Fundamental Theorem of Quantum Measurements

[Demystifier]

Even in this sense position is a continuous quantity. What you are talking about is the binning in the sense of the resolution, but that doesn't make the measured quantity continuous or discontinuous.

Binning and resolution are different things. Binning is a statistical tool, resolution is a property of a measuring apparatus.

Also you don't make a quantitiy continuous only when binning it when measuring it: If you measure the angular momentum of a quantum system you can also bin it within wide bins. Angular momentum is discrete according to QT, and the binning doesn't make it continuous.

Did somebody say that binning makes a quantity continuous?

 

[Dadface]

https://www.nist.gov/sites/default/files/documents/pml/div683/museum-length.pdf

I'm not sure if this article is relevant to part of the discussion here but if so it may give some idea of the errors associated with certain length measurements made by using current state of the art technology.

 

[Danny Boy]

@Demystifier Thanks. One last quick query:
If we consider the measurement $$\rho \mapsto \tilde{\rho} := \frac{\int_{\mathcal{M}}A_c \rho A_c^{\dagger}}{\text{Pr}(c \in \mathcal{M})}$$
would I be correct in stating that if $\rho$ is a pure state then $\tilde{\rho}$ would be a mixed state? Is this correct or is there not enough info to state this definitively?

 

[Demystifier]

@Demystifier Thanks. One last quick query:
If we consider the measurement $$\rho \mapsto \tilde{\rho} := \frac{\int_{\mathcal{M}}A_c \rho A_c^{\dagger}}{\text{Pr}(c \in \mathcal{M})}$$
would I be correct in stating that if $\rho$ is a pure state then $\tilde{\rho}$ would be a mixed state? Is this correct or is there not enough info to state this definitively?

That's correct.

 

[vanhees71]

Binning and resolution are different things. Binning is a statistical tool, resolution is a property of a measuring apparatus.Did somebody say that binning makes a quantity continuous?

No, to the contrary, you claimed binning or resolution of an apparatus makes a continuous quantity discrete, which is of course wrong. I guess I simply don't understand what you want to say :-((.

 

[Demystifier]

No, to the contrary, you claimed binning or resolution of an apparatus makes a continuous quantity discrete, which is of course wrong. I guess I simply don't understand what you want to say :-((.

I said that experimental resolution (not a binning) maps a theoretical continuous observable into an experimental discrete observable, which of course is trivially true.

 

[Danny Boy]

@Demystifier Okay so you think that $\tilde{\rho}$ produced after measurement $$\rho \mapsto \tilde{\rho} := \frac{\int_{\mathcal{M}}A_c \rho A_{c}^{\dagger}}{Pr( c \in \mathcal{M})}$$ is always a mixed state? But why can't $\tilde{\rho}$ possibly be a superposition of states hence a pure state?

 

[Demystifier]

@Demystifier Okay so you think that $\tilde{\rho}$ produced after measurement $$\rho \mapsto \tilde{\rho} := \frac{\int_{\mathcal{M}}A_c \rho A_{c}^{\dagger}}{Pr( c \in \mathcal{M})}$$ is always a mixed state? But why can't $\tilde{\rho}$ possibly be a superposition of states hence a pure state?

It can be pure, but then there is no integration over ${\cal M}$ and $A_c$ is a projector. It is a special case. It is possible mathematically, but as I argue in other posts above it is impossible in a realistic laboratory when $c$ is continuous.

 

[Danny Boy]

@Demystifier Yes I understand that if we don't have the integral and it is a projection $A_c$ then we get a pure state again. But if we have an integral is it always a mixed state or does this also depend on how we define $A_c$? For example what about an integral where $A_c$ are projectors?

 

[kith]

For example what about an integral where $A_c$ are projectors?

Have you tried to calculate how the expression looks like for a simple example of this type?

 

[Demystifier]

@Demystifier Yes I understand that if we don't have the integral and it is a projection $A_c$ then we get a pure state again. But if we have an integral is it always a mixed state or does this also depend on how we define $A_c$? For example what about an integral where $A_c$ are projectors?

If there is an integral, then it is always a mixed state, even if $A_c$ are projectors.

 

[akvadrako]

I said that experimental resolution (not a binning) maps a theoretical continuous observable into an experimental discrete observable, which of course is trivially true.

Isn't this the essence of quanta? That no matter how you setup your experiment, you'll only measure discrete outcomes.

 

[Danny Boy]

@Demystifier Thanks this is what I was interested in discussing. Consider the following counter-example to your statement: If we consider an initial pure state $| \psi \rangle$ then after position measurement we would get a superposition of the form $$| \psi' \rangle = \int_{\mathcal{M}}c(x)|x\rangle dx$$ where $\langle x' | x \rangle = \delta(x'-x)$. Thus in terms of density matrices we have that the measurement does the following: $$| \psi \rangle \langle \psi | \mapsto | \psi' \rangle \langle \psi' |$$ which is a pure state to pure state measurement, hence this is an example of having the integral and projectors which produces a pure state from a pure state after measurement.

