Schrödinger's cat: what's the state?

In summary: Hilbert space is non-separable. Still, the algebra of observables is commutative. (And the so-called "classical states" are just the points of the phase space, and they are what correspond to the 1D projections in this construction.) Now, as a matter of logic, if we limit ourselves to commuting observables, no QM state can assign different eigenvalues to them. So, sure, we could consider a QM state that distributes probability over the states of the classical algebra of functions on the phase space you're thinking of, but that's not a quantum state—you can't prepare it (I'm assuming we're not talking about mixed states here). So, in particular, a quantum state
  • #1
Peter Morgan
Gold Member
274
77
A physicist prepares a box and tells us that in the box there is a cat that is in a superposition of being alive and being dead. How can we be sure whether they're telling the truth? Is the state a superposition or a mixture?
If we open the box and measure only whether the cat is alive, using the projection operator ##\hat A=\left(\begin{array}{cc}1 & 0\\0 & 0\end{array}\right)##, we can't tell, because for example for two different density matrices, a mixture ##\hat M_\alpha=\left(\begin{array}{cc}\alpha & 0\\0 & 1{-}\alpha\end{array}\right)## or a pure state ##\hat S_\alpha=\left(\begin{array}{cc}\alpha\!\! & \!\!\!\!\sqrt{\alpha(1{-}\alpha)}\!\\ \!\!\sqrt{\alpha(1{-}\alpha)}\!\!\!\! & \!\!1{-}\alpha\end{array}\right)##, we obtain precisely the same probability, ##\alpha##, that the cat is alive. To tell whether the physicist is lying, we have to use other observables, such as what I'll call the Lewis Carroll operator, ##\hat C=\left(\begin{array}{cc}0 & 1\\1 & 0\end{array}\right)##, which takes a live cat and kills it and takes a dead cat and resuscitates it. With this operator as well as ##\hat A##, we can construct a different projection operator, ##\hat P_{\!+}=\frac{1}{2}\left(\begin{array}{cc}1 & 1\\1 & 1\end{array}\right)## and measure a probability such as ##\mathsf{Tr}\left[\hat P_{\!+}\hat\rho\right]##, where ##\hat\rho## is whatever the state is, and discriminate at least between ##\hat M_\alpha## and ##\hat S_\alpha## (to differentiate between all possible states requires more projection operators, but the above is enough for the discussion that follows).
Quantum Mechanics, however, tells us that measurements in Classical Mechanics are always mutually commutative, and, indeed, that's perhaps the fundamental difference between QM and CM. In that case, CM measurements cannot tell whether the alleged QM preparation of a superposition is what the quantum preparer says it is or not. If CM accepts that all its measurements are and must be mutually commutative, a classical physicist can just say, "Huh, it's just a mixture, which I understand well enough", you're just lying that it's a superposition. In fact, however, the Lewis Carroll operator is well-enough-defined, as above, for a classical physicist, it's just not easy to implement for a classical cat. QM is just making a straw man for itself.
If the classical physicist is not prevented from using the Lewis Carroll and similar operators, then they can tell whether the state is a pure state, and they can confirm the quantum physicist's claims about the state, but with that expansion of what a classical physicist can do, to what I call CM+ in my paper Unary Classical Mechanics, a quantum physicist is much less different from a classical physicist than has usually been asserted.
So: what flaws are there in this argument, what subtleties do people feel should be developed, and is this argument in the literature already?
 
  • Like
Likes Greg Bernhardt
Physics news on Phys.org
  • #2
An observable of the total system (cat + environment) will tell whether it is in superposition or mixture.

See for examples chapter 8 of "Quantum Mechanics and Experience" by David Albert and pages 373-376 of 'Sneaking A Look At God's Cards' by GianCarlo Ghirardi.
 
  • Like
Likes Peter Morgan
  • #3
StevieTNZ said:
An observable of the total system (cat + environment) will tell whether it is in superposition or mixture.

