A Fundamental Theorem of Quantum Measurements

[Danny Boy]

The Fundamental Theorem of Quantum Measurements (see page 25 of these PDF notes) is given as follows:

Every set of operators $\{A_n \}_n$ where $n=1,...,N$ that satisfies $\sum_{n}A_{n}A^{\dagger}_{n} = I$, describes a possible measurement on a quantum system, where the measurement has $n$ possible outcomes labeled by $n$. If $\rho$ is the state of the system before the measurement, $\tilde{\rho}_n$ is the state of the system upon obtaining measurement result $n$, and $p_n$ is the probability of obtaining result $n$, then $$\tilde{\rho}_{n} = \frac{A_n \rho A_{n}^{\dagger}}{p_n}~~~~\text{where}~~~~p_n = \text{Tr}[A_{n}^{\dagger}A_n \rho].$$We refer to the operators $\{ A_n \}_n$ as the measurement operators for the measurement.

Query: Consider measurement operators $\{A_c \}_c$ where now the measurement outcomes $c$ is considered a continuous index rather than discrete as above. For example a Gaussian measurement of the form $$A_{c} = \int_{-\infty}^{\infty}\frac{e^{-(x'-c)^2}/(4V)}{(2 \pi V)^{1/4}}|x' \rangle \langle x' |dx'.$$
In this case of continuous measurement outcomes $c$, it would seem based on the theorem above that $p_c$ now gives the probability density rather than the probability. Hence am I correct in stating that the probability of getting an outcome from some subset $\mathcal{M}$ is given by $$Pr(c \in \mathcal{M}) = \int_{\mathcal{M}} p_c dc = \int_{\mathcal{M}} \text{Tr}[A_{c}^{\dagger}A_{c} \rho]dc$$ and that the final state should then be given by $$\tilde{\rho} = \int_{\mathcal{M}}p_c [A_c^{\dagger}\rho A_c]dc = \int_{\mathcal{M}}\text{Tr}[A_{c}^{\dagger}A_{c} \rho]A_c \rho A_c^{\dagger} dc $$

Please advise on this, thanks.

 

[Demystifier]

Your equations are not wrong, but you didn't explain their physical interpretation. Your equations correspond to a situation in which measurement of a continuous observable $c$ is sharp, but, for some reason, the observer does not know what that sharp value is. That's the meaning of the integration over $c$ in your last equation. By contrast, if the observer knows the value of $c$, then there is no integration over $c$ and the final state is given by the first equation with $n\rightarrow c$.

 

[Danny Boy]

@Demystifier Thanks for your response, this is what I was interested in discussing. My thinking was as follows: From the Fundamental Theorem of Quantum Measurement, for the discrete case, we obtain a probability mass function $Pr(n) = p_n = \text{Tr}[A_n A_n^{\dagger} \rho]$ with random variable $n$ (which is our measurement outcome). Whereas for the continuous case, where we have continuous outcomes $c$, the corresponding probability density function is $Pr(c) = p_c = \text{Tr}[A_c A_c^{\dagger} \rho]$, hence in the same way that we consider the probability in normal distribution by taking the integral over some subset (since the probability of a single value of normal distribution is zero, by definition of being a probability density function) we do the same here by taking the integral over some subset which I call $\mathcal{M}$. What do you think, does this make sense?

 

[Demystifier]

Yes, it makes perfect sense mathematically. But physically, I would say that possible measurement outcomes are never continuous in the first place.

 

[Danny Boy]

Are you saying that physically (or experimentally) we can consider the probability $p_c = \text{Tr}[A_c A_c^{\dagger} \rho]$ as the probability of obtaining measurement outcome $c$ regardless of whether $c$ is continuous or not?

 

[Demystifier]

Are you saying that physically (or experimentally) we can consider the probability $p_c = \text{Tr}[A_c A_c^{\dagger} \rho]$ as the probability of obtaining measurement outcome $c$ regardless of whether $c$ is continuous or not?

The equation $p_c = \text{Tr}[A_c A_c^{\dagger} \rho]$ is always correct, but the interpretation of $p_c$ depends on whether $c$ is discrete or continuous. In the former case $p_c$ is probability, while in the latter case $p_c$ is probability density.

