteq U_{\tau}^{-1}PU_{\tau}## (all operators here are bounded, so there are no domain issues). An observable ##A## is represented by a projector valued measure ##P^{(A)}## over the real line. So, we define the dual action on observables by: ##P^{(A_{H}(\tau))}(E)\doteq\gamma_{\tau}^{*}(P^{(A)}(E))##, for any ##E## in the Borel algebra of the real line. We call ##A_{H}(\tau)## the observable ##A## in the Heisenberg picture, characterizaed by the PVM ##P^{(A_{H}(\tau))}##.

It's easy to show that: ##tr(\rho\gamma^{*}(P))=tr(\gamma(\rho)P)##. But, since ##tr(\rho P^{(A)}(E))## represents the probability that the measuring of ##A## in state ##\rho## falls in ##E##, the mentioned identity shows the equivalence of both pictures in a rigorous way.

And that's it.

By the spectral theorem, the dual action is obviously extended to self-adjoint operators and we get: ##A_{H}(\tau)=\gamma_{\tau}^{*}(A)= U_{\tau}^{-1}AU_{\tau}##. Also: ##\sigma(A_{H}(\tau))=\sigma(A)## and, defining the expectation value as ##<A>_{\rho}=\int_{\mathbb{R}}\lambda\mathrm{d}\mu_{\rho}^{(A)}(\lambda)## (where ##\mu_{\rho}^{(A)}## is the probability measure ##\mu_{\rho}^{(A)}(E)=tr(\rho P^{(A)}(E))##), we also get ##<A>_{\rho_{_{t}}}=<A_{H}(t)>_{\rho}##, which are the usual standard results. (notice that we only need the trace identity mentioned above for proving this result, there's no need to invoke the operators ##A##, that's the advantage of introducing this probability measure).

We could start to introduce the (most likely, unbounded) self-adjoint operators associated with the PVM ##P^{(A)}## and also to write the time evolution in infinitesimal or differential form, but in doing that you introduce a lot of spurious technical problems related to the domain of the operators involved.

The above definitions and results also hold when there's an explicit time dependence. In that case, the definition is: ##P^{(A_{Ht}(\tau))}(E)\doteq\gamma_{\tau}^{*}(P^{(A_{t})}(E))##. This can also be extended to non-homogeneous time evolution (i.e., for "time dependent hamiltonians"), though I guess some of the details may need some work, since the time evolution is not given by the one used here.

 

[atyy]

Regarding my statement about not being sure about whether the Schroedinger and Heisenberg pictures being rigourously equivalent, that was probably made in the context of QFT. Are they equivalent there too? I know some results like http://arxiv.org/abs/0806.1079, but I don't know whether that covers all of QFT.

 

[TrickyDicky]

This can also be extended to non-homogeneous time evolution (i.e., for "time dependent hamiltonians"), though I guess some of the details may need some work, since the time evolution is not given by the one used here.

Thanks for chiming in.
This is the extension and time evolution I'm actually interested in, as commented in my previous posts, and that I think is obstructed by the lack of isomorphism referred to in this thread.
Maybe you know of some reference where this is worked out and can share it, that would be great.

 

[TrickyDicky]

Regarding my statement about not being sure about whether the Schroedinger and Heisenberg pictures being rigourously equivalent, that was probably made in the context of QFT. Are they equivalent there too?

They are not equivalent as per the Haag's theorem that declares the interaction picture in QFT inexistent. I'm basically trying to trace the origin of this inequivalence back to QM, where the two pictures are traditionally considered equivalent, but I'm having a lot of trouble finding a proof that includes non-conserved systems, and this is the dynamical time dependent situation I suspect might have not a rigorous equivalence between Heisenberg and Schrodinger pictures due to the well known issues with commutation relations as summarized for instance in this excerpt from "Visual quantum mechanics" by Thaller in page 151:'The canonical commutation relations follow from the Weyl relations, but the converse is not true: The canonical commutation relations do not imply the Weyl relations. There are examples of self-adjoint operators A and B, that satisfy the canonical commutation relations on an invariant dense domain D. (The invariance means that A : D → D and B : D → D, so that the products AB and BA and the commutator are well defined on D.) The operators A and B in these examples are both essentially self-adjoint on D, and still the unitary groups do not fulfill the Weyl relations.'
This lack of isomorphism would allow rigorous equivalent pictures of unitary time evolution with a time invariant Hamiltonian, but not with an explicitly time-dependent one the way I understand it.