See for examples chapter 8 of "Quantum Mechanics and Experience" by David Albert and pages 373-376 of 'Sneaking A Look At God's Cards' by GianCarlo Ghirardi.
Thanks for the particular reference, which I've now looked at, admittedly at some speed (someone on Facebook also suggested that book by Albert, though without suggesting a particular chapter, so I was well motivated to look). As far as I can tell, however, Albert's Chapter 8 doesn't much speak to the account I give above. In particular, AFAICT he works only with pure states. He asserts that a pure state properly describes the observer and the measurement apparatus and the measured system (the cat) [which we could reduce to a mixed state over just the cat's degrees of freedom, which in general will not be diagonal in whether the cat is alive or dead]. Fair enough, but I suppose (1) the argument as I give it above applies just as much to Observer+Apparatus+System in a box as it does to just a cat in a box (albeit there's not such a cute name as "the Lewis Carroll operator"); and (2) the argument as I give it above applies to any claim that a mixed state that properly describes the cat has any off-diagonal elements in the alive/dead basis determined by the ##\hat A## projection operator.
For the Ghirardi, it looks like I'll have to borrow the book IRL.
 
  • #4
StevieTNZ said:
pages 373-376 of 'Sneaking A Look At God's Cards' by GianCarlo Ghirardi
I've had a look at that as well, and again it seems there's the assumption that the QM experimenter can tell that they've produced a superposition and not a mixture because they can carry out mutually noncommutative measurements, which I don't contest. The CM experimenter, however, assuming they accept what the QM exceptionalist tells them, that CM can't measure mutually incompatible measurements (because the algebra of CM measurements is commutative), can't verify that claim. To repeat the suggestion in the OP, if the CM experimenter says, "sure, I can check that", then they are very much closer to being a QM experimenter than they are usually allowed to be.
 
  • #5
To the OP above, I would add:
"Insofar as resuscitation of a long-dead cat is in practice impossible for either a classical or a quantum physicist, of course no-one can prepare an eigenstate of the Lewis Carroll operator. If we can physically implement such reversals for a given real system, however, it can equally be modeled by a classical or a quantum physicist."​
 
  • #6
Peter Morgan said:
Quantum Mechanics, however, tells us that measurements in Classical Mechanics are always mutually commutative

How does QM tell us this?
 
  • #7
PeterDonis said:
How does QM tell us this?
The observables in classical mechanics are usually said to be the commutative algebra of functions on whatever the phase space is. Right? A Gibbs state over that algebra allows us to construct a Hilbert space, say H, which is what Section V of the Unary Classical Mechanics I link to in the OP does, loosely following Koopman's 1931 construction. The algebra of observables is still commutative on the usual convention, despite this construction of a Hilbert space.
It is perhaps a whimsy to say that it's QM that tells us that CM must have a commutative algebra of observables, but if no-one is telling CM that its algebra of observables must be commutative, then I guess CM can and perhaps should start using the Lewis Carroll operator and any operator that acts on the Hilbert space H that we can construct using the Gibbs state, whenever it seems convenient or empirically necessary to do so. Which, so the story goes, makes CM very much closer to QM than it's usually thought to be.
Sorry if that's too much for your short question! I have to keep tamping down my enthusiasm for this mathematics.
 
  • #8
Peter Morgan said:
It is perhaps a whimsy to say that it's QM that tells us that CM must have a commutative algebra of observables

No, I would say it's just wrong unless you can show how "CM must have a commutative algebra of observables" can be deduced from QM.

Peter Morgan said:
if no-one is telling CM that its algebra of observables must be commutative

What does "telling" have to do with it? You already said earlier in your post that when you construct a Hilbert space from an algebra of functions on a classical phase space, the algebra of observables is still commutative. So nobody had to "tell" CM that the algebra of observables "must" be commutative: it just is commutative, because that's what the math says.

Peter Morgan said:
I guess CM can and perhaps should start using the Lewis Carroll operator and any operator that acts on the Hilbert space H that we can construct using the Gibbs state

But you said earlier in your post that the construction using the Gibbs state leads to a commutative algebra of observables. So how does this "CM" get closer to QM?
 
