# What counts as a measurement in QM?

1. Nov 7, 2005

### evk

What counts as a "measurement" in QM?

"Measurements" in quantum mechanics seem to have an important role. They collapse the Schrodinger wave equation, for example (please correct me if that's no longer the modern view). The state of a system BETWEEN MEASUREMENTS is what is inherently probabilistic and therefore subject to interference phenomena (as in the two-slit experiment).

My question is, what counts as a "measurement"? Is it ANY interaction of the system being analyzed with the outside world? Surely it can't matter whether or not a conscious entity is involved in the observation, right? In Schrodinger's thought experiment with the cat in the box, can the system only remain in a "fuzzy" state so long as absolutely no interaction takes place between the box and the outside world? Suppose a physicist with measurement devices was inside the box instead of a cat, are you really telling me the analysis would then be different? How must one draw the dividing line between the system being analyzed and "the rest of the world"?

Thanks

2. Nov 7, 2005

### jcsd

The postulates of quantum mechnics tell us the possible results of a measuremnt and they tell us the state the system will be in immediately after a measurment, but they never bother to tell us what a measurement is!

The simple answer is that nobody has given an entirely satisfactory answer to your question, that's why it's know as "the measurement problem".

3. Nov 7, 2005

### vanesch

Staff Emeritus
I second that

If you want to see some discussion about it here, have for instance a look at the thread

Happily that we have this measurement problem. It is what keeps discussion forums on quantum physics alive

4. Nov 8, 2005

### lalbatros

interaction with a macroscopic system

isn't that a measurement ?

Macroscopic means a practically continuous set of possible states.
This brings the possibility for irreversible transitions.

This goes back to a trivial fact: discrete fourier series cannot represent an exponential decay, but fourier transforms can because frequencies in the fourier transform expansion can be as close as necessary to produce to long-term behaviour of the exponential. Similarly, a discrete system in a mixed state can decay when interacting with a continuum. There is still micro-reversibility, but the 'small' system has been 'measured' irreversibly.

Developping the maths for that doesn't bring more than this simple conclusion.
It is nevertheless interresting to read the discussions on coherence and decoherence.
There, the point of view is complementary: how is it possible to isolate a small system as long as possible from external perturbations is such a way as to keep it in a coherent state!
Obviously, by studying this, one aknowledge that interactions can easily 'decohere' a system, leaving it in a 'incoherent' state. An incoherent state cannot be represented by a wave function, of course, but it can be represented by probabilities.

Is there really a 'measurement problem' ?

Last edited: Nov 8, 2005
5. Nov 8, 2005

### vanesch

Staff Emeritus
It depends on whether you think that there is contained, in QM, an ontological state description (the only thing I see that qualifies is the state vector in Hilbert space) or whether you simply consider the "state description" in QM as epistemological (a summary of your knowledge of the system under study).
In the last case, there is no measurement problem ; however, there is a "real world out there" problem . You've lost the link between your mathematical objects in your theory and "the world out there" ; as such, the association between laboratory experiments and calculations seems impossible to me: at a certain moment you HAVE to associate physical objects (in the lab) with mathematical objects in your theory (this association is what I'd call ontological), and then we're back to case 1.
To try to make you see what I mean, consider classical physics. You may define a probability distribution over phase space of your system. That's a purely epistemological description. It is sufficient to LEARN something about the system (for instance, by LOOKING at it), and the probability distribution changes (because your knowledge of it changes). At no point (except a few weirdos around here :grumpy: ) you would consider that the system by itself IS the probability distribution. But there's no harm. You know that the state of the system (the ontological state, which exists, independent of what you know of it) is a point in phase space. You know that your epistemological description is a complete or incomplete knowledge of that point in phase space.
If you go to QM, the equivalence of your point in phase space is your statevector in hilbert space, and the equivalence of your probability distribution over phase space is your density operator.
However, QM has a property that classical physics hasn't, and that is entanglement: you cannot decide to assign a quantum state in hilbert space to subsystems. If a system interacts with another, the quantum state is only given in the product space, and the individual quantum state of the subsystem does not exist. You can now "coarse grain" and trace out the part of hilbert space that doesn't interest you, and you'll end up with a reduced density matrix. But I don't think you can call that the ontological state of the system, because it does WRONGLY describe correlations with the other system you decided to ignore (EPR style correlations). It DOES give you a good epistemological description of what you decided to consider (to know) of the system.