 

[A. Neumaier]

Yes, it makes perfect sense mathematically. But physically, I would say that possible measurement outcomes are never continuous in the first place.

So you doubt the validity of Born's rule for continuous observables?

 

[Danny Boy]

@Demystifier Yes agreed (this is what I stated in my original post). What I am asking is what is the most sensible way of representing the final state $\tilde{\rho}$. I have already suggested weighting by the probabilities $p_c$:
$$\tilde{\rho} := \int_{\mathcal{M}} p_{c}[A_{c}^{\dagger}\rho A_c]dc.$$ Another idea is $$\tilde{\rho} := \frac{\int_{\mathcal{M}}A_{c}^{\dagger}\rho A_c}{pr(c \in \mathcal{M})} = \frac{\int_{\mathcal{M}}A_{c}^{\dagger} \rho A_c}{\int_{\mathcal{M}}\rho_cdc}.$$

 

[Demystifier]

So you doubt the validity of Born's rule for continuous observables?

No, I doubt the physical existence of continuous observables.

 

[Demystifier]

@Demystifier Yes agreed (this is what I stated in my original post). What I am asking is what is the most sensible way of representing the final state $\tilde{\rho}$. I have already suggested weighting by the probabilities $p_c$:
$$\tilde{\rho} := \int_{\mathcal{M}} p_{c}[A_{c}^{\dagger}\rho A_c]dc.$$ Another idea is $$\tilde{\rho} := \frac{\int_{\mathcal{M}}A_{c}^{\dagger}\rho A_c}{pr(c \in \mathcal{M})} = \frac{\int_{\mathcal{M}}A_{c}^{\dagger} \rho A_c}{\int_{\mathcal{M}}\rho_cdc}.$$

The first one is the correct one. But for practical purposes the second one can also be a good approximation, assuming that $\mathcal{M}$ is sufficiently small so that $p_c$ does not vary much within $\mathcal{M}$.

 

[A. Neumaier]

No, I doubt the physical existence of continuous observables.

So position does not exist as an observable?

 

[bhobba]

So position does not exist as an observable?

Of course it does - but what does not exist is an eigenstate of position - its not even square integrable. That's where Rigged Hilbert Spaces are required.

I have a my own view on this I know you do not agree with - so will leave it at the above comment. These thing's are just mathematical fictions introduced for mathematical convenience.

Thanks
Bill

 

[vanhees71]

I've no clue what you are debating about ;-)). It's my understanding that the issue of continuous spectra of essentially self-adjoint operators on the separable Hilbert space is well formulated both in the traditional way a la von Neumann as well as in the modern formulation of rigged Hilbert space, which is more to the taste of physicists' "robust approach" to this issue. It's clear that the position of a particle can never be sharply determined since there is no position eigenvector. In other words (within non-relativistic physics!) you can localize a particle only within a finite volume, and that volume can be made arbitrarily small. Further, relativistic QT even weakens the localizability further, because to confine a particle in an ever smaller volume you need to apply ever stronger fields, leading to creation-annihilation processes of more quanta. A very readable and physical discussion of this can be found in vol. IV of the textbook series by Landau and Lifshitz.

 

[Demystifier]

So position does not exist as an observable?

Nobody ever measured position with an infinite precision, ergo physical position observable is not continuous.

 

[A. Neumaier]

Nobody ever measured position with an infinite precision, ergo physical position observable is not continuous.

But it is modeled as such in classical mechanics, in Bohmian mechanics, and (according to Born's original probability interpretation of the wave function) also in orthodox quantum mechanics.

 

[Demystifier]

But it is modeled as such in classical mechanics, in Bohmian mechanics, and (according to Born's original probability interpretation of the wave function) also in orthodox quantum mechanics.

So? Theoretical physics (or any theoretical science) deals with idealizations that only approximately correspond to actual observations. Theoretical physics is ... see my signature.

 

[A. Neumaier]

Theoretical physics is ... see my signature.

But so is all language, including that used in our discussion. We can only think in idealized terms. All notions of experimental physics, including the notion of an ''actual observation'' are idealizations, too, that only approximately corresponding to reality. Thus your comments empty everything from having logical force.