  • #9
PeterDonis said:
What does "telling" have to do with it? You already said earlier in your post that when you construct a Hilbert space from an algebra of functions on a classical phase space, the algebra of observables is still commutative. So nobody had to "tell" CM that the algebra of observables "must" be commutative: it just is commutative, because that's what the math says.
I can agree with your comments above. Whimsy or not, no-one is telling anyone anything. I had already backpedaled in the way I had written this argument into a new section of my Unary Classical Mechanics, but I've just edited it to excise any hint of "telling". Thanks!
The math in, say, a Hamiltonian presentation of classical mechanics, amongst the many presentations that have been constructed over the last three centuries and more, certainly presents the algebra of observables as commutative. Even in a Hamiltonian presentation, however, there is a noncommutative algebra of transformations of those observables, which is generated by the Poisson bracket. That transformation algebra does not result in a noncommutative algebra of observables, but in a Koopman presentation of classical mechanics a noncommutative algebra of observables is very natural, albeit certainly an extension. To me, commutative CM starts to look like a straw man. It's not that CM has to be extended to be noncommutative in a Koopman formalism, it's just that someone using CM can extend the algebra of observables to be noncommutative if they find it convenient to do so. If CM has to be commutative, I think it's clear that a classical mechanic can't use the Lewis Carroll operator ##\hat C## as well as ##\hat A##, even though it kinda makes classical sense, albeit not for a cat, right? Then the classical physicist can reasonably say that on the basis of their measurements the cat is just a mixture, right?
Anyway, I will understand if you don't want to spend any more time on this, but it would be counterproductive for us both for me to try to reproduce the six pages of "Unary Classical Mechanics" here. If you decide to have a look at that, which some academics have found very interesting, FWIW, though others have not, and want to say or ask anything else, obviously I'm at your disposal. Otherwise, thanks again for pushing me into a corner above.
PS: The community that works with this kind of formalism usually calls it the Koopman-von Neumann formalism, which AFAICT is for the von Neumann name drop rather than because von Neumann did much to push the formalism forward. It's often abbreviated to just KvN. To me —though, yeah, I could be wrong— there's something of a sense that the published literature on KvN has been making progress since about 2000, with me as just a small cog in that process. I was especially encouraged to discover last year that Carlton Caves, a big cog, did some work in 2012 that makes a fairly robust connection between Quantum Non-Demolition measurements and KvN.
 
  • #10
Peter Morgan said:
The math in, say, a Hamiltonian presentation of classical mechanics, amongst the many presentations that have been constructed over the last three centuries and more, certainly presents the algebra of observables as commutative.

I don't see this as a matter of "presentation". The observables are what they are; either they commute, or they don't. You find that out by constructing a specific model for a specific purpose. Even in QM there are sets of observables that commute (for example, linear momentum components in different directions); if those observables are sufficient to solve the problem you're trying to solve, then QM could just as easily be said to have a commutative algebra of observables for that case.

Peter Morgan said:
Then the classical physicist can reasonably say that on the basis of their measurements the cat is just a mixture, right?

It seems to me that the question of which state the cat is in is an objective question, which should not depend on whether we call ourselves "classical" or "quantum" physicists, or how we label our observables; and similarly for the question of what observables we would need to measure to distinguish a particular pair of possible cat states.
 
  • #11
PeterDonis said:
Even in QM there are sets of observables that commute (for example, linear momentum components in different directions); if those observables are sufficient to solve the problem you're trying to solve, then QM could just as easily be said to have a commutative algebra of observables for that case.
Certainly. Equally, however, if a commutative algebra of observables is not enough to model an experiment, as for Bell-EPR experiments, et cetera, but CM can be extended so that it has a noncommutative algebra of observables, and with that extension CM then can model such experiments, that is perhaps OK, too.
PeterDonis said:
It seems to me that the question of which state the cat is in is an objective question, which should not depend on whether we call ourselves "classical" or "quantum" physicists, or how we label our observables;
If the CM state space is isomorphic to the QM state space, then yes, perhaps up to a messy discussion about ontology and interpretation of the mathematics. Otherwise, I think the question runs deep.
[Sorry, what I think is a bug just lost a longer comment that I don't have time to type in again. I accidentally tried to edit the preview, and PF appears to have instantly just deleted all of it. Probably no loss to anyone else, of course.]
 