 

[rubi]

There is no problem with the Heisenberg picture (not in QM and also not in QFT). The Stone-von Neumann theorem is actually not really relevant at all, since it just says that the Schrödinger representation that everybody is using is actually the unique (strongly continuous) representation of the Weyl algebra. That is a nice additional fact (and it's false for the infinite dimensional version by the way), but it doesn't really matter, since if there were another representation, we probably wouldn't be using it anyway. (Why should we? The Schrödinger representation is working fine and there seems to be no need for a different representation.) Haags theorem is not related to this. The reason for the non-existence of the interaction picture for Wightman QFT's is that the interacting Hamiltonian cannot be written as the sum of the free Hamiltonian and an interaction term. (Haag tells us that ##H_{int}:=H-H_0## is necessarily not self-adjoint and hence doesn't generate the unitary one-parameter group that we use in naive calculations.)

 

[TrickyDicky]

There is no problem with the Heisenberg picture (not in QM and also not in QFT).

Who said there is a problem with the Heisenberg picture? This is about the equivalence of the the pictures.

The Stone-von Neumann theorem is actually not really relevant at all, since it just says that the Schrödinger representation that everybody is using is actually the unique (strongly continuous) representation of the Weyl algebra.

Certainly is not relevant if what you think is being discussed here is some problem with the Heisenberg problem. But actually this thread has nothing to do with that, and all sources coincide that the S-vN theorem is relevant wrt the equivalence of representations in QM.

That is a nice additional fact (and it's false for the infinite dimensional version by the way), but it doesn't really matter, since if there were another representation, we probably wouldn't be using it anyway. (Why should we? The Schrödinger representation is working fine and there seems to be no need for a different representation.)

This is quite confusing, again this is about equivalence of representations, what difference does it make if the Schrodinger rep. works fine? The equivalence is a necessary feature per the mathematical model.

Haags theorem is not related to this.

At this point I don't know what you are saying the Haag's theorem is not related to. Certainly is not related to any problem with the Heisenberg picture, on the contrary the Heisenberg picture is preferred in QFT. Precisely the chage of time from parameter to dimension, while spacetime position is not an operator makes manifest the problem with the interaction Hamiltonian.

 

[rubi]

Who said there is a problem with the Heisenberg picture? This is about the equivalence of the the pictures.

That's what I mean. There is no problem with the equivalence of the Heisenberg picture and the Schrödinger picture. Once you have a self-adjoint Hamiltonian (which you have, by the axioms of QM), you automatically get a unitary time-evolution operator for free and you can define ##A(t) := U^\dagger(t) A U(t)##, which defines your Heisenberg observables. You can recover the Schrödinger picture by doing the inverse transformation, which is also well-defined, since ##U(t)## is invertible (##U(t)^{-1}=U(-t)##). The Stone-von Neumann theorem is only relevant at the point, where you don't yet have a Hilbert space and operators on it, but rather some abstract *-algebra. Once you have a Hilbert space and operators, you are not working with the abstract *-algebra anymore, so the Stone-von Neumann theorem becomes useless.

Certainly is not relevant if what you think is being discussed here is some problem with the Heisenberg problem. But actually this thread has nothing to do with that, and all sources coincide that the S-vN theorem is relevant wrt the equivalence of representations in QM.

The equivalence of different representations of the CCR or the Weyl algebra has nothing to do with the equivalence of the Schrödinger picture and the Heisenberg picture. You are probably confusing these two things. Once we have chosen a representation of the CCR or Weyl algebra, we don't care about their uniqueness anymore. The uniqueness is relevant before we make a choice of representation. But historically, we have always been working with the Schrödinger representation already and there is no demand for another representation. (In fact, in loop quantum cosmology, people are working with non strongly continuous representations of the Weyl algebra, which are not excluded by the Stone-von Neumann theorem.)

This is quite confusing, again this is about equivalence of representations, what difference does it make if the Schrodinger rep. works fine? The equivalence is a necessary feature per the mathematical model.

