Who Can Apply the Operator on a Wave Function to Measure an Observable?

[Joscha]

I was wondering about a question in qm:
Who can apply the operator on a wave function to measure an observable? Is it only granted to human beings, which can't apparently not right, because we measure with particular devices or can "everything" measure? If so, wouldn't it implies that there is a permanent measurement?
As you can see, I'm a kind of confused. I'd be glad about some help!

 

[Fredrik]

Operators are just functions. They're mathematical objects, not things that someone has to "apply". The operators you're interested represent measuring devices. There are many other threads that discuss what sort of physical system can be considered a "measuring device".

 

[tom.stoer]

Who can apply the operator on a wave function to measure an observable?

I think this is missleading: An operator representing an observable (like the Hamilton operator H for the energy E) does not tell you how to measure the observable experimentally. Measuring devices and operators have nothing to do with each other.

 

[Fredrik]

I think this is missleading: An operator representing an observable (like the Hamilton operator H for the energy E) does not tell you how to measure the observable experimentally. Measuring devices and operators have nothing to do with each other.

I think the claim in the second sentence is too strong. Observables can (and should) be defined as equivalence classes of measuring devices, as explained on the first three pages of Araki. Those equivalence classes are the real-world "things" that are represented mathematically by self-adjoint operators in the theory.

Joscha: Here's another way of saying what you need to hear: If f is a state vector (=wavefunction), and A an operator representing an observable, Af doesn't represent the state after a measurement of A. So "applying the operator" isn't the mathematical representation of performing the measurement.

 

[arkajad]

Applying the operator can be related to the collapse. But collapse is usually not treated in textbooks yet. Textbooks still rely on the old "axioms" due to von Neumann, and then refined in axiomatic algebraic approaches such as in Jauch and Araki. But this situation may well change in the coming years.

 

[tom.stoer]

I think the claim in the second sentence is too strong. Observables can (and should) be defined as equivalence classes of measuring devices, as explained on the first three pages of Araki. Those equivalence classes are the real-world "things" that are represented mathematically by self-adjoint operators in the theory.

A measuring device measures an observable - fine. But neither does the measuring device tell you how the operator looks like, nor does the operator tell you how to construct and use the measuring device!

Example: the momentum operator is just -id/dx; how do you construct a measuring device based on this information?

 

[arkajad]

$p=-id/dx$ is the canonical momentum. Yes, it is a hermitian operator, so it is usually called an "observable". But, when, for instance, a vector potential is present, it is gauge-dependent, therefore it is not an observable. Kinetic momentum $m\dot{\mathbf{x}}=\mathbf{p}-(e/c)\mathbf{A}$ is an observable. It can be measured, for instance, by a ballistic pendulum.

 

[Fredrik]

A measuring device measures an observable - fine. But neither does the measuring device tell you how the operator looks like, nor does the operator tell you how to construct and use the measuring device!

Example: the momentum operator is just -id/dx; how do you construct a measuring device based on this information?

The only option I see to having a separate axiom for each observable (an axiom that explicitly specifies the corresponding operator) is to only allow observables that correspond to symmetries of spacetime. I don't have all the details clear in my head, so I can't elaborate well enough to leave you satisfied with the answer. I'll just say that I think that all the commutation relations can be obtained from the theory of representations of Lie groups and Lie algebras, and that the question of (for example) what specific operator should represent momentum, is completely determined by a choice of a representation of the relevant algebra. (I still have a lot to learn before I can explain all the details, or even be sure that I'm making sense).

 

[arkajad]

When there are external fields, for instance a vector potential, there is no translational symmetry. And yet kinetic momentum (but not the canonical momentum that relates to "presymmetry") is an observable that has a well defined operation meaning.

 

[tom.stoer]

$p=-id/dx$ is the canonical momentum. Yes, it is a hermitian operator, so it is usually called an "observable". But, when, for instance, a vector potential is present, it is gauge-dependent, therefore it is not an observable. Kinetic momentum $m\dot{\mathbf{x}}=\mathbf{p}-(e/c)\mathbf{A}$ is an observable. It can be measured, for instance, by a ballistic pendulum.

All what you are writing is well-known but it misses the essential point.

The construction of an operator (i.e. using space-time symmetry considerations) and the construction of a measuring device do not anything in common. Whether there is a physical observable, whether there are gauge fields present and one has to distinguish between canonical and kinematical momentum does by no means tell you anything regarding the construction of the measuring device.

If you have a physical observable (in the strict Dirac sense i.e. w/o further gauge degrees of freedom - which can be decided purely within the formalism of QM) you knwo that there is an observable which you can measure in principle, but you don't know how to do it in practice.

Again: the momentum -id/dx is a well-defined physical observable (in the absence of gauge fields) but writing down -id/dx doesn't mean that you know how the device will look like.

 

[arkajad]

You have missed the essential point: when gauge fields are present, and they are present everywhere around us and all the time, then the canonical momentum is not an observable (no operational procedure exists) , while kinetic momentum is, it can be measured by a ballistic pendulum.

 

[Demystifier]

If so, wouldn't it implies that there is a permanent measurement?

