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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJul 25th 2011

I have slightly expanded, reorganized and polished the entry state. (Added definition of classical state in Heisenberg picture, added pointer to the entry classical state, moved pointers to quasi-state and state in AQFT and operator algebra to the paragraph on quantum states and added at the very end a list of “related concepts” in an attempt to organize what used to be somewhat of a mess here). But this entry deserves to be polished and organized and expanded still more.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeJul 25th 2011

I added some stuff to classical state and wrote an idea section for quantum state.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJul 25th 2011

Thanks!

I have created a Definition-section at quantum state and very briefly added the definition in the Heisenberg picture. Need to expand on this. But am in a rush now.

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeJul 25th 2011

At both quantum state and state, you seem to say that the Schroedinger picture only treats the case where a quantum mechanical system is given by a Hilbert space, while the Heisenberg picture treats the general case where the system is given by a $C^*$-algebra. This makes no sense to me.

If the difference between the pictures is simple the question of whether states or observables evolve with time, then this is completely orthogonal to the distinction that you’re writing about. The best justification that I can think of is that people working on general axiomatic formulations of quantum mechanics will be drawn to both the $C^*$-algebra approach and a static notion of state, but for different reasons. It’s certainly possible to use a $C^*$-algebra of static observables and a space of evolving states on that algebra, or else to use a Hilbert space of evolving (pure) states and an algebra of static operators on that Hilbert space.

Perhaps you intend more than this when you refer to the pictures, since there were many other differences between Schroedinger and Heisenberg. But both of them gave formulations that von Neumann described using a Hilbert space, and both only later were treated in terms of $C^*$-algebras. The best justification that I can think of is that Heisenberg made the algebra of observables more prominent that Schroedinger did, but of course both had an algebra of observables and a space of states. And neither, strictly speaking, used either a Hilbert space or a $C^*$-algebra originally.

I do think that your two-stage description can be useful. But its use is that people often learn quantum mechanics using Hilbert spaces before learning the operator algebra approach. So I’ve edited the pages to reflect this.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeJul 25th 2011

I guess you are right, maybe the way I wrote it is a distortion of the history to some extent.

But not much, is it? Of course both observables and states exist in both pictures, but in one picture the observables are there by definition and the states are derived, while in the other picture it is the other way round. Don’t you agree?

In the Schrödinger picture the Hilbert space is given a priori. So you say: a state is a ray in there. In the Heisenberg picture on the other hand we do not need to mention any Hilbert spaces at all (even if maybe historically people did not keep the two notions apart so carefully) and to produce states we first need to go through some yoga.

To me that seems to be a very good and useful way of looking at it. Given that you have already further edited the entry, do you think we need to do more to clarify this, or is it okay now?

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeJul 27th 2011
• (edited Jul 27th 2011)

Of course both observables and states exist in both pictures, but in one picture the observables are there by definition and the states are derived, while in the other picture it is the other way round. Don’t you agree?

No, I don’t. It certainly is possible to start with one and derive the other, but this is a separate issue from which one we regard as evolving through time. If both of these issues are referred to in literature as the issue of “picture”, then the literature is conflating two separate issues (which would be understandable, given the etymology).

I think that it’s historically accurate to say that Schrödinger began with states while Heisenberg began with operators, but I don’t think that either of them derived one systematically from the other. Certainly neither mentioned Hilbert spaces (that was John von Neumann, years later) or C*-algebras (that was Irving Segal, decades later). I should read the primary sources, but secondary sources will say things like:

• in Schrödinger’s wave mechanics, the state of a quantum system (at any given time) is given by a complex-valued function on configuration space, while an observable is given by a self-adjoint differential operator;
• in Heisenberg’s matrix mechanics, the state is given by an infinite-dimensional complex vector, while an observable (at any given time) is given by a self-adjoint infinite-dimensional matrix.

It’s easier to make this description of wave mechanics mathematically precise if you start with the states (operators on what space of functions?), but it makes no difference to matrix mechanics. And if you swap the locations of the parentheticals “(at any given time)”, then while it no longer accurately reflects what Schrödinger or Heisenberg did, it’s still a perfectly sensible system; that’s separate from every other issue. Yet that is the one thing that I know as the “picture” issue today.

None of this is to deny the important difference between starting with a Hilbert space vs starting with a C*-algebra. It’s just that this is:

• independent of whether states or observables evolve through time,
• not the issue that I know of as “picture”,
• not attributable to Schrödinger and Heisenberg.

