Position Eigenstates: Measurement & Imperfect Performance

[dextercioby]

You must be stating this accidentally, while not having read my posts properly.

If the psi has the form \psi:\mathbb{R}^4\to\mathbb{C}, and if the rho was defined by \rho=|\psi|^2, then also the rho would have the form \rho:\mathbb{R}^4\to [0,\infty[. However, the rho, about which I have been asking about, has the form \rho:\mathbb{R}^2\to [0,\infty[. Therefore it makes no sense to attempt to define it by \rho=|\psi|^2.

What is \rho ?

 

[Jano L.]

For most problems it doesn't matter, as the forward evolution in time of the delta function leads immediately to more reasonable states (such as that gaussian spike). You don't have to ascribe physical significance to "the wave function at the exact moment of collapse" (phrasing it this way makes it easier to understand the question as an interpretational issue) to get sensible results for any time after that... and that delta function does make for a simple initial condition.

This was particularly helpful.

Actually, time evolution of delta distribution leads to oscillating function of position and time $G(x,t)$ whose modulus squared is the same everywhere and decreases in time as $1/t$, which does not conform to the Born interpretation of the regular wave functions $\psi$. $G$ is just a propagator of the Schroedinger equation (a kernel function in an integral giving the evolution operator).

 

[atyy]

I've been trying to understand Ozawa's model for a sharp position measurement and state reduction. Although at least the measurement part is agreed on by Busch, it is very formal that I don't know whether there is consensus about whether sharp position measurements are possible.

I have found another reference supporting the claim of Distler and Paban that sharp position measurements are not possible.

Caves & Milburn, Quantum-mechanical model for continuous position measurements, Phys Rev A, 1987
"In the standard formalism of quantum mechanics the change in the state of a system produced by an instantaneous, precise measurement may be calculated by using projection operators. In the case of position measurements such an approach is inappropriate. Formally this is because there are no normalized position eigenstates. Physically it is because an arbitrarily precise measurement of position requires arbitrarily strong coupling to the system and an arbitrarily large amount of energy. Thus in the case of position (or momentum) measurements one must generalize the standard formalism to include imprecision in the measurements. ..."

 

[Dale]

Thus in the case of position (or momentum) measurements one must generalize the standard formalism to include imprecision in the measurements. ..."

I assume that this imprecision is not from the uncertainty principle, but is an additional imprecision above and beyond that.

 

[atyy]

I assume that this imprecision is not from the uncertainty principle, but is an additional imprecision above and beyond that.

Yes, in Caves and Milburn's and Distler and Paban's papers, the additional imprecision is due to the measuring device.

Having glanced at Caves and Milburn's paper, what they do actually seems very similar to Ozawa. They have a measurement apparatus which interacts briefly with the system, and the give an explicit Hamiltonian for the interaction. After the interaction, they do a projective measurement of the position of the apparatus.

What I wasn't sure about was whether the projective measurement of the apparatus is justified. Caves and Milburn explain that although it is impossible, taking the projective measurement of position on the apparatus instead of the system itself is enough to introduce noise into the measurement, so they allow themselves a projective measurement on the apparatus as an approximation.

Interestingly, Wiseman makes a similar comment in http://arxiv.org/abs/quant-ph/0302080 "If one attempts to apply a projection postulate directly to the atom, one will obtain nonsensical predictions. However, it can be shown that assuming a projective measurement of the field will yield results negligibly different from those assuming a projective measurement at any later stage, because of the rapid decoherence of macroscopic objects [48]. For this reason, it is sufficient to consider the field as being measured. Because the field has interacted with the system, their quantum states are entangled. If the initial state of the field is assumed known (which is not unreasonable in practice, because it is often close to the vacuum state), then the projective measurement of the field results in a measurement of the atom. Such a measurement however is not projective. Instead, we need a more general formalism to describe such measurements."

So I think within a formalism that allows projective measurements of position on the apparatus but not the system, Ozawa does correctly present a Hamiltonian that will execute sharp position measurements, and I think his state reduction to a wave function that isn't a delta function is probably also correct.

At any rate, it is definitely correct that a delta function is not a physical state. But it seems much more involved to answer the question about what the post measurement state is (when there is one), and one way of doing that seems to be to explicitly propose a measurement model and see what happens.

 

[tom.stoer]

I assume that this imprecision is not from the uncertainty principle, ...