In a sense, you are right. Different subsystems interact most of the time, and interaction is described by an interaction Hamiltonian which is an OPERATOR. In most cases the interaction involves a large number of the degrees of freedom, in which case the interaction is usually referred to as DECOHERENCE.

Humans play a secondary role here. They have an apparent ability to "choose" the interaction they want, which corresponds to a measurement of the observable they want.

 

[arkajad]

Hamiltonian interaction is just an interaction, not a measurement. With Hamiltonian interactions nothing "decoheres". You have just an ordinary, reversible evolution

"Choice" is also not important. The universe is measuring itself all the time without human beings. What human beings do is: they interpret.

 

[Demystifier]

Hamiltonian interaction is just an interaction, not a measurement. With Hamiltonian interactions nothing "decoheres". You have just an ordinary, reversible evolution

"Choice" is also not important. The universe is measuring itself all the time without human beings. What human beings do is: they interpret.

I agree, but it does not contradict what I said.

 

[tom.stoer]

You have missed the essential point: when gauge fields are present, and they are present everywhere around us and all the time, then the canonical momentum is not an observable (no operational procedure exists) , while kinetic momentum is, it can be measured by a ballistic pendulum.

I know that very well, but that is not the essential point!

You can take any observable you like, e.g. p, x, H, J as an example in order to avoid the problem with gauge fields.

The essential point is that the observable (p, x, H, J, ...) never tells you how to construct a measuring device. That's the point menitioned in the original post. An operator does not "measure" the observable in a physical sense; it does not represent the measuring device.

 

[arkajad]

An operator can represent a measuring device. For instance the kinetic momentum represents pretty well the ballistic pendulum for certain measurements. An appropriate function of the position operator can represent pretty well a particle location detector.

An operator doesn't tell us how to construct anything, because linear operators don't speak. But looking at a given operator sometimes we can find out what kind of a physical operation it may correspond to. For some operators it may be straightforward, for other operators difficult or impossible. But usually we have a converse need: we have an experimental arrangement and we want to find operators that can be used to simulate this arrangement mathematically. Usually not so many options are there to chose from.

 

[tom.stoer]

Say you have a free electron, a charged particle in a Pauli trap or something like that. Now using the position operator x how do you measure the position? Using p² how to you measure the kinetic energy?

 

[arkajad]

I am not measuring the position with an operator. I need a physical device. Tell me which is your physical device that measures the position of your electron, describe how it does it, and I will try to find the operator that will simulate the behavior of this device when coupled to your electron.

 

[Careful]

Say you have a free electron, a charged particle in a Pauli trap or something like that. Now using the position operator x how do you measure the position? Using p² how to you measure the kinetic energy?

Hehe, good remark First of all, I am against the viewpoint of observable algebra's like C* or Von Neumann algebra's since they make it appear as if when an operator is self adjoint, we should be able to ''measure it''. I also happen to be convinced that $x,p$ are actually never ever measured (certainly not p). Going into this issue in depth would lead me to several pages of writing (which I have no time for), but I definitely feel that people like Wigner, Pauli and more recently Penrose have said the most intelligent things about this. Physicists are actually very dismissive about these considerations since they all feel it is metaphysical while I claim it is very physical: it constitutes a theory for the direct observations we make all the time.

 

[Careful]

I am not measuring the position with an operator. I need a physical device. Tell me which is your physical device that measures the position of your electron, describe how it does it, and I will try to find the operator that will simulate the behavior of this device when coupled to your electron.

Now, that's the second time you do this! If you do somehow think that your measurements are not the results of the ordinary born postulate applied to well defined operators: please, do tell us what they are and how you compute them in ordinary quantum theory. Or if you are saying that you have other operators in mind : do define them (which will be terribly difficult since they are all nonlocal) and compute the consequences. Show me one paper where you did this.

I mean your type or reasoning is of that of an engineer who wants a ''hands on'' explanation why his apparatus is going to measure some nonlocal observable without actually understanding in detail the local microscopic physics behind it (your only justification is that it ''works''). You are correctly convinced that that an apparatus ''preselects'' an observable for you: but that begs the question, WHY is that?? Nothing in QM tells you that some physical system selects it's observables... so there is -as you suggest- a preferred local observable in nature from which one ought to construct composite non-local observables. But this is a daunting task! I begs the question: by what intrinsic procedure are these observables constructed?? It must be something outside quantum mechanics which does this for you.

 

[arkajad]

If have, say an electron, and if you have a photographic plate that is sensitive to passing electrons, you can approximately simulate the behavior of this plate using a function of a position operator. I will not refer you to any paper because you insist on "how you compute them in ordinary quantum theory". I do not know how to do it in "ordinary quantum theory". And going beyond "ordinary quantum theory" is not being encouraged on this particular forum. All I can say is that in the not so realistic limit of "instant activation of the sensitivity of the plate" the usual Born's prescription for calculating the probabilities of registering the passing electron is approximately correct.