Again, I am already talking about the equivalence of the pictures. The Heisenberg picture and the Schrödinger picture are both using the Schrödinger representation (which is the unique strongly continuous representation of the Weyl algebra, but that is irrelevant, since we are not planning to use a different representation anyway). If we chose a different representation, we could still talk about the Schrödinger and Heisenberg picture within that representation and all you need to switch between them is a self-adjoint Hamiltonian.

At this point I don't know what you are saying the Haag's theorem is not related to. Certainly is not related to any problem with the Heisenberg picture, on the contrary the Heisenberg picture is preferred in QFT. Precisely the chage of time from parameter to dimension, while spacetime position is not an operator makes manifest the problem with the interaction Hamiltonian.

I was referring to your reply to atyy, where you started to mention the interaction picture, which is neither the Schrödinger picture nor the Heisenberg picture. Since Haags theorem (which you mentioned in that post) refers to the interaction picture only, it is not relevant to the equivalence of the Heisenberg picture and the Schrödinger picture. Also, the problem with the interacting Hamiltonian in QFT is not due to the time variable. Actually Haags theorem requires only spatial translation invariance in its proof.

 

[TrickyDicky]

That's what I mean. There is no problem with the equivalence of the Heisenberg picture and the Schrödinger picture. Once you have a self-adjoint Hamiltonian (which you have, by the axioms of QM), you automatically get a unitary time-evolution operator for free and you can define ##A(t) := U^\dagger(t) A U(t)##, which defines your Heisenberg observables. You can recover the Schrödinger picture by doing the inverse transformation, which is also well-defined, since ##U(t)## is invertible (##U(t)^{-1}=U(-t)##). The Stone-von Neumann theorem is only relevant at the point, where you don't yet have a Hilbert space and operators on it, but rather some abstract *-algebra. Once you have a Hilbert space and operators, you are not working with the abstract *-algebra anymore, so the Stone-von Neumann theorem becomes useless.

Ok, this is clearer. But I thought I had made explicit enough in my previous posts that for pedagogical sake I'm actually placing my argument at that point, before having the Hilbert space by axiom, precisely for the reason you highlight about the relevance of the S-vN theorem.

The equivalence of different representations of the CCR or the Weyl algebra has nothing to do with the equivalence of the Schrödinger picture and the Heisenberg picture. You are probably confusing these two things. Once we have chosen a representation of the CCR or Weyl algebra, we don't care about their uniqueness anymore. The uniqueness is relevant before we make a choice of representation. But historically, we have always been working with the Schrödinger representation already and there is no demand for another representation.
Again, I am already talking about the equivalence of the pictures. The Heisenberg picture and the Schrödinger picture are both using the Schrödinger representation (which is the unique strongly continuous representation of the Weyl algebra, but that is irrelevant, since we are not planning to use a different representation anyway). If we chose a different representation, we could still talk about the Schrödinger and Heisenberg picture within that representation and all you need to switch between them is a self-adjoint Hamiltonian.

See above. I can see why you would think they have nothing to do if the axiomatic starting point is taken.

I was referring to your reply to atyy, where you started to mention the interaction picture, which is neither the Schrödinger picture nor the Heisenberg picture. Since Haags theorem (which you mentioned in that post) refers to the interaction picture only, it is not relevant to the equivalence of the Heisenberg picture and the Schrödinger picture.

I mentioned the interaction picture because this picture implies the equivalence of the the other two.

 

[rubi]

Ok, this is clearer. But I thought I had made explicit enough in my previous posts that for pedagogical sake I'm actually placing my argument at that point, before having the Hilbert space by axiom, precisely for the reason you highlight about the relevance of the S-vN theorem.