As I said, the proof of the uncertainty principle fails for generalized states

the well-know proof for the uncertainty relation does not work for position eigenstates b/c it uses expressions like |ψ|², <ψ|ψ> and similar expressions which are ill-defined for ψ(x) ~ δ(x); I do not know whether this proof can be generalized, e.g. using rigged Hilbert spaces

 

[vanhees71]

I cannot repeat this often enough: The uncertainty principles holds for all states a quantum system can have. Generalized eigenstates for spectral values of self-adjoint operators in the continuous part of the spectrum do not represent physical states of the system!

 

[Jano L.]

I cannot repeat this often enough: The uncertainty principles holds for all states a quantum system can have. Generalized eigenstates for spectral values of self-adjoint operators in the continuous part of the spectrum do not represent physical states of the system!

I think everybody here agrees with that. The sad thing is Dirac made the impression that $|x\rangle$ (or $|p\rangle$) is just another ket - that is to be interpreted as a "state of the quantum system". I am afraid most teachers and textbooks still leave this impression on the students.

 

[tom.stoer]

You cannot ignore that fact that these states are used in calculations! My statement is not about "states a system can have" but about calculations we make!

 

[atyy]

What about the square-root of a delta function? It's discussed at the start of Gagen et al, Physical Review A, 48, 132-142 (1993) http://www.millitangent.org/pubs/q_meas/08_double_well_zeno.pdf.

Is the square-root of a delta function a position eigenstate, or is only the delta function a position eigenstate (but not in the Hilbert space)?

Is the square-root of a delta function normalizable?

I found another thread on this https://www.physicsforums.com/showthread.php?t=344902.

 

[atyy]

There are valid formulations without collapse and never in the theory are measuring devices supposed to be imperfect.

If I understand dextercioby correctly, maybe this is the most elegant solution - to deny that the question makes sense. One way of getting rid of collapse, even with Copenhagen is to couple the system to measuring ancillae, and push all measurements (something that yields a "classical" definite outcome) to the end of the experiment ("Principle of Deferred Measurement" http://en.wikipedia.org/wiki/Deferred_Measurement_Principle). Since there is only one measurement, there is no need for a post-measurement state.

There are two disadvantages, but I think they are not relevant for this question. The first is that measurement cannot be used as state preparation, but this can be argued to be not a general principle. The second is that in relativity, if a set of measurements is simultaneous in one frame, they will not be simultaneous in another frame (unless we also add that spatially separated measurements are impossible, ie. we do the measurements at the end of the experiment in the same place at the same time, so no Bell tests are allowed). However, there is no relativistic position operator, so that solves the problem, ie. the question of an exact position measurement only makes sense in non-relativistic quantum theory.

Within this framework of deferred measurements, although there is no post-measurement state, there is a post-interaction state. I think Ozawa's Hamiltonian for a sharp position measurement allows that calculation with nothing that is equivalent to a collapse to a delta function. (I use the term "equivalent" because Ozawa's original presentation doesn't use state reduction as a fundamental postulate, but I think the postulate can be derived from his equations).

 

[jostpuur]

What is \rho ?

I explained it in post #20 first. In post #25 I linked a related picture, which cannot be missed when glancing through the thread.

What is the probability density

<br /> \rho:\mathbb{R}^2\to [0,\infty[<br />

that describes the particle's hitting point? What principles imply what this probability density ends up being?

Here I put forward an interesting and relevant question, and it still remains unanswered. Is this because the physicists of modern times do not know the answer to this extremely basic question of the most basic quantum mechanics?

The wave function has the form \psi:[0,\infty[\times\mathbb{R}^3\to\mathbb{C}, and the probability density \rho:\mathbb{R}^2\to [0,\infty[. It makes no sense to "set" \rho=|\psi|^2.

Both the wave function and the probability density depend on time; so it's simply

\rho(x,t) = |\psi(x,t)|^2;\;\forall x \in \mathbb{R}^3\,\wedge\,\forall t \in \mathbb{R}

If the wave function is a time-dependent wave packet, then the probability density is time-dependent as well.

If the psi has the form \psi:\mathbb{R}^4\to\mathbb{C}

Where do you get R^4 from - you talked about R^3 previously?

You are deliberately not understanding what I write. This is probably because you don't want to admit to yourself that you don't know the answer to my question.

 

[vanhees71]

You cannot ignore that fact that these states are used in calculations! My statement is not about "states a system can have" but about calculations we make!