6. Nov 8, 2005

### lalbatros

Vanesh,

If I understood well, you don't like the idea that the state of the 'measured' system depends on the action (point of view?) of the observer. But I think this is not the case, after the measurement-interaction, there is only one big universe wavefunction, just as before the measurement. The only difference is that the interaction has made it irreversibly different: very close to ideally collapsed states but not exactly. In principle, the correlations have not vanished and could be recovered by time-inversion. Practically, there is no way to perform this time-inversion (because all evolution in the universe would ultimately need to be time-inverted). The strange object we need to get familiar with is a large scale wavefunction.

This is just -for me- the same problem than for classical mechanics and the second principle. It is a difficult subject, but there is agreement that micro-reversibility is not contradictory to the second principle. And the second principle is compatible with the classical micro-description of the world.

Of course, there are differences too. Like the EPR question. Like the fact that in QM the irreversibility deals with a micro-system interacting with a macro system, while in CM it is macro-macro.

My feeling is that these two question are the same and should have a common answer. This makes QM measurement less a new problem, and more like an old problem.

7. Nov 8, 2005

### vanesch

Staff Emeritus
Why is everybody here forgetting the "c" in my name :grumpy: ?
(my full name is Patrick Van Esch, and Esch is a little town in Luxemburg which seems to be the genealogical origin of it).
I agree with that ! There has been a measurement interaction (the unitary time evolution operator attached to the physics of the measurement apparatus, the observer body and all that). I just have difficulties considering that an "observation" has *killed off* a big part of the entire wavefunction of the universe which is what standard state reduction prescribes.
Yes, that's also my POV. Except that I consider that *before* measurement, that large scale wavefunction was in a product state (or at least the relevant part of it was in a product state) of an "ignorant observer" and the "pure system", while the measurement interaction entangled both.
Well, it is slightly more subtle, in that, contrary to classical systems, there IS no individual state description of an entangled system, while you can always consider the projected sub-phase space in classical mechanics as describing de state exactly of the subsystem.
Sounds like music in my ears
To me, the essential part of the measurement problem remains: how do you go from a SUM of states (this time, entangled and all that), to ONLY ONE with a certain PROBABILITY ? This cannot be done with a unitary operator.
With a unitary operator you can mix states, have a Schmidt decomposition with classical-looking states and everything, but at the end of the day, YOUR BODY ends up entangled if all is unitary. Each of these body states is in a product state with a classically looking universe if we are optimistic (using environmental decoherence) but it is STILL in SEVERAL states. YOU, as a conscious being, only PERCEIVE ONE of these states. *that* is puzzling. You ought to be seeing all of them, because your body sees all of them. It is the essence of the measurement problem, to me.
If, in classical physics, say, green light is coming from the left and red light is coming from the right, you are aware, you perceive, you are conscious of BOTH the green light and the red light, because that will be your body state (the ontological one, in the phase space of your body).
In quantum theory, in that big wavefunction of the universe, your one and the same single body (brain) HAS seen a click in one term, and HASN'T seen a click in the other. Both state vectors are there in your body state. But you only PERCEIVE one of them, with a certain PROBABILITY. How come ?

8. Nov 8, 2005

### lalbatros

Patrick, sorry for the 'c' !

, since you admit that the all-whole-wavefunction evolves as usual, in an unitary way.