If you start with the *-algebra ##\mathfrak{A}## of observables, the argument is still exactly the same, except that you need to specify some representation ##\rho:\mathfrak{A}\rightarrow\mathcal{L}(\mathcal{H})## of your algebra on some Hilbert space ##\mathcal{H}## (either by specifying an algebraic state, specifying the representation directly or appealing to the Stone-von Neumann theorem). The Stone-von Neumann theorem tells you that you have exactly one choice for ##\rho## (up to isomorphism), if you want to have a strongly continuous representation of the Weyl algebra on your Hilbert space. If you don't care about strongly continuous representations of the Weyl algebra, then you just pick a different ##\rho## (like the people in LQC). Now you can construct your unitary one-parameter group just as before (##U(t):=e^{-i t \rho(H)}##) and your Heisenberg observables ##A(t) = U^\dagger(t)\rho(A)U(t)##. So if you start with a *-algebra, you just add an additional step. The Stone-von Neumann theorem is still not related to the equivalence of the pictures. It's purpose is only to argue for the uniqueness of the Schrödinger representation. If there was no such uniqueness argument, we would just be faced with the problem of such a choice that can't be motivated from the theory and would require input from experiments, but no mathematical problems would arise from that. (And as I said, LQC is an example, where a different representation is chosen.)

See above. I can see why you would think they have nothing to do if the axiomatic starting point is taken.

What do you mean by 'axiomatic starting point'? Starting with a *-algebra is also axiomatic. Maybe I should phrase the question differently: What do you think is wrong with the formula ##A(t) = U^\dagger(t)\rho(A)U(t)##?

I mentioned the interaction picture because this picture implies the equivalence of the the other two.

You can already prove the equivalence from the existence of a Hamilton operator alone (be it as an element of a *-algebra or as an operator on a Hilbert space). Using the interaction picture would just make it more complicated.

 

[atyy]

@rubi, does the Schroedinger picture also work in rigourous QFT? My understanding is that rigourous QFT usually uses the Wightman axioms which are Heisenberg picture, and then the actual construction is via a Euclidean path integral, which constructs a Wightman QFT if the Osterwalder-Schrader conditions are met.

 

[rubi]

@rubi, does the Schroedinger picture also work in rigourous QFT? My understanding is that rigourous QFT usually uses the Wightman axioms which are Heisenberg picture, and then the actual construction is via a Euclidean path integral, which constructs a Wightman QFT if the Osterwalder-Schrader conditions are met.

Yes, it also works for Wightman QFT's. The Wightman axioms guarantee the existence of a strongly continuous unitary representation of the Poincare group. That includes a unitary representation of time-translations. It's true that the Wightman axioms are formulated in the Heisenberg picture, but you can switch so the Schrödinger picture by defining ##\phi(x,0) := U(-t)\phi(x,t)U(-t)^\dagger## and ##\left|\Psi(t)\right> := U(t)\left|0\right>##, where ##\left|0\right>## is the vacuum state, which also exists according to the Wightman axioms. Then ##\left|\Psi(t)\right>## satisfies a Schrödinger equation corresponding to the Hamilton operator, which exists by Stone's theorem, since ##U(t)## is strongly continous. It's true however, that this can be very cumbersome if you have a theory constructed via path integrals in the Osterwalder-Schrader framework, since the reconstructed theory might be hard to work with in general. (The reconstruction theorem gives you a Wightman theory, but that theory can take a very complicated form.)

 

[atyy]

Thanks rubi. Another thing one has in the non-rigourous QFT literature is that the Schroedinger picture can be formulated using a Schroedinger functional, where the wave function is a function of the field configuration. Is this also valid in the Schroedinger picture for rigourous QFT?

 

[rubi]

This is a useful way to look at it, but it's not 100% accurate. The state ##\left|\Psi(t)\right>## from my previous post would take the role of this functional, but it's not strictly speaking a functional on the set of wave-functions in the same way as the wave-function of QM is a function of positions. This is due to the fact that the objects ##\phi(x,0)## are not really operators but rather operator-valued distributions that need to be smeared appropriately to yield properly defined operators, so you can't expand ##\left|\Psi(t)\right>## in an eingenbasis of ##\phi(x,0)##. However, you can expand it in a (generalized) eigenbasis of of "##\phi[f]:=\int\phi(x,0)f(x,t)d^4x##", where ##f## can be any sharply peaked Schwarz function. So the idea that ##\left|\Psi(t)\right>## is a wave-functional is approximately true in this approximate sense. But this is just a mathematical subtlety and it's fine to think of ##\left|\Psi(t)\right>## as a wave-functional at least intuitively.