Of course not. I love the \delta distribution, particularly when I have to do an integral over it ;-)).

Nevertheless you must keep in mind what's representing something in the real world you describe in physics, and a \delta distribution or a plane-wave solution of the momentum-eigenvalue problem do not represent a pure state in quantum mechanics. If you assume this, you run into many contradictions within the theory itself, and if you claim that such generalized eigenvectors represent a state you should also be able to describe how you prepare it (at least in principle) for particles in the real world. I don't see, how you would do that.

 

[vanhees71]

What about the square-root of a delta function? It's discussed at the start of Gagen et al, Physical Review A, 48, 132-142 (1993) http://www.millitangent.org/pubs/q_meas/08_double_well_zeno.pdf.

Is the square-root of a delta function a position eigenstate, or is only the delta function a position eigenstate (but not in the Hilbert space)?

Is the square-root of a delta function normalizable?

I found another thread on this https://www.physicsforums.com/showthread.php?t=344902.

Where precisely is the square root of a \delta distribution (NOT FUNCTION!) described in that paper? This has no sense in standard distribution theory.

Even if you can define something like this in a reasonable way by extending distribution theory somehow, the \delta distribution would not be normalizable, because the integral over a \delta distribution without multiplying it with an appropriate test function doesn't make sense either.

The easiest way to see this is to keep in mind that the position operator as an essentially self-adjoint operator is defined on a dense subset in Hilbert space (the "test functions"). The generalized eigenvalues are in the dual of this subset, which is much larger than Hilbert space. It's containing the \delta distribution, plane waves, etc. But only true Hilbert-space vectors represent pure states!

 

[atyy]

Where precisely is the square root of a \delta distribution (NOT FUNCTION!) described in that paper? This has no sense in standard distribution theory.

Even if you can define something like this in a reasonable way by extending distribution theory somehow, the \delta distribution would not be normalizable, because the integral over a \delta distribution without multiplying it with an appropriate test function doesn't make sense either.

The easiest way to see this is to keep in mind that the position operator as an essentially self-adjoint operator is defined on a dense subset in Hilbert space (the "test functions"). The generalized eigenvalues are in the dual of this subset, which is much larger than Hilbert space. It's containing the \delta distribution, plane waves, etc. But only true Hilbert-space vectors represent pure states!

I don't know either and had many of the same questions as you - the brief mention in the Gagen et al paper was the first time I saw it. The only thing I don't know is whether one really cannot extend the theory so that the the square root of a delta function is in some sense a "normalizable pure state".

 

[tom.stoer]

I explained it in post #20 first. In post #25 I linked a related picture, which cannot be missed when glancing through the thread.

...

Here I put forward an interesting and relevant question, and it still remains unanswered. Is this because the physicists of modern times do not know the answer to this extremely basic question of the most basic quantum mechanics?

...

You are deliberately not understanding what I write. This is probably because you don't want to admit to yourself that you don't know the answer to my question.

The picture you are presenting are spots on a 2-dim. surface; but ρ = |ψ|² is defined over 3-space. I still don't understand your problem.

 

[atyy]

Of course not. I love the \delta distribution, particularly when I have to do an integral over it ;-)).

I think this is the reason why formal collapse of the ancilla (but not the system) to a position eigenstate makes sense (as eg. in Caves and Milburn, and effectively in Ozawa), if we assume that successive measurements are made on the system but not the ancilla. In that case, the ancilla can give a classical result, and collapse formally to a delta function. When the partial trace is taken after collapse of the ancilla and system, the delta function falls under an integral, and the state of the ancilla falls within the Hilbert space.

 

[jostpuur]

The picture you are presenting are spots on a 2-dim. surface; but ρ = |ψ|² is defined over 3-space. I still don't understand your problem.

The problem is that I don't know what formulas or principles imply what the probability density on the two dimensional surface is. Surely you can see that the "spots on a 2-dim. surface" appear to obey some probability density. Something like f(x_1,x_2)\approx C(1+\cos(\frac{2\pi x_1}{\lambda})) in the small region depicted by the pictures. Where does this probability density on the two dimensional surface come from?

I have explained my problem very clearly now, and people here should be able to understand it.

 

[Nugatory]

Where does this probability density on the two dimensional surface come from?

I have a probability density in three dimensions. The set of points that make up the surface of the screen is a subset of that three-dimensional space, so I can evaluate the probability density at those points...