Indeed, assume that after the measurement (analyser) you perform a second analysis (like with a crossed-analyser). You could model the outcome of the second experiment in two ways:
• either start from a density operator representing the collapsed state of the measured system
• or use the all-whole-wavefunction that includes the small system and the macro-measuring device entangled wavefunctions
The two models will give exactly the same result.

They could give different results only if your second analyser is able to 'unfold' the smeared correlations from the first measurement. And this is probably impossible for real-world measurements. However, this is may not be impossible for nano-scale interactions (is it called meso-scale maybe?), a scale where correlations are long-lived, not too much perturbed.

Sorry also for the musics, which sounds like poor knowledge. I don't know much more than classical mechanics, the Boltzmann H-theorem, and its holy proof. This is an uncomfortable region of physics for me. I hadly see a link between the H-theorem and the second principle. I hardly can understand the proof. And I cannot see any generality in this theorem. Nevertheless, this H-theorem is taken as the cornerstone for understanding irreversibility from reversibility.

In a sense, I see an easier track in QM, since I would analyse the density operator and think in terms of the correlations. And irreversibility would only be an excellent approximation of a state vector by a state density matrix.

Why should we be perturbed by two models that agree so easily?

Concerning you final remark:

Why do you imagine such complicated scenarios?
You will never be able to look at the universe from the outside!
Maybe this could be useful to develop new physics, but the predictions will only be compared to what we can see from the inside! If the model looks strange on the cover should not be important!?

Last edited: Nov 8, 2005
9. Nov 8, 2005

### straycat

You might be interested in Saunders' treatment of this subject in his paper [1] "Time, Quantum Mechanics, and Probability", section 4.2 . He argues that you should *not* expect to "see all of them," but rather you should expect (prior to measurement, at time t_1) to see *one* of them (after measurement, e.g. at time t_2), you just don't know which one.

David

[1] http://xxx.lanl.gov/abs/quant-ph/0111047

10. Nov 8, 2005

### vanesch

Staff Emeritus
No problem but about EVERYBODY here writes vanesh (probably imagining I'm of Indian origin ?).

I know ! But you're cheating here: you're USING the Born rule to calculate probabilities. You do that already when doing the partial trace for the reduced density matrix: WHY do you think that you can use the partial trace ? Because you sum then over the Born probabilities of the outcomes (in any basis, trace is basis-independent) of the outcomes for the environment or the system you're not interested in. But that means you ALREADY use the Born rule, so you cannot use that to DERIVE the Born rule (circular reasoning).

I only mentionned that because I had the impression that you considered the (reduced) density matrix as the *ontological* state of your subsystem (in which case these correlations would be unexplainable!). If you accept that the ontological state of the system is always given by a vector in Hilbert space, then we are in agreement on this. (the epistemological description "state" can be a density matrix: it writes down what you KNOW of the system, not what IS the system).

I don't have to look from the outside. In classical physics, you don't have to "look from the outside" to consider that there is a big phase space of all stuff in the universe, that you, as part of that universe, are also in a "state" (the chunk of phase space that corresponds to the degrees of freedom of your body). And THAT state of that body of yours will then determine what you will consciously experience (how that happens is then left as an exercise for philosophers and doesn't concern any physics as such). The FULL STATE of your body will determine what you will experience. But that's so evident in classical physics that nobody writes that down. You're dead when the ENTIRE STATE of your body satisfies certain conditions (like being splattered all over because you've been hit by a truck or so). You're alive when your body is in another state...

However, the ENTIRE STATE of your body, in the quantum description of the universe, is made up of different terms in the wavefunction of the universe. In certain terms there is a contribution of your bodystate which corresponds to looking at some birds in the field, in other terms you're splattered all over because while looking at the birds you didn't see the truck, in still other terms you decided not to go for a walk and you're chatting on PF...
This is the entire, ontological, state of your body: ALL these different terms are present, some with an important Hilbert norm, others with less. Apparently you only consciously observe ONE of these terms. With a probability given by the Born rule. And it is because of this probability given by the Born rule for your mind to see ONE of the states of your body who are present, that you ALSO find the Born rule for experiments in the lab and everything.