 

[TrickyDicky]

If you start with the *-algebra ##\mathfrak{A}## of observables, the argument is still exactly the same, except that you need to specify some representation ##\rho:\mathfrak{A}\rightarrow\mathcal{L}(\mathcal{H})## of your algebra on some Hilbert space ##\mathcal{H}## (either by specifying an algebraic state, specifying the representation directly or appealing to the Stone-von Neumann theorem). The Stone-von Neumann theorem tells you that you have exactly one choice for ##\rho## (up to isomorphism), if you want to have a strongly continuous representation of the Weyl algebra on your Hilbert space. If you don't care about strongly continuous representations of the Weyl algebra, then you just pick a different ##\rho## (like the people in LQC). Now you can construct your unitary one-parameter group just as before (##U(t):=e^{-i t \rho(H)}##) and your Heisenberg observables ##A(t) = U^\dagger(t)\rho(A)U(t)##. So if you start with a *-algebra, you just add an additional step. The Stone-von Neumann theorem is still not related to the equivalence of the pictures. It's purpose is only to argue for the uniqueness of the Schrödinger representation. If there was no such uniqueness argument, we would just be faced with the problem of such a choice that can't be motivated from the theory and would require input from experiments, but no mathematical problems would arise from that. (And as I said, LQC is an example, where a different representation is chosen.)

What I'm trying to get across is that I' not referring to the case you are considering with ##U(t):=e^{-i t \rho(H)}##, indeed with that evolution operator it doesn't make any difference if one wants to change representation because the result doesn't vary. With this time evolution the CCR are equivalent to the Weyl relations and the Schrodinger and Heisenberg pictures are indeed unitarily equivalent.
But let's look at the scenario I was stressing in the this thread. Non-conserved Hamiltonian operator, that doesn't commute with itself at different times. The time evolution operator in this case is not the one written above, but something like
##U(t)={\rm T}\exp\left(-\,{{\rm i} \over \hbar}\int_{t'}^{t}H\left(t''\right)\,{\rm d}t''\right)\left\vert\Psi\left(t'\right)\right\rangle\, \qquad t > t'## , and with this operator the CCR clearly don't imply the Wey relations, but the latter are the only ones that ensure unitarily equivalent representations, and under rigorous mathematical definitions in this case the Schrodinger and Heisenberg pictures are not strictly unitarily equivalent descriptions of time evolution and a change of representation(basis change i.e. unitary transformation) is not rigorously valid.
What you were saying about the use of the Schrodinger representation in practice in QM is true, in the sense that in practical terms my previous paragraph doesn't have grave consequences, the Schrodinger representation works with active transformations of the state vectors, that is the basis is fixed from the start and there is no need to perform further changes of representation as you commented.
Do you have any disagreement with the above?

Maybe I should phrase the question differently: What do you think is wrong with the formula ##A(t) = U^\dagger(t)\rho(A)U(t)##?

As commented above I think nothing is wrong with the formula, it all depends on what specific U(t) is used for the specific experimental input.

 

[rubi]

[...]and with this operator the CCR clearly don't imply the Wey relations

The CCR and the Weyl algebra don't know anything about time-evolution. They are the same for all physical system with ##n## particles for any given ##n##, independent of the concrete Hamiltonian.

[...] but the latter are the only ones that ensure unitarily equivalent representations, and under rigorous mathematical definitions in this case the Schrodinger and Heisenberg pictures are not strictly unitarily equivalent descriptions of time evolution and a change of representation(basis change i.e. unitary transformation) is not rigorously valid.

If you have a time-dependent Hamiltonian, you can still define unitary operators ##U(t,t')## (for instance by the time-ordered exponential you wrote above, if it converges) and you can still define the Heisenberg picture as ##A(t) := U(t,t_0)^\dagger A U(t,t_0)##. I have used time-independent Hamiltonians in my post only for convenience. The only thing that matters is that ##U(t,t_0)## is a unitary operator, which it is, since the Schrödinger equation conserves probability. The Heisenberg picture and the Schrödinger picture are still exactly equivalent. It's actually just a consequence of the law of associativity:
##(\left<\Psi_0\right|U(t,t_0)^\dagger) A (U(t,t_0) \left|\Psi_0\right>) = \left<\Psi_0\right|(U(t,t_0)^\dagger A U(t,t_0)) \left|\Psi_0\right>##
If the Schrödinger picture and the Heisenberg picture were not equivalent, it would imply that the associative law was wrong. (Of course, ##\left|\Psi_0\right>## must be such that ##U\left|\Psi_0\right>## is in the domain of ##A##, but that has no influence on the equivalence of the pictures.)