But this seems so obvious that I must not be understanding your question...

 

[jostpuur]

The final answer, which is a probability density on the two dimensional surface, does not depend on time. So an attempt to define it with

<br /> f(x_1,x_2)=|\psi(t,x_1,x_2,R)|^2<br />

doesn't work. We would also like to have the condition

<br /> \int\limits_{\mathbb{R}^2}f(x_1,x_2)dx = 1<br />

in the end.

 

[atyy]

The final answer, which is a probability density on the two dimensional surface, does not depend on time. So an attempt to define it with

<br /> f(x_1,x_2)=|\psi(t,x_1,x_2,R)|^2<br />

doesn't work. We would also like to have the condition

<br /> \int\limits_{\mathbb{R}^2}f(x_1,x_2)dx = 1<br />

in the end.

I think it's tricky to do it the way you would like. I would propose that the fully proper way to do it (I've never seen it done) is to include the interaction of the screen in the Hamiltonian, ie. we consider the particle and the screen a single quantum system. As the screen is taken to be composed of an increasingly large number of particles, it should increasingly decohere the system. Up to this point everything in the calculation is fully quantum and the evolution is completely unitary. In the large t limit (or maybe large but finite t), we projectively measure the system and the screen in an appropriate basis.

The tricky part is writing the interaction term for the screen and evaluating it. I've seen it done for simpler setups, let me look for some examples.

 

[tom.stoer]

The final answer, which is a probability density on the two dimensional surface, does not depend on time.

I think this is wrong. For each individual particle (described by a time dependent wave function) the probability density for the particle (to be absorbed by the screen) is time-dependent as well. Using one single particle in your experiment you will not observe a spot on the screen before you have started the experiment.

 

[dextercioby]

Alright, let's get down to business: observables with continuous spectrum are called 'unquantized'. The free particle is, because of that, not welcome in quantum mechanics. There's nothing quantized about it. :D

I don't buy collapse, even though school teaches it to you. I think the only evolution of a state is that given by Schrödinger's equation. The Hamiltonian in there should describe the system's evolution and that's it.

Now, the position observable has a continuous spectrum, for just about any non-specially relativistic system. Its eigenstates lie outside of the Hilbert space (which the theory needs and couldn't be formulated without), so one may ask: are they useful ? Well, apparently not.

 

[atyy]

Alright, let's get down to business: observables with continuous spectrum are called 'unquantized'. The free particle is, because of that, not welcome in quantum mechanics. There's nothing quantized about it. :D

I don't buy collapse, even though school teaches it to you. I think the only evolution of a state is that given by Schrödinger's equation. The Hamiltonian in there should describe the system's evolution and that's it.

Now, the position observable has a continuous spectrum, for just about any non-specially relativistic system. Its eigenstates lie outside of the Hilbert space (which the theory needs and couldn't be formulated without), so one may ask: are they useful ? Well, apparently not.

I commented on your approach in post #41, and I think it works - but did I understand it correctly?

 

[atyy]

I think this is wrong. For each individual particle (described by a time dependent wave function) the probability density for the particle (to be absorbed by the screen) is time-dependent as well. Using one single particle in your experiment you will not observe a spot on the screen before you have started the experiment.

What I think joostpuur means is that when we look at the plate which is placed at z=L from the slit at the end t=T of the experiment, the distribution of spots on the plate ρ(x,y) is clearly not changing. So if we use ψ(x,y,z,t=0) at the start of the experiment, what time and what observables are we choosing to measure in order to get ρ(x,y)? The answer is clearly not |ψ(x,y,z=L,t=T)|², unless the screen is placed at L and T are both very large. For example, if the screen is placed near to the slit so that L is small, we expect to image the slit. Usually the problem is treated by having the initial wave function f(x,y,t), where the momentum in the z direction is large and definite, so we know that the particles reach the screen at a definite time calculated classically. If we don't want to make the approximation, I think we have to explicitly model the interaction of the screen with the particle, so that even at large T for small L, the wave function will be decohered and "localized" near L.

 

[tom.stoer]

I don't think that this is the problem jostpuur has.

Let's consider a time-independent setup first.

