# I The Heisenberg cut demystified

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1. Apr 17, 2016

### A. Neumaier

That doesn't follow.

Selecting a system to study is necessary for most of physics. But it is always a subjective act - since Nature isn't divided into system and the rest of the universe. Most of the time we don't want to consider the universe as a whole but only the particles in a particular beam, or particular materials, etc., and in each case it is the physicist that makes the selection, not Nature.

The cut, and the associated reduced description is of precisely this kind - specify precisely which system you are studying and you have specified the corresponding cut.

2. Apr 17, 2016

### A. Neumaier

Physicists are part of Nature, but their descriptive choices are not part of the physical description of Nature. We choose what to describe by being selective. Without selection no concept formation and no physics.

3. Apr 17, 2016

### vanhees71

Well, physcists have a choice to construct their devices, but that's not a subjective act. A given setup in the lab is objective. Everybody can rebuild it anywhere and check the results.

4. Apr 17, 2016

### A. Neumaier

But to say ''I want to study this particular setup in terms of these and these observables and ignore all the rest'' - a necessity for doing physics - is a subjective choice. Indeed, different people (and the same people at different times) make different choices.

Both the classical space and the quantum Hilbert space representation depend on this choice. In the quantum case, this is called the Heisenberg cut.

In a classical system one has precisely the same situation but few people are interested in discussing these extensively. But an microcopic classical system also has a different dynamics when observed (coupled to a detector described by classical mechanics) than an unobserved system not coupled to anything. Only in the latter case is the dynamics conservative.

Thus the analogy to the quantum case is very close.

5. Apr 17, 2016

### atyy

Is there a wave function of the universe? If there is no wave function of the universe, then how "big" you choose to make your Hilbert space is subjective.

Should the experimental apparatus be included in the Hilbert space?

Should the experimentalist be included in the Hilbert space?

6. Apr 18, 2016

### A. Neumaier

There must be a state (not nececcarily a wave function) of the universe since there is no limit of size to which quantum mechanics applies, and the universe as a whole is the only truly closed system in Nature.

Thus there is no question of existence, only one of its interpretation. Both Copenhagen and ensemble interpretations cannot interpret it since they must assume that there is some (nonexistent) measurement apparatus outside the system to observe it and to give an interpretation in terms of probabilities. MWI interprets it in a completely incomprehensible way. My thermal interpretation gives a meaningful interpretation of the state of the universe consistent with experimental practice.

7. Apr 18, 2016

### vanhees71

For me the idea of a wave function of the entire universe doesn't make sense. It's not empirically testable whether an assumed state describes the universe, because I cannot observe ensembles of equally prepared universes.

In practice the measurement apparatus is usually not included in theoretical calculations. In my field, the only place where it enters is when we compare particle spectra to experiment we have to include the various cuts made by the experiment and/or apply an acceptance filter of the detector, which is empirically given by the calibration the experimentalists did for their detector.

8. Apr 18, 2016

### A. Neumaier

This doesn't need to be testable. It is enough that an assumed state is consistent with the observations about the universe. Testing this is possible in principle.

How do you test the state of a glass of water in equilibrium? You cannot - you can test only a few observables, and these take essentially sharp values. But this is enough to convince you that the cup of tea has a state.

From experience one assigns to it a corresponding statistical operator comprising all information one can measure with some confidence at a single cup of tea. Using equally prepared multiple cups of tea doesn't improve this description. Thus measurements on a single cup of tea gives all the information needed to write down its (grand canonical) state. This is the state used by engineers to work predictively with water and other macroscopic substances. Nobody ever tested empirically whether a wave function describes the glass of water - this is completely irrelevant for practical prediction.

One can do the same with the universe. All facts known (or obtainable in principle) about the universe are conceivable as measurements (or summaries of measurements) of certain observables, many of them with essentially sharp values, and one can construct from it a reasonable state for the universe.

9. Apr 18, 2016

### vanhees71

Sure, but here you tacitly use a macroscopic description. You assign the (grand-)canonical Statistical Operator to a cup of tea but not a pure state of all microscopic details, because that's simply not possible. You cannot even store the information needed to write down this state, but indeed it's not necessary, beause all we are interested in are macroscopic properties encoded in a few macroscopic observables like the amount of tea and its temperature in your cup.

10. Apr 18, 2016

### A. Neumaier

The point is that since one can test only a macroscopic description one describes the glass of water by a density operator - like any macroscopic system. One never invokes the wave function (even should it exist - one doesn't care, as it is not testable). Nevertheless this is good quantum physics - we can predict many properties of the glass of water from its density operator.

What is good quantum physics of a glass of water is also good quantum physics of the universe. The universe is just another macroscopic quantum system - namely the biggest one. We know that it is approximately in local equilibrium. We even know its approximate temperature, and we know the values of many of its observables (e.g., those of the glass of water under discussion before). We also have a fairly good knowledge about its mass density, especially in the neighborhood of our solar system.

All this makes sense only if we assign to the universe a density operator that describes its state. Though we do not know this density operator $\rho$ very accurately, we know very much about $\langle A\rangle =\mbox{tr}~\rho A$ for many observables $A$ localized on the Earth.

Last edited: Apr 21, 2016