What you were saying about the use of the Schrodinger representation in practice in QM is true, in the sense that in practical terms my previous paragraph doesn't have grave consequences, the Schrodinger representation works with active transformations of the state vectors, that is the basis is fixed from the start and there is no need to perform further changes of representation as you commented.
Do you have any disagreement with the above?

It's not only true that your paragraph doesn't have consequences. It also has exactly no consequences at all. The mathematical subtlety you are thinking of just isn't there and I think the misunderstanding is that you think the time evolution is somehow related to the Weyl algebra. The Weyl algebra is concerned only with position and momentum shifts and not with time evolution.

As commented above I think nothing is wrong with the formula, it all depends on what specific U(t) is used for the specific experimental input.

The Schrödinger picture and the Heisenberg picture are equivalent whenever you have a unitary time-evolution operator ##U(t,t_0)##. That is independent on how ##U## was handed to you. It may be given by a time-independent Hamiltonian, a time-dependent Hamiltonian or even a some folium-preserving *-automorphism of your algebra of observables if you want to be totally general.

 

[TrickyDicky]

The CCR and the Weyl algebra don't know anything about time-evolution. They are the same for all physical system with ##n## particles for any given ##n##, independent of the concrete Hamiltonian.

Right, but I think you are missing how the time evolution fits in the picture(npi). The algebras may "know nothing about time evoution" by themselves but let's remember why the asymmetry in the uniqueness of the representations of the commutation relations comes about. The problem is that for the uniqueness of the representation the operators must be bounded, like is the case for the Weyl relations. In the CCR they can't both be bounded, now this is in practice dodged by the fact that this operators form a conjugate pair that cannot be measured simultaneously but this imposes something relevant in relation with time, since the operators can't be there at the same time, you must be either in the position or the momentum basis and this implies that for any unitary transformation which tranforms one basis into another, the time evolution Hamiltonian must commute with itself at different times, that is it must be a time independent hamiltonian in order to preserve the uniqueness of the unitary representation like the Weyl relations. So in practice there are no empirical differences between the two pictures, and in that sense they are equivalent, but they can't be said to be equivalent in a rigurous mathematical sense when using the CCR, because you need to introduce some kind of non-unitary time evolution(call it collapse in the case of measurements, or truncating the Dyson series) when these unbounded operators are involved in the presence of [H(t),H(t')]≠0.

If you have a time-dependent Hamiltonian, you can still define unitary operators ##U(t,t')## (for instance by the time-ordered exponential you wrote above, if it converges) and you can still define the Heisenberg picture as ##A(t) := U(t,t_0)^\dagger A U(t,t_0)##. I have used time-independent Hamiltonians in my post only for convenience. The only thing that matters is that ##U(t,t_0)## is a unitary operator, which it is, since the Schrödinger equation conserves probability.

Yes, well, you should mention also that one must do something to make it converge as the exponential is divergent(Dyson series), and this something involves nonunitary operations. ##U(t,t_0)## is unitary by convention of the formalism, it is just basically demanded that the operator takes unit normalized states to unit normalized states, but the norm of the wavefunction is not observable so one could demand anything convenient to the situation at hand, in this case unit normalized norms are convenient. But this is taking us far from the equivalence issue.

The Heisenberg picture and the Schrödinger picture are still exactly equivalent. It's actually just a consequence of the law of associativity:
##(\left<\Psi_0\right|U(t,t_0)^\dagger) A (U(t,t_0) \left|\Psi_0\right>) = \left<\Psi_0\right|(U(t,t_0)^\dagger A U(t,t_0)) \left|\Psi_0\right>##
If the Schrödinger picture and the Heisenberg picture were not equivalent, it would imply that the associative law was wrong. (Of course, ##\left|\Psi_0\right>## must be such that ##U\left|\Psi_0\right>## is in the domain of ##A##, but that has no influence on the equivalence of the pictures.)

I'd say it does have influence, this is after all originated by a problem of domains of unbounded operators.

It's not only true that your paragraph doesn't have consequences. It also has exactly no consequences at all. The mathematical subtlety you are thinking of just isn't there and I think the misunderstanding is that you think the time evolution is somehow related to the Weyl algebra. The Weyl algebra is concerned only with position and momentum shifts and not with time evolution.