So the question is: why do you not experience the full quantum state of your body as it is present in the wavefunction of the universe ?

I answer this simply by POSTULATE, but you should be aware that this is a strange postulate, saying something about the conscious connection between a body state and a mind. And if you DON'T do that, I really, really don't see why I'm only aware of part of the state of my body and how this can be *derived* from the unitary evolved state (the wavefunction of the universe).

11. Nov 8, 2005

### jcsd

At a more proletarian level: Bog standard quantum mechanical measurement theory does not consider the measuremnt appartus and the particle (or whatever) as a single quantum mechanical system. However when you do you still have the same basic problem, for example:

Consider a quantum mechanical system consisting of an electon in a superpostion of states (spin up, spin down) and some mesuremnt appartus; when the measuremnt apparatus measures the electron decoherence occurs. As now a measuremnt of the measuremnt apparatus-electron system will always correspond to a measuremnt having occured and the electron not being in a superpostion of states, you may well be tempted to say "problem solved!" There are now two quite distinct 'branches' to the wavefunction one corresponding to spin up and one corresponding to a spin down measurement, however the system is still in a superpostion of states, so we still have the same basic problem: "Why when we measure the electron do we only get a single state?"

12. Nov 8, 2005

### ZapperZ

Staff Emeritus
What if you measure a non-commuting observable and deduce the superposition that way? I.e. [A,B] != 0, so to see the superposition of observable A, you measure B and deduce from the "weird" property of B that you have just detected the superposition of A. Example: energy gap measured in Stony Brook/Delft experiment.

Zz.

13. Nov 8, 2005

### jcsd

But surely, though in the formal sense we have performed a measurement of some sort, we have not (again in the strictly formal sense) performed a measuremnt of A as the result of the measuremnt is not an eignevalue of the operator A.

14. Nov 8, 2005

### ZapperZ

Staff Emeritus
Correct. But if we want to preserve the superpostion of A, this is one way to "observe" it. I didn't say we were going to measure A. I simply pointed out how to detect the superpostion of A.

Zz.

15. Nov 8, 2005

### -Job-

Can't the collapse of a particle's wave function collapse the wave function of another particle? By observing B are you collapsing both B and A?

16. Nov 8, 2005

### vanesch

Staff Emeritus
This is the problem (?) with quantum mechanics as we know it: ALL interaction is described by a UNITARY operator. And with such an operator, you cannot do the collapse thing which is a projector. No matter how big and complicated the system is. That's the fundamental difficulty: the "collapse" operation is mathematically not compatible with a unitary operator. The best we can do is environmental decoherence, but that is STILL a superposition. It explains why we don't see INTERFERENCE but it doesn't explain why we only see ONE term.

A naive analogy: imagine, that after a lot of calculation in a new theory about light boxes, you find that out of a certain box with two holes must come red light out of the left hole and green light out of the right hole. When you do the experiment, however, sometimes there comes green light out of the right hole, and in other instances of the experiment, there comes red light out of the left hole.
So it seems that your calculation only predicted some STATISTICS of what was to happen. But no, the thing that corresponds to the "state of the box" said "red AND green light". It wasn't meant to be a description of an ensemble of boxes.
Then people say: hoho, but if you have red and green light, that must give you funny interference patterns! And then others do detailled calculations and show that for all practical purposes, you will NOT see interference patterns between the red and the green light, because of the influence of the environment. THIS is the equivalence of "environmental decoherence".

So, problem solved because you won't observe interference between red and green light ? I wouldn't think so: your theory said that you should see BOTH, and you only see one at a time, with probability 50%. So you introduce an extra postulate: whenever I OBSERVE a box which sends out red and green light, I only observe ONE of these states, probabilitically. But the box is in the state "red + green" allright. You can even check that, because when you do a careful experiment AVOIDING the environment, you DO see interference between red and green light. So it really WASN'T a statistical mixture which you erroneously took to be the state of a system.