 

[Demystifier]

@Demystifier Thanks this is what I was interested in discussing. Consider the following counter-example to your statement: If we consider an initial pure state $| \psi \rangle$ then after position measurement we would get a superposition of the form $$| \psi' \rangle = \int_{\mathcal{M}}c(x)|x\rangle dx$$ where $\langle x' | x \rangle = \delta(x'-x)$. Thus in terms of density matrices we have that the measurement does the following: $$| \psi \rangle \langle \psi | \mapsto | \psi' \rangle \langle \psi' |$$ which is a pure state to pure state measurement, hence this is an example of having the integral and projectors which produces a pure state from a pure state after measurement.

You are mixing apples and oranges (vectors in Hilbert space and density operators in the same space). Your second equation, which is written in terms of density matrices, does not contain an integral. You first equation is not written in terms of density matrices, so it does not belong to the class of equations we were discussing so far.

Nevertheless, you have a point. The first equation can be written in terms of density matrices. But then some of the $A_x$ operators will not be projectors. It is still true that if all $A_c$ are projectors, then the integration gives a mixed state. You can get a pure state with integration (I was wrong about that in a previous post), but then some $A_c$ are not projectors.

 

[Demystifier]

Isn't this the essence of quanta? That no matter how you setup your experiment, you'll only measure discrete outcomes.

Yes, but we were talking about a different issue.

 

[akvadrako]

Yes, but we were talking about a different issue.

Maybe I misunderstand what the issue is. What I wanted to say is in an experiment with a resolution of $\Delta x$, you can image $x_0$ being chosen continuously. But after it's fixed, results of a single measurement will come from a finite set of options, $x_0 + N \Delta x$. In the pre-setup of your experiment the choices of $x_0$ are also fixed by whatever it's starting conditions are. Continuing to the pre-pre-setup and beyond one eventually ends up at their current state, which already pre-determines a finite set of possible $x_0$ values and hence discrete outcomes for your experiment.

 

[Zafa Pi]

No, I doubt the physical existence of continuous observables.

This is amusing. About a year or so ago you said something to the effect that reality was continuous (I am not able to find it). I replied, "No it's not, it's discrete. Prove me wrong." @Dale then deleted my comment because "Prove me wrong." wasn't up to PF standards. I actually meant it as a joke.

Nice to see you now know the TRUTH.
BTW, the term "an observable" is almost always used in the context of an operator in QM, which may be what is bothering @Neumaier.

 

[bhobba]

About a year or so ago you said something to the effect that reality was continuous (I am not able to find it).

I think its just a confusion of context.

So far we have not had to move away from calculus, differentiation and all that which imply's continuity - in our models. They all work. But is that what is really going on - I to have my doubts on that - but physics is a mathematical model (recently some have pulled me up on that one - please not here - start a new thread if you want to discuss it) and that's what the model says so our best guess is - yes it is. But a guess is just that a guess - my gut, like Dymstifyers gut tells him, and likely yours as well, - its likely wrong. Still as you correctly say there is no proof - experimental that is - one way or the other. It may even be an inherently undecidable proposition.

Here is a series of lectures that looks at, mathematically, how discrete behavior that often does occur in QM comes about - amongst other interesting things such as how to define weird infinite sums like 1 - 1 + 1 - 1 ....


Thanks
Bill

 

[Zafa Pi]

but physics is a mathematical model

See page 42 of your signature reference.

 

[bhobba]

See page 42 of your signature reference.

Its been a while since I read a mathematical modelling textbook eg
https://www.amazon.com/dp/0821836501/?tag=pfamazon01-20

But page 42 looks pretty much like the definition of a mathematical model to me, but I accept opinions could vary.

Thanks
Bill

 

[Demystifier]

This is amusing. About a year or so ago you said something to the effect that reality was continuous (I am not able to find it). I replied, "No it's not, it's discrete. Prove me wrong." @Dale then deleted my comment because "Prove me wrong." wasn't up to PF standards. I actually meant it as a joke.

Nice to see you now know the TRUTH.
BTW, the term "an observable" is almost always used in the context of an operator in QM, which may be what is bothering @Neumaier.

Only a fool never changes his mind. My views evolve, see e.g. https://www.physicsforums.com/insights/stopped-worrying-learned-love-orthodox-quantum-mechanics/

That being said, I have to say that I still think that objective reality may be continuous, but it just can't be measured.