 

[Danny Boy]

@Demystifier Given that I consider the idea above: $$\rho \mapsto \tilde{\rho} := \frac{\int_{\mathcal{M}} A_{c}^{\dagger} \rho A_c}{Pr(c \in \mathcal{M})} = \frac{\int_{\mathcal{M}} A_{c}^{\dagger} \rho A_c}{\int_{\mathcal{M}}Tr[A_c^{\dagger}A_c \rho]}$$ What do you think I should consider $\mathcal{M}$ to be given some choice of $c$ (hence $\mathcal{M}$ would serve as some neighborhood of $c$ which is an open interval containing $c$)? I assume that this would be apparatus dependent? But maybe there is a general idea that seems reasonable, what do you think?

 

[Demystifier]

@Demystifier Given that I consider the idea above: $$\rho \mapsto \tilde{\rho} := \frac{\int_{\mathcal{M}} A_{c}^{\dagger} \rho A_c}{Pr(c \in \mathcal{M})} = \frac{\int_{\mathcal{M}} A_{c}^{\dagger} \rho A_c}{\int_{\mathcal{M}}Tr[A_c^{\dagger}A_c \rho]}$$ What do you think I should consider $\mathcal{M}$ to be given some choice of $c$ (hence $\mathcal{M}$ would serve as some neighborhood of $c$ which is an open interval containing $c$)? I assume that this would be apparatus dependent? But maybe there is a general idea that seems reasonable, what do you think?

Yes, that's apparatus dependent.

 

[Demystifier]

But so is all language, including that used in our discussion. We can only think in idealized terms. All notions of experimental physics, including the notion of an ''actual observation'' are idealizations, too, that only approximately corresponding to reality.

I agree.

Thus your comments empty everything from having logical force.

I don't even understand what is that supposed to mean.

 

[vanhees71]

Nobody ever measured position with an infinite precision, ergo physical position observable is not continuous.

This argument doesn't make sense. The possible outcomes of measurements of position are continuous since the spectrum of the position operator is continuous. Of course, there's no state where any observable is precisely determined to have a value in the continuous part of the spectrum of the corresponding representing operator, but that doesn't mean that the observable is not continuous.

 

[Demystifier]

This argument doesn't make sense. The possible outcomes of measurements of position are continuous since the spectrum of the position operator is continuous. Of course, there's no state where any observable is precisely determined to have a value in the continuous part of the spectrum of the corresponding representing operator, but that doesn't mean that the observable is not continuous.

By "physical observable", I didn't mean a mathematical object of theoretical physics. I meant an object that has a physical existence in the real laboratory. In that sense, even though a continuous observable exists as a mathematical object, it does not exist in a real laboratory.

But I am a theoretical physicist, so why do I take such an experimental perspective in this context? Because I think that quantum-mechanical observable is nothing more but a tool in instrumental interpretation of QM, which does not make any sense outside of the laboratory. This should be contrasted with the concept of beable, which is an ontic object supposed to be there even if we don't measure it.

 

[vanhees71]

Well, then it's even easier to argue: There's no empirical hint at any discreteness of position, at least not down to subatomic scales covered by the Standard Model. At least the argument that only because you cannot determine accurately the position position is not a continuous observable, is in no way convincing.

 

[Demystifier]

Well, then it's even easier to argue: There's no empirical hint at any discreteness of position, at least not down to subatomic scales covered by the Standard Model. At least the argument that only because you cannot determine accurately the position position is not a continuous observable, is in no way convincing.

I think you haven't understood what do I claim. You make valid arguments against a non-convincing thesis, but this non-convincing thesis is not a thesis that I defend. I defend a different thesis, with which you would agree if you only understood what this thesis claims.

My thesis is very trivial. Experiments cannot determine position (or any other observable) with perfect precision, therefore there is some finite resolution $\Delta x$ in the experiments, therefore the measured position is of the form $x=n\Delta x$ with integer $n$. I don't see how could anybody find this argument controversial.