This is an analogy of what we do when we consider the wavefunction of the universe, and then say that we "see only one term".

17. Nov 9, 2005

### ZapperZ

Staff Emeritus
If you look at the mathematics carefully, you will see that if A and B do not commute, a measure of the observable A says nothing about the outcome of B. Only when A and B commute are you able to determine both quantities in a single measurement (for non-degenerate states). For example, Lz does not commute with Lx and Ly. A measure of Lz will STILL leave Lx and Ly values undetermine, preserving their superposition.

This is a very fundamental aspect of QM. In fact, the commutator relation [A,B] has often been called as the First Quantization. It is the origin of the uncertainty relations in QM, and the source of the so-called "measurement problem".

Zz.

18. Nov 10, 2005

### lalbatros

Vanesch

I suggested that a pure-state POV or a mixed state POV may lead to exactly the same results when correlations are 'smeared'.

The interpretation of the Born rule, for me, is just that a pure-state and a density matrix can represent very closely the same system and predict its evolution with very much the same precision. In such a situation, the Born rule is a useful language.

But is the Born rule really necessary ? Since the approximate equivalence has the same meaning and is easily integrated in the formalism without the Born rule. Maybe the Born rule is a simple matter of convenience ...

Could we not try without the Born rule?

Last edited: Nov 10, 2005
19. Nov 10, 2005

### lalbatros

Vanesch,

I almost agree with your remark:
But then, I think that classical space is shaped by particles (GR, Mach principle). And how could I assign a coordinate to a particle alone ? In a sense, this is already a kind of entanglement: it is an impossibility to talk about a part without considering the whole. But this is less a problem in CM, maybe because of the type of systems/interactions being studied.

So, then, what is non-classical with QM entanglement?

One aspect is for sure linked to the EPR paradox. And I don't see another.
The non-locality is the difficulty here. The Born postulate is not the difficulty.
That is to say that the 'measurement postulate' is not shaken by the question of entanglement.
But maybe the classical concept of space-time will be shaken sometimes by this question.

Last edited: Nov 10, 2005
20. Nov 11, 2005

### Careful

You cannot do without the Born rule:
In copenhagen and in consciouness it is explicitely there (this is the AND formalism).
In environmental decoherence (the OR formalism) there is nothing in the density matrix which tells you how to interpret it. There is for example no a priori reason why the different pure states (which span the density matrix as a convex combination) should be orthogonal; this is an extra postulate (and even if you would adapt this convention you still don't have a unique decomposition when equal weights are present). Moreover, as far as I know, all these so called decoherence proofs show is that *IF* the environment decoheres, *THEN* the environment + system decoheres given a reasonable interaction Hamiltonian (I may be wrong here, but I do not think so).

Concerning your second comments: coordinates in GR have nothing to do with physics, they are a mathematical artifact. Now, you might wonder about assigning physical coordinates to matter; this is an act of measurement and depends upon a priori hypotheses about (a) your measurement apparatus and (b) space time structure.
However, it has nothing to do with entanglement !! On the contrary, it is a famous and painfully correct statement of A. Einstein that if everything were entangled in the universe, then it would be impossible to find out any laws for it´´ and indeed our ability to construct laws of nature depends upon the assumption that isolated systems exist in nature (something which is not possible in QM). This is another translation´´ of the measurement problem: show why the world disentangles at reasonable scales of 10^{-8} meters.

To shake the classical concept of space time (which has a priori nothing to do with entanglement) is one of the most dangerous and uncontrollable games in town. People should have more and deeper respect for the principle of LOCALITY since the alternative is a 80 years old untamed monster which hasn't been shown by any means to be NECESSARY, albeit its apllications to microphysics are succesful.

Cheers,

Careful

Last edited: Nov 11, 2005