 

[Careful]

If have, say an electron, and if you have a photographic plate that is sensitive to passing electrons, you can approximately simulate the behavior of this plate using a function of a position operator. I will not refer you to any paper because you insist on "how you compute them in ordinary quantum theory". I do not know how to do it in "ordinary quantum theory". And going beyond "ordinary quantum theory" is not being encouraged on this particular forum. All I can say is that in the not so realistic limit of "instant activation of the sensitivity of the plate" the usual Born's prescription for calculating the probabilities of registering the passing electron is approximately correct.

See, my previous post, I have extended it meanwhile. I am rather sympathetic to what you say here and people who think this is not worthwhile discussing can go and f*ck off I think we would both agree that you need to solve this problem outside ordinary QM but where we disagree is on how to do it: you seem to be thinking about extending it in Piron's way by generalizing away from Hilbert space to include classical systems, right?

 

[arkajad]

No. Not the Piron's way. Something along the lines of http://arxiv.org/abs/hep-th/9407157" way.

 

[Careful]

No. Not the Piron's way. Something along the lines of http://arxiv.org/abs/hep-th/9407157" way.

Ok, I will read it carefully this evening although I am convinced that considering classical apparati is certainly not the way to go. So I will respond somewhere tomorrow, I have different things to do now.

 

[tom.stoer]

I am not measuring the position with an operator. I need a physical device.

Sometimes it's helpful to take note what oothers have written in their posts. If you look at my post you will find numerous claims that an operator (observable) neither is nor describes a measuring device. Unfortunately you decided to ignore this completely and to start an irrelevant discussion regarding gauge fields.
Then you claimed that "an operator can represent a measuring device. For instance the kinetic momentum represents pretty well the ballistic pendulum for certain measurements. An appropriate function of the position operator can represent pretty well a particle location detector".
After my question how to measure the position using the operator x you told me that this does not work (of course it doesn't work but I never claimed this). You now want me to "tell you a physical device that measures the position of your electron, describe how it does it, and I will try to find the operator". It is ridiculous to modify the problem an solve something that has not been asked for.

Coming back to the question asked in post #1 "Who can apply the operator on a wave function to measure an observable?"

As I said in my first posts, applying an operator to a wave function does not measure the observable. Multiplying a wave function in position space with x does not measure the position. This procedure does not say anything regarding a measurement. OK?

 

[arkajad]

I was simply replying to your comments.

As I said in my first posts, applying an operator to a wave function does not measure the observable.

Of course it does not. What measure is a physical device, not an operator. So, we are in agreement.

Multiplying a wave function in position space with x does not measure the position.

Here we are in agreement again.

This procedure does not say anything regarding a measurement. OK?[/QIOTE]

Other procedures do. OK.

 

[Careful]

No. Not the Piron's way. Something along the lines of http://arxiv.org/abs/hep-th/9407157" way.

Ok, I read the first 8 pages which is I guess sufficient to get the idea (actually I immediatly understood what you did on page 3 which is enough since you claim that's the main result). Right, so what do I think about this kind of generalization of GRW? First of all, you violate a property I hold very dear in physics, which is locality. Indeed for an inhomogeneous medium the Lambda term and the Hamiltonian do not commute anymore so that your born rule becomes nonlocal. That is, the probability of a measurement at a depends upon amplitudes which correspond to an event at b. That is pretty bad although it would not pose any practical problem if you restrict your setup to sufficiently small length scales. For the universe however, it would be a disaster. Furthermore, you *have* to consider an inhomogeneous medium since you know you cannot explain the measurement problem if you start from microscopic physics. Hence, what I missed was an equation for the medium ! Clearly, it must depend upon matter sources and so on. But then, this would provide a kind of classical background noise, which must probably persist to exist in the particle vacuum. Does that not give you an extra cosmological constant problem?? Also, would your medium not contradict the michelson morely experiment? All these realistic measurement theories run into terrible diffuculities when you try to generalize this even to special relativity, let along general relativity. These are the only comments I make from a physical point of view; from the philosophical point of view I of course totally disagree with what you try to do, but everybody has the right to his own opinions.

 

[tom.stoer]

No. Not the Piron's way. Something along the lines of http://arxiv.org/abs/hep-th/9407157" way.

I started to read that paper; it looks non-standard so it requires some care to comment it here. Anyway - I have the feeling that there's no way to save "reality" in qm, even if we accept violation of locality. I can't proof this, but I think we are simply fooled by the idea to transport our classical concepts to the quantum world.

But of course new ideas are always welcome.

 

[Careful]

Anyway - I have the feeling that there's no way to save "reality" in qm, even if we accept violation of locality. I can't proof this, but I think we are simply fooled by the idea to transport our classical concepts to the quantum world.

I think your feeling is wrong but your conclusion is right. There is no way you can prove such thing, but that is not really the issue. The point is that you cannot formulate a natural theory which would lead to the same predictions. So Ockham's razor decides here, not some silly no-go theorem.

 

[tom.stoer]

I think that there are some indications that a proof could be possible. The Kochen-Specker theorem shows that QM is not simply a "classical statistical theory". The Bell theorem rules out a certain class of "local hidden variable theories".

In the end you are right: once one has proved that a certain class of "simple" theories has been ruled out, more complicated theories are not forbidden but become unreasonable due to Ockhams razor. The question is what is the border line between "simple" and "too complicated" theories.