 

[vanhees71]

No, as I said, this argument doesn't make any sense, and I understood your argument indeed as you've just detailed. The measured position is of the form $x = x_0 \pm \Delta x$ (with $\Delta x$ defined as the standard deviation with some given confidence leven and $x_0$ the average value). There's no discreteness for $x_0$, at least not according to standard QT. There's also not absolute limit for $\Delta x$.

Of course, there are some theoretical ideas on "quantized spacetime", but that's not very convincing to me yet since there's no empirical evidence for it nor a convincing solution of the notorious problem of quantizing gravity using such an approach either.

 

[Demystifier]

No, as I said, this argument doesn't make any sense, and I understood your argument indeed as you've just detailed. The measured position is of the form $x = x_0 \pm \Delta x$ (with $\Delta x$ defined as the standard deviation with some given confidence leven and $x_0$ the average value). There's no discreteness for $x_0$, at least not according to standard QT. There's also not absolute limit for $\Delta x$.

Of course, there are some theoretical ideas on "quantized spacetime", but that's not very convincing to me yet since there's no empirical evidence for it nor a convincing solution of the notorious problem of quantizing gravity using such an approach either.

You still don't understand what I am talking about. You define $\Delta x$ as standard deviation, but I define $\Delta x$ as a resolution of the measuring apparatus. That's not the same. Standard deviation is obtained from a statistics of many measurements. Resolution is a property of a single measurement. For instance, if one performs a series of measurements and obtains the values: 3.01, 3.12, 2.88, 3.22, 2.91, ... then it is clear that the resolution is 0.01, even though the standard deviation is significantly larger. With resolution 0.01, the possible measurement outcome of a single measurement is $n\cdot 0.01$ with integer $n$.

 

[Demystifier]

Related to this, I have an intriguing question for experimentalists. Suppose that a measuring apparatus has the resolution 0.01. And suppose that one performs a series of measurements with the results: 3.17, 3.17, 3.17, 3.17, 3.17, 3.17, 3.17, 3.17. What is the final result, is it
$$3.17\pm 0.00$$
or is it something like
$$3.17\pm 0.005$$
?

 

[vanhees71]

You still don't understand what I am talking about. You define $\Delta x$ as standard deviation, but I define $\Delta x$ as a resolution of the measuring apparatus. That's not the same. Standard deviation is obtained from a statistics of many measurements. Resolution is a property of a single measurement. For instance, if one performs a series of measurements and obtains the values: 3.01, 3.12, 2.88, 3.22, 2.91, ... then it is clear that the resolution is 0.01, even though the standard deviation is significantly larger. With resolution 0.01, the possible measurement outcome of a single measurement is $n\cdot 0.01$ with integer $n$.

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.

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.

 

[stevendaryl]

Related to this, I have an intriguing question for experimentalists. Suppose that a measuring apparatus has the resolution 0.01. And suppose that one performs a series of measurements with the results: 3.17, 3.17, 3.17, 3.17, 3.17, 3.17, 3.17, 3.17. What is the final result, is it
$$3.17\pm 0.00$$
or is it something like
$$3.17\pm 0.005$$
?

It seems to me that the notation x \pm \Delta x is used both for standard deviation and for resolution limitations. I guess you have to read the report to know which is meant. According to this paper: http://www2.ece.rochester.edu/courses/ECE111/error_uncertainty.pdf, if you are measuring positions with a resolution of 0.01 meters, then you should report it as 3.17 \pm 0.001. The error can never be less than 1/10 the resolution.

The use of x \pm \Delta x when \Delta x is a standard deviation is kind of misleading, because a measured value can be off from the true value by more than one standard deviation.

 

[vanhees71]

Related to this, I have an intriguing question for experimentalists. Suppose that a measuring apparatus has the resolution 0.01. And suppose that one performs a series of measurements with the results: 3.17, 3.17, 3.17, 3.17, 3.17, 3.17, 3.17, 3.17. What is the final result, is it
$$3.17\pm 0.00$$
or is it something like
$$3.17\pm 0.005$$
?

Of course neither the former nor the latter. You cannot use the rules of statistical-error analysis to deal with sytematical errors. If the resolution of an apparatus (systematic uncertainty!) is 0.01, the systematic error is 0.01, not 0.000 or some smaller value due to averaging as for independent measurements with purely statistical errors.

 

[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