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In discussions around a recent Stack answer which covered axiomatic quantum mechanics, Bell's Theorem and Random Fields (related to axiomatic QFT) the phrase "Heisenberg Cut" was used. The basic question is what does this mean?

Having reviewed this discussion I can provide some background to set a context for the question. In (Von Neumann) QM a distinction is made between : OBSERVER [Cut] QUANTUM OBJECT

This simplistic distinction has been critized later in that article and elsewhere such that the term "Heisenberg Cut" has fallen into disuse. If one wishes to make distinctions in a QM experiment then here is a larger proposal:


here O-O is any hypothetical observer-observer as discussed in the "Wigner's Friend" concept.

So has the concept of "Heisenberg Cut" been re-introduced in modern axiomatic QM/QFT and if so what is its formalisation (ie how can one make experimental or Bell Theorem-like deductions from it?)

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I take the heart of the Heisenberg cut to be the way we calculate probabilities and expected values in QM. For elementary QM, for some measurement described by an operator $\hat M$, the expected value in a state described by a density matrix $\hat\rho$ is given by the trace $\left<\!M\right>=\mathsf{Tr}\left(\!\hat M\hat\rho\!\right)$. What we put in the $\hat\rho$ is what is in our model universe. The measurement operator describes our measurement apparatus, which is not in the model universe, but instead describes how our measurement apparatus gets information out of the model universe. There's an almost-symmetry between the ways $\hat M$ and $\hat\rho$ appear; it's almost as if there's a measurement apparatus universe as well as the model universe. Different measurement apparatuses can affect each other in the measurement apparatus universe without changing the model universe, which is called measurement incompatibility (did I just create an interpretation of QM? Do I know this one? I guess it's too glib, sadly.)

Edit: We can extend the mathematics in many different ways, but one deserves mention because it's of great practical value. We can introduce transformations $\hat T_i$ that operate between the preparation apparatus and the measurement apparatus, in which case we have $\left<\!M\right>=\mathsf{Tr}\left(\!\hat M\hat T_n\cdots\hat T_2\hat T_1\hat\rho\!\right)$. We can at any time say that $\hat M\hat T_n\cdots\hat T_5$, say, or any other part of this list (without changing the order), is our measurement.

Anyway, to some extent we can move stuff from the model universe into the measurement apparatus universe and vice versa, although we may have to get into technical stuff like POVMs to do it. Once we go to quantum field theory, there's a tight relationship between the measurement apparatus universe and the model universe, because we use the same Lego blocks to build measurements and states.

The separation into states and measurements is absolutely fundamental in QM. It's how the relationship between Hilbert spaces and experimental results works, which causes trouble when people want to do cosmology, with everything in the model universe. Although I hadn't previously thought about calculating where one should put the separation, I can see that if one chooses a particular accuracy that one wants one's model of an experimental apparatus to achieve relative to one's real apparatus, that might put a limit on where one can put the Heisenberg cut. I'm not sure, however, that one can't always improve the sophistication of one's description of a measurement, particularly if one is willing to go to POVMs. I suppose, however, that putting people inside your model universe is always going to be in the realm of toy models. The separation into states and measurement famously comes under the microscope in Bell's article `Against "Measurement" '.

Incidentally, I see you went to Willem de Muynck, who is perhaps a little idiosyncratic, but I've often found his a good counterpoint of view.

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Thanks for this Answer +1. I think this site works best with Questions, then Answers rather than discussion comments. Comments: calculating is the term used by decoherence; there are "degrees of classicality" in these POVMs however, which I need to understand further; there were so many Universes in this answer I need to work out which universe we actually inhabit. – Roy Simpson Feb 1 '11 at 18:50

The term "Heisenberg cut" has only been used in the philosophical discussions about the character of quantum mechanics. It's the classical-quantum boundary that should be placed somewhere in between the observed quantum objects and observer's perceptions.

It's never a mistake to place the cut closer to the observer - to treat a larger set of phenomena using the machinery of quantum mechanics because at the end, quantum mechanics applies everywhere. On the other hand, it could be a mistake to treat some systems or their properties classically.

The "minimal" location of the Heisenberg cut - one that treats a maximum fraction of the world classically - may be calculated by decoherence. In modern quantum mechanics, decoherence is what defines the classical-quantum boundary - but the term "Heisenberg cut" is rarely used for this boundary in modern physics.

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I like the shortness and sharpness of this answer. However the last paragraph tells us that its (minimal) location can be calculated, per experiment I presume. So the term (perhaps under another name) does appear in calculations. This suggests too that it has been formalised somewhere and maybe has mathematical properties we could prove supplementary theorems about... – Roy Simpson Feb 1 '11 at 13:15
Yes, Roy, as I say, the calculation has been formalized under the term decoherence, see e.g. Zurek's important paper about it - the very cut is referred to as the classical-quantum boundary or something else, check a few papers... – Luboš Motl Feb 1 '11 at 18:37
What is maximal location of the Heisenberg cut? – Anixx Dec 18 '12 at 9:15

in addition to what Lubos said about QM, the philosophy of axiomatic QFT is to construct/describe QFTs without any reference to concepts of classical physics. As far as I know there is not even a concept for a "classical limit" in axiomatic QFT. Especially "macroscopic classical" devices like detectors are modeled in AQFT via an observable, that is a selfadjoint operator, and that is a pure quantum concept. Therefore I don't think that you'll find a formalization of something like the "Heisenberg cut" in axiomatic QFT.

As for the interpretation of the measurement process etc., this is usually left to the philosophical interpretation of QM, from a pure philosophical point of view there is no conceptual difference of the interpretation of QM and that of QFT, which is the reason why most people working on this topic concentrate on the technically much simpler QM.

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Would you say that this concept has a formalisation in some other variant of QM? After all Lubos is claiming that it can be calculated: a formula or set of axioms is what answers this question affirmatively; it is answered negatively by saying something like "QM has no such formalisation and here is the proof that none can exist..." – Roy Simpson Feb 1 '11 at 14:21
I suppose "without any reference to concepts of classical physics" is a little too strong, since AQFT does have a concept of states and measurements, which did exist classically before QM, albeit without the issue of measurement incompatibilty. – Peter Morgan Feb 1 '11 at 16:02
@Roy: I'd draw the line and say that everything that phycisists calculate in QM with a formalized, mathematical framework does not have any connection to any concept like the Heisenberg cut. This concept is addressed, however, by people working on the philosophical interpretation of QM like Omnes. – Tim van Beek Feb 1 '11 at 16:12
@Peter Morgan: The "state" in AQFT is an appropriate state of a net of operator algebras, it is not a state in the sense of pre-QM physics. The measurement process itself is not formalized in AQFT, AQFT stops with the statement "the expectation value of an observable x in a state y is...", which is a purely quantum concept. Of course the concepts of states and measurements did exist in pre-QM physics, but AQFT is not a "quantized" classical theory like e.g. Lagrangian QFT. There is no a priori given classical theory. – Tim van Beek Feb 1 '11 at 16:15
@Tim van Beek Agreed that a state is something mathematically rather different in AQFT than it is in classical deterministic mechanics. I suppose, however, that "the expectation value of an observable x in a state y is..." would make reasonable sense to Boltzmann or Maxwell. For AQFT, though not for QG, there's also the shared conceptual starting points of Minkowski space or curved space-time. I guess I see all this as constant evolution of concepts, not of clean breaks that disconnect us from earlier Physicists. Sorry my way of approaching your phrasing annoyed. – Peter Morgan Feb 1 '11 at 16:29

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