 

[rubi]

Continuity in the context of observables doesn't mean that we mustn't bin the measurement outcomes, but rather that for every choice of binning, we can achieve a finer binning by using a more accurate measurement device, i.e. there is no physical limitation on the possible accuracies of future measurement devices. The limitations are only of technological nature and can be overcome by using more advanced technology. At the moment, we have no experimental evidence for the existence of a physically finest possible binning of position or momentum, so the state of our knowledge is that position and momentum are continuous observables.

 

[Zafa Pi]

But page 42 looks pretty much like the definition of a mathematical model to me, but I accept opinions could vary.

Ballentine makes it clear that physics consists of both a mathematical model and a correspondence to the physical world.
Claiming "physics is a mathematical model" is likely to hurt the feelings of experimentalists.

 

[Zafa Pi]

Only a fool never changes his mind.

You've held that position for quite a while now. Perhaps it's time for a change.

I still think that objective reality may be continuous, but it just can't be measured.

there is no physical limitation on the possible accuracies of future measurement devices.

Those positions are far from universally accepted (e.g., Planck length limitation), and sufficiently untestable to be "not even wrong".

With a slightly amended set theory beyond ZFC all of analysis can be done with infinitesimals and thus countable. If you were raised with that orientation, then continuous would mean something different.
The usual real numbers are merely "a thinking tool", and should be kept from reality with likes of the tooth fairy.

 

[rubi]

Those positions are far from universally accepted (e.g., Planck length limitation), and sufficiently untestable to be "not even wrong".

Planck length is just a unit of length and not a smallest length scale. There's no problem with having half a Planck length or things like this and there is no indication today that there is such a thing as a smallest length scale. The currently accepted view is that observables such as position and momentum are indeed continuous. Of course, this may change in the future, like all scientific knowledge.

With a slightly amended set theory beyond ZFC all of analysis can be done with infinitesimals and thus countable. If you were raised with that orientation, then continuous would mean something different.
The usual real numbers are merely "a thinking tool", and should be kept from reality with likes of the tooth fairy.

I don't see what you mean by that. Non-standard analysis can already be done in ZFC and any extension of ZFC (and also super weak subsystems like second order arithmetic) will have uncountably many real numbers as well (hyperreals may have an even larger cardinality). But I don't see what your argument is. Already the rationals have the property that any interval can be split up into even smaller intervals.

 

[Zafa Pi]

Planck length is just a unit of length and not a smallest length scale. There's no problem with having half a Planck length or things like this and there is no indication today that there is such a thing as a smallest length and the currently accepted view is that observables such as position and momentum are indeed continuous.

The above is about QM, which is theory.

there is no physical limitation on the possible accuracies of future measurement devices.

This is about reality. You are mixing metaphors.

Already the rationals have the property that any interval can be split up into even smaller intervals.

The rationals are not continuous in the usual sense (Dedekind).

 

[rubi]

The above is about QM, which is theory.

It's true in all our accepted scientific theories.

This is about reality. You are mixing metaphors.

It is also about theories, because of course we propose only theories that model reality. If we have no experimental evidence for a limitation of possible measurement accuracy, we wouldn't arbitrarily include such a limitation in our models.

The rationals are not continuous in the usual sense (Dedekind).

Again, I don't see your point. The question is whether there is a smallest length scale and even if we were to reject the real numbers for whatever reason and resort to the rationals, there would still not be a smallest length scale in a model based on the rationals. But I don't understand your criticism of the reals anyway.

 

[Zafa Pi]

I don't know how to make my position clearer.

there is no physical limitation on the possible accuracies of future measurement devices.

Would you please give a reference to justify the above.

I don't understand your criticism of the reals anyway.

You're in good company, my girl friend doesn't understand my criticism of the tooth fairy.

 

[rubi]

I don't know how to make my position clearer.

Probably, because it cannot be made clear in the first place.

Would you please give a reference to justify the above.

How about you quote me correctly? This sentence was part of a definition, not a claim about reality: "Continuity in the context of observables doesn't mean that we mustn't bin the measurement outcomes, but rather that for every choice of binning, we can achieve a finer binning by using a more accurate measurement device, i.e. there is no physical limitation on the possible accuracies of future measurement devices." Please don't try to hide your lack of arguments behind this kind of dishonesty. The claim in my post was: "At the moment, we have no experimental evidence for the existence of a physically finest possible binning of position or momentum." Apparently, you can't prove me wrong here without resorting to dishonest methods.

You're in good company, my girl friend doesn't understand my criticism of the tooth fairy.

Well, I have explained, why your reasoning is wrong and you haven't responded to my criticism, so the situation is quite different.

 

[bhobba]

Ballentine makes it clear that physics consists of both a mathematical model and a correspondence to the physical world.
Claiming "physics is a mathematical model" is likely to hurt the feelings of experimentalists.

Get your issue now - yes that's right. But I don't think those into mathematical modelling as a discipline are that unconcerned about experimental verification of their models

Thanks
Bill