  • #12
Peter Morgan said:
CM can be extended so that it has a noncommutative algebra of observables

What would be the difference between "CM extended" in this way and QM?
 
  • Like
Likes Peter Morgan
  • #13
PeterDonis said:
What would be the difference between "CM extended" in this way and QM?
There can be issues about complex structure:
  • There is an isomorphism between an EM random field and quantized EM (which uses the Hodge dual as a complex structure: one can see arXiv:1709.06711, "Classical states, quantum field measurements", which is perhaps about to be published if a referee decides the changes I've made are acceptable);
  • and there is one between the Klein-Gordon random field and the quantized complex KG field (one can see arXiv:0905.1263, "Equivalence of the Klein-Gordon random field and the complex Klein-Gordon quantum field", published in EPL, or Appendix B of arXiv:1709.06711);
  • but there is no such isomorphism, as far as I know, between a classical random field and the quantized real Klein-Gordon field (which, however, I take to be not physical).
For the quantized Dirac field, I give a construction in section 3 of arXiv:1709.06711 that I find of significant interest, however that is perhaps more for reasons that go too far into my principal research interest, which is the mathematically rigorous construction of interacting QFTs (also, the construction is a little heavily into my own notation, which may take time for you to decode).
If you like a bit of math, try Vladimir Kisil's discussion of real and complex representations of the Heisenberg group, arXiv:1611.05650, published in Geometry, Integrability, and Quantization.
A further difference, which shows up in various places through the literature, is that the Hamiltonian function in CM is positive semi-definite, however the Liouvillian operator that is constructed by using the Poisson bracket to generate time-like translations in CM is not positive semi-definite. In contrast, the Hamiltonian operator in QM that is constructed by using the Correspondence Principle to generate time-like translations in QM is positive semi-definite. This, I think, goes to the heart of the relationship between CM and QM in KvN approaches being about isomorphisms between Hilbert spaces and operator algebras whereas the relationship between CM and QM in the traditional approaches is about the much messier business of quantization. The difference is very telling, IMO, and I think everyone can get something from understanding what the difference is, but I'm sadly not a good enough mathematician to characterize the difference with the elegance it deserves.
 
Last edited:
  • #14
Peter Morgan said:
There can be issues about complex structure

Isn't this just another way of saying that QM uses probability amplitudes, whereas CM just uses probabilities, so in QM you can have interference effects that don't occur in CM?
 
  • #15
PeterDonis said:
Isn't this just another way of saying that QM uses probability amplitudes, whereas CM just uses probabilities, so in QM you can have interference effects that don't occur in CM?
The structure of random field EM and of quantized EM are identical, for example, so no. Isomorphisms! For both random fields and quantum fields, the expected value for an observable ##\hat A## in a state with density matrix ##\hat\rho## can be written as the linear map ##\hat A\mapsto\mathsf{Tr}\!\left[\hat A\hat\rho\right]##. For pure states, supposing ##\hat\rho=|\psi\rangle\langle\psi|##, we can write the trace as $$\hat A\mapsto\mathsf{Tr}\!\left[\hat A\hat\rho\right]=\mathsf{Tr}\!\left[\hat A|\psi\rangle\langle\psi|\right]=\langle\psi|\hat A|\psi\rangle.$$ If, as well, the measurement is ##\hat A=|\phi\rangle\langle\phi|##, then we obtain "probability amplitudes", $$|\phi\rangle\langle\phi|\mapsto\mathsf{Tr}\!\left[|\phi\rangle\langle\phi|\psi\rangle\langle\psi|\right]=\langle\psi|\phi\rangle\langle\phi|\psi\rangle=|\langle\psi|\phi\rangle|^2,$$ which is the same for both QM and CM in a KvN formalism, but IMO it's best to think of the linear Trace formula as more fundamental than the Born formula.
 