See above.

 

[rubi]

Right, but I think [...] presence of [H(t),H(t')]≠0.

There is so much wrong with this paragraph that I don't even know how to respond to this. It really doesn't make any sense. You are just stringing together words without understanding their meaning.

Yes, well, you should mention also that one must do something to make it converge as the exponential is divergent(Dyson series), and this something involves nonunitary operations. ##U(t,t_0)## is unitary by convention of the formalism, it is just basically demanded that the operator takes unit normalized states to unit normalized states, but the norm of the wavefunction is not observable so one could demand anything convenient to the situation at hand, in this case unit normalized norms are convenient. But this is taking us far from the equivalence issue.

1. Whenever the Cauchy problem of the Schrödinger equation is well-posed and ##H(t)## is self-adjoint, there exists a family of unitary operators ##U(t,t_0)##. If the Cauchy problem is not well-posed, there is no time-evolution at all and it doesn't make sense to talk about the Schrödinger picture or the Heisenberg picture to begin with.
2. The convergence of the Dyson series is not required for ##U(t,t_0)## to exist.
3. Whether the norm of the wavefunction is observable or not is completely irrelevant for the existence of ##U(t,t_0)##.

I'd say it does have influence, this is after all originated by a problem of domains of unbounded operators.

The point of math is not to say something, but to prove it. If you knew anything about unbounded operators at all, you would know that the expressions on both sides of the equation are only defined if the domains are restricted appropriately and that in this situation, the law of associativity holds exactly. The multiplication of operators is precisely defined to be associative. The proof for the equivalence of the Schrödinger picture and the Heisenberg picture is exactly the application of the law of associativity. The left hand side is the Schrödinger picture and the right hand side is the Heisenberg picture. If the equal sign between them holds, which it does by the very definition of operator multiplication, then both sides are equal and therefore the pictures are equivalent. Yes, it is really that trivial.

Here is one last thing that should provoke your thought: If there was anything wrong with the equivalence of the Schrödinger and the Heisenberg picture, then certainly, someone would have noticed by now and it would have been mentioned in the literature. It should appear very unlikely to you that a layperson with no clear understanding of the basics discovers a problem with something that has been studied by thousands of highly intelligent physicists and mathematicians, many of them who dedicated their lives to the subject, for almost 100 years, especially in those parts of the theory that are taught to every undergrad. So if someone tries to explain to you where your mistake is, you should try to understand where you are wrong, rather than justify your bold claims with incoherent use of language.

 

[aleazk]

Thanks for chiming in.
This is the extension and time evolution I'm actually interested in, as commented in my previous posts, and that I think is obstructed by the lack of isomorphism referred to in this thread.
Maybe you know of some reference where this is worked out and can share it, that would be great.

The argument in my previous post only asks the time evolution to be given by a bounded operator (in this way, the different actions can be defined on the whole Hilbert space and, since ##\rho## is of trace class, the cyclic property of the trace holds and is used to prove the trace identity, which is the relevant result).

In the case of non-homogeneous time evolution, the time evolution is not characterized anymore by a (strongly continuous) one parameter unitary group, but by a family of unitary operators that depend on the initial time, ##\left\{ U(t_{1},t_{2})\right\} _{t_{1},t_{2}\in\mathbb{R}}## (if the time dependent hamiltonian is bounded, these operators can be rigorously calculated via the Dyson series). But they are still unitary, therefore bounded, and therefore the proof is the same from this point until the end if we define ##\rho_{t}=V_{t}\rho V_{t}^{\dagger}## and ##P^{(A_{H}(t))}=V_{t}^{\dagger}P^{(A)}V_{t}##, where ##V_{t}=U(t,0)## (which are unitary for each ##t## and therefore bounded; the operators ##V_{t}## for different values of ##t## do not form a unitary one parameter group, though).

 

[TrickyDicky]

Instead of displaying snide remarks you could just ask for clarification of the specific passages you didn't understand.

 

[aleazk]

TD: the equivalence has nothing to do with the Stone-von Neumann theorem, the CCR, etc. In fact, it can be proved in any arbitrary Hilbert space because it's a more primitive property, related to how you apply symmetries on this Hilbert space (the good old passive vs. active).

