# Equivalence between Weyl relations and CCR

1. Apr 25, 2015

### 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?

2. Apr 26, 2015

### 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.

3. Apr 26, 2015

### TrickyDicky

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?

4. Apr 26, 2015

### dextercioby

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

5. Apr 26, 2015

### TrickyDicky

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).

6. Apr 28, 2015

### 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)?

7. May 3, 2015

### TrickyDicky

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

8. May 3, 2015

9. May 3, 2015

### TrickyDicky

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?

10. May 3, 2015

### 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.

Last edited: May 3, 2015
11. May 3, 2015

### 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.

12. May 4, 2015

### TrickyDicky

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.

13. May 4, 2015

### TrickyDicky

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.

Last edited: May 4, 2015
14. May 4, 2015

### 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.)

15. May 4, 2015

### TrickyDicky

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

16. May 4, 2015

### rubi

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.

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.)

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.

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.

17. May 5, 2015

### TrickyDicky

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

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

18. May 5, 2015

### rubi

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 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 $A(t) = U^\dagger(t)\rho(A)U(t)$?

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.

19. May 5, 2015

### 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.

20. May 5, 2015

### rubi

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.)