Possibly we could name these after von Neumann and Segal (and use another word than “picture”), but I’d rather just mention Hilbert spaces and C*-algebras.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeJul 31st 2011
• (edited Jul 31st 2011)

Hi Toby,

not sure how we should resolve this.

Incidentally, I just noticed that in this blog here they are also currently discussing this.

Some Theo (I guess it’s MO’s Theo) there voices an opinion that seems to be more in line with mine. Not that this means anything.

He also amplifies the TQFT=Schrödinger and AQFT=Heisenberg identification. I should charge a copyright license fee for that.

• CommentRowNumber8.
• CommentAuthorTobyBartels
• CommentTimeJul 31st 2011

Although the blog post discusses the Schrödinger picture only in the context of a Hilbert space (a Hilbert space of the form $L^2(X)$, in fact, although that’s no loss of generality if you assume the axiom of choice), later the author still writes

In the Schrödinger picture, we keep our algebra of observables $A$ invariant and modify the state $\psi$ (or equivalently the expectation $\mathbb{E}$), but in the Heisenberg picture, we keep the expectation $\mathbb{E}$ invariant and modify the algebra of observables $A$

which indicates how the Schrödinger picture applies to the more general case of an arbitrary algebra of observables.

I’m afraid that I don’t really understand Theo’s comment. He writes (paragraph 2) ‘If some conditions are satisfied (they aren’t, but let’s pretend they are)’, and I’m not sure what to make of that or his equivalence of categories in the next paragraph (which depends on these unfulfilled conditions). I don’t know why he puts the Schrödinger picture in $Vect$ (why not at least $Hilb$?) or why he needs a vector space $V$ before he can say what category the Heisenberg picture is in. (With Schrödinger in $Vect$, I anticipated Heisenberg in $Alg$, but really it’s in $End(V) Mod$ for some reason. I don’t usually think of the Heisenberg picture as dependent on the Schrödinger picture in this way; if anything, it’s the reverse.) I agree that constructing $End(V)$ from a vector (Hilbert) space $V$ is important (especially to understand the Heisenberg picture of a quantum mechanical system given in a Hilbert space formalism), and therefore (in the categorified case) constructing ‘something like’ an $E_n$-algebra $End(V)$ from a TQFT $V$, but I don’t see where AQFT comes into it. Nowhere does he mention time evolution, which I guess is a point against me (since I see whether states or observables evolve with time as the essential difference between the pictures), but I still don’t understand him.

I agree that work on FQFT usually uses the Schrödinger picture, while AQFT usually uses the Heisenberg picture. Nevertheless, if you assume that spacetime is globally hyperbolic and pick a foliation by Cauchy surfaces, then AQFT has a Schrödinger picture (or more generally, one Schrödinger picture for each such foliation). I don’t know how to make a Heisenberg picture of FQFT, but on the other hand, I don’t know how to make ordinary quantum mechanics (including Schrödinger’s wave mechanics) into an FQFT either, so I tend to think of FQFT as a rather different subject.

• CommentRowNumber9.
• CommentAuthorTobyBartels
• CommentTimeJul 31st 2011
• (edited Jul 31st 2011)

In any case, both state and quantum state should have the structure that you gave them by separating out these cases (Hilbert space vs algebraic). Also, state often had information that was lacking in quantum state! (and directly about quantum mechanics). So I reorganised both of them.

I also added some material to the Idea section of state.

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeAug 1st 2011
• (edited Aug 1st 2011)

I don’t know how to make a Heisenberg picture of FQFT,

That’s what I tried to answer in this article for extended FQFT on cobordisms with Lorentzian structure.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeAug 1st 2011

That’s what I tried to answer in this article for extended FQFT on cobordisms with Lorentzian structure.

So if you haven’t been aware of that argument yet, I’d be interested in hearing your comments. (I took the liberty of adding this as a reference to picture of mechanics, too.) I have been thinking that there would be a whole lot of interesting things still to be said along these lines, but didn’t really get around to much, yet. One observation that I would like to eventually expand on is that if we regard a Lorentzian manifold as a smooth poset, then the path $n$-groupid that underlyies the extended FQFT description on that manifold can be nicely understood as being the $n$-category of paths in that poset in generalization of the notion of paths in an orbifold. Some notes on this are in the section The path (oo,1)-category of a Lorentzian space at Lorentzian manifold.

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeSep 8th 2018
• (edited Sep 8th 2018)

since the main content of this entry for the time being remains to be a pointer to the details given at state in AQFT and operator algebra, I have made that pointer more easily recognizable now.