 

[rubi]

Instead of displaying snide remarks you could just ask for clarification of the specific passages you didn't understand.

I am not misunderstanding anything. I can tell exactly that you don't really understand the meaning of the words you are using in that paragraph and I find it inappropriate to respond this way if someone tries to help you. It is literally impossible that someone who understood QM, representation theory and the Stone-von Neumann theorem would write such a paragraph. For example, what does it even mean that "operators can't be there at the same time"? Operators are either there or they are not there. From this nonsensical statement, you deduce an implication that just doesn't follow. Then you claim that this implies something about empirical equivalence (whatever that means), which is also a non sequitur. Then suddenly you argue about how collapse is supposed to be relevant and remark that it could have something to do with a truncated Dyson series (how in the world could you get this idea?).

The whole whole paragraph is a big non sequitur and the vague, imprecise formulation makes it impossible to even identify the mistake in your thinking.

 

[aleazk]

The Heisenberg picture and the Schrödinger picture are still exactly equivalent. It's actually just a consequence of the law of associativity:
##(\left<\Psi_0\right|U(t,t_0)^\dagger) A (U(t,t_0) \left|\Psi_0\right>) = \left<\Psi_0\right|(U(t,t_0)^\dagger A U(t,t_0)) \left|\Psi_0\right>##
If the Schrödinger picture and the Heisenberg picture were not equivalent, it would imply that the associative law was wrong. (Of course, ##\left|\Psi_0\right>## must be such that ##U\left|\Psi_0\right>## is in the domain of ##A##, but that has no influence on the equivalence of the pictures.)

Hi, rubi, I agree with everything you said in this thread. I also know that you said what I quoted in order to show how the equivalence is trivially shown and that it has nothing to do with the Stone-von Neumann theorem, etc. But, let me make the small correction that this argument doesn't necessarily show the equivalence of both pictures in the most general sense in which it's possible, or, most importantly, required. And that's because the definition of expectation values as ##<A>_{\psi}=(\psi\mid A\psi)##, with ##\psi\in D(A)##, is, though correct, not the most general one.

Given an observable ##A## characterized by the PVM ##P^{(A)}## and a state ##\rho## (a positive trace class operator of trace 1), one can define the following probability measure associated to the pair (observable, state): ##\mu_{\rho}^{(A)}(E)=tr(\rho P^{(A)}(E))##, for every Borel set ##E## of the real line. By its definition, it's clear that the interpretation of the measure ##\mu_{\rho}^{(A)}## is that it represents the probability that the measuring of ##A## in state ##\rho## falls in ##E##. In this way, we can define the expectation value of the observable ##A## in state ##\rho## by: ##<A>_{\rho}=\int_{\mathbb{R}}\lambda\mathrm{d}\mu_{\rho}^{(A)}(\lambda)##.

The expectation value will exist when ##\mathbb{R}\ni\lambda\longmapsto\lambda## is in ##L^{1}(\mathbb{R},\mu_{\rho}^{(A)})##. This will depend on the state ##\rho##. Now, let's restrict to pure states only; ##\rho_{\psi}## will denote the pure state determined by the normalized vector ##\psi\in\mathcal{H}##. It can be shown that: i) ##<A>_{\rho_{\psi}}## exists iff ##\psi\in D(\mid A\mid^{\frac{1}{2}})##; ii) if ##\psi\in D(A)## then the expectation value clearly exists since ##D(A)\subset D(\mid A\mid^{\frac{1}{2}})##, in this case we also get ##<A>_{\rho_{\psi}}=(\psi\mid A\psi)##.

Thus, the problem with that argument is that it only shows the equivalence of both pictures for only a proper subset from the set of all pure states for which the expectation value can actually be defined. It could turn out that for these other pure states the equivalence is not valid. Just to exemplify how tricky these things can be, there are examples in which two unbounded operators commute in a common domain in which both are essentially self-adjoint but, nevertheles, the associated (to the self-adjoint extensions) PVMs do not commute.

The solution to this problem is easy, one simply defines everything in terms of the PVMs. After doing this, the equivalence can be rigorously proved for all states in which the expectation value exists, as I mentioned in a previous post.

 

[berkeman]

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