We should distinguish between the probability density σ on the screen, describing the spots, and the probability density ρ=|ψ|². The reason is simple: the probability density σ is calculated based on ρ, but using the fact the we already know with certainty that the particle has been absorbed on (or in) the screen. So we get

\sigma(x,y) = c \int_L^{L+\epsilon}dz\,\rho(x,y,z)

We know that the particle has been absorbed, therefore it's allowed to normalize σ such that

\int_\text{screen}dx\,dy\,\sigma(x,y) = 1

For a non-stationary setup and with less then 100% absorption it's more difficult, but I think that it's still a problem how to interpret probabilities and w/o any need to study the interaction Hamiltonian.

 

[jostpuur]

I think it's tricky to do it the way you would like. I would propose that the fully proper way to do it (I've never seen it done) is to include the interaction of the screen in the Hamiltonian, ie. we consider the particle and the screen a single quantum system. As the screen is taken to be composed of an increasingly large number of particles, it should increasingly decohere the system. Up to this point everything in the calculation is fully quantum and the evolution is completely unitary. In the large t limit (or maybe large but finite t), we projectively measure the system and the screen in an appropriate basis.

The tricky part is writing the interaction term for the screen and evaluating it. I've seen it done for simpler setups, let me look for some examples.

Sounds good. I'm sure that in the end the entagelment with the measurement device and the decoherence is the way to the right solution. Also, for years I have thought that somekind of "effective collapse" and perhaps even an "effective collapsing density with respect to time" could be derived from the decoherence interpretation, but I never managed to figure out any details, so to me it has remained only as a vague idea.

However, my overall understanding of PDEs has improved a little bit during the years, so now I was able to write down a nice guess. I'll use the notation defined in my original problem statement, so the detector plane is described by the equation x_3=R. How about the following. The particle's state is described by two functions:

<br /> \psi(t,x_1,x_2,x_3)<br />

for all x_3 &lt; R, and

<br /> \phi(t,x_1,x_2,R)<br />

for x_3=R. Then we define the time evolution of the state by

<br /> i\hbar\partial_t \psi = -\frac{\hbar^2}{2m}\nabla_x^2 \psi<br />

and

<br /> \partial_t\phi = \frac{\hbar}{m}\textrm{Im}(\psi^*\partial_3\psi)<br />

Or to be more precise:

<br /> \partial_t \phi(t,x_1,x_2,R) = \lim_{\epsilon\to 0^+}\frac{\hbar}{m}\textrm{Im}\big(\psi^*(t,x_1,x_2,R-\epsilon)\partial_3\psi(t,x_1,x_2,R-\epsilon)\big)<br />

Then we give these functions the following probability interpretations. The probability that the particle is in \Omega\subset \mathbb{R}^2\times ]-\infty,R[ at time t is

<br /> \int\limits_{\Omega} |\psi(t,x)|^2 dx<br />

(three dimensional integral) and the probability that the particle is in \Omega\subset \mathbb{R}^2\times \{R\} at time t is

<br /> \int\limits_{\Omega} \phi(t,x)dx<br />

(two dimensional integral.) I have no other justification for this guess but the fact that the function

<br /> P(t) = \int\limits_{\mathbb{R}^2\times ]-\infty,R[}|\psi(t,x)|^2dx + \int\limits_{\mathbb{R}^2\times \{R\}} \phi(t,x)dx<br />

remains a constant under the time evolution. In other words

<br /> \partial_t P(t) = 0<br />

This can be verified by changing the order of the derivatives and integrals, using the defined time evolution equations, and the divergence theorem.

Doesn't it look interesting at least? If the probability currect is aimed at the detector plate, it would naturally get absorbed and get frozen into in. So the time evolution would be somewhat unitary only for the x_3&lt;R part, and on the plane x_3=R the wave function somehow changes into an approximation of something else. I'm not sure what conditions would need to be satisfied for \partial_t \phi \geq 0 to be true though. If \partial_t \phi &lt; 0 happens, it ruins the absorption interpretation.

 

[bhobba]

The problem is that I don't know what formulas or principles imply what the probability density on the two dimensional surface is.

You are not talking about the systems state, you are talking about a REPRESENTATION of the state. Just like in linear algebra a representation of a vector in terms of a particular basis is not the vector.

Representations are freely chosen to help in solving problems. You chose to represent the state expanded in terms of position eigenstates in three dimensional space, which by definition is called the wave-function. You are free to do that - but whether it's of value in solving a particular problem is another matter.

In the situation with a two dimensional screen where the observable is positions on the screen you should expand the state in eigenstates