  • Like
Likes Mentz114
  • #16
Peter Morgan said:
random field EM

What is "random field EM"? Is there a reference that develops this theory?
 
  • #17
PeterDonis said:
What is "random field EM"? Is there a reference that develops this theory?
I think I can really only cite my own arXiv:1709.06711, "Classical states, quantum field measurements", which, as I say above, is perhaps about to be published if a referee decides the changes I've made are acceptable. FWIW, the referee in question said of the paper that "While I didn't check every formula in detail, the results presented in the paper seem sound and even elegant. I do have issues with the presentation, however", followed by details, mostly saying that the text doesn't explain very well what I'm doing and why, which I think is reasonably fair. The way I have constructed for working with random fields, particularly for the EM field, is, I think, fairly elegant.

In the math literature, a random field is just an indexed set of random variables. I give a couple of old math-y references in my "Bell inequalities for random fields", J. Phys. A: Math. Gen. 39 (2006) 7441-7455, arXiv:cond-mat/0403692, but the math literature on random fields was always too mathematical for my purposes in physics, and I haven't tried to keep up with it. For classical random field theories, that index set typically also has some additional structure, such as Lorentz invariance. All one needs is that a random field can be presented as a commutative algebra of operators, with a state over that algebra giving you a joint probability density, which can be made to look so like a quantum field that it's kinda comfortable, which is kinda the point, the better to try to see precisely what the differences are.

Again, I'll understand if you decide you don't want to put the work into understand this just now. I've been working with this way of thinking for almost 20 years, so it all seems so natural to me that I don't have as much sense as I should of why everyone doesn't see how natural it is. Even so, I haven't yet managed and I may never manage to convince anyone serious that this is a path to really understanding QM/QFT, and it's a good rule that how long someone has been working on something is no guarantee at all that anything they say will be of any interest to anyone else. There's a lot of rubbish out there, and here's mine.
 
  • #18
Peter Morgan said:
I think I can really only cite my own arXiv:1709.06711, "Classical states, quantum field measurements", which, as I say above, is perhaps about to be published if a referee decides the changes I've made are acceptable.

PF isn't really intended for discussing original research, which it looks like this is. So I think it's time to close this thread.
 
  • Like
Likes Peter Morgan

Related to Schrödinger's cat: what's the state?

1. What is Schrödinger's cat thought experiment?

Schrödinger's cat is a thought experiment proposed by physicist Erwin Schrödinger in 1935. It is used to illustrate the concept of quantum superposition, where an object can exist in multiple states simultaneously until it is observed or measured.

2. What is the state of Schrödinger's cat?

In the thought experiment, Schrödinger's cat is placed in a sealed box with a vial of poison, a radioactive substance, and a Geiger counter. The state of the cat is unknown because it is both alive and dead until the box is opened and the cat is observed.

3. How does Schrödinger's cat relate to quantum mechanics?

Schrödinger's cat is a way to explain the principles of quantum mechanics, specifically the concept of superposition and the role of the observer in determining the state of a particle or object. It highlights the strange and counterintuitive nature of quantum mechanics and the limitations of our classical understanding of the world.

4. What is the significance of Schrödinger's cat?

Schrödinger's cat is significant because it challenges our understanding of reality and raises questions about the role of consciousness and observation in shaping the physical world. It also highlights the limitations of our current scientific theories and the need for further exploration and understanding of quantum mechanics.

5. Has Schrödinger's cat experiment ever been conducted in real life?

No, Schrödinger's cat experiment has never been conducted in real life. It is a thought experiment used to illustrate a concept and is not meant to be carried out in reality. However, the principles and ideas behind the experiment have been tested and confirmed through various experiments in quantum mechanics.

Similar threads

  • Quantum Physics
5
Replies
143
Views
6K
Replies
16
Views
1K
Replies
11
Views
1K
  • Quantum Physics
Replies
17
Views
1K
Replies
1
Views
780
Replies
9
Views
1K
  • Quantum Physics
Replies
13
Views
792
  • Quantum Physics
Replies
2
Views
1K
Replies
2
Views
472
Replies
11
Views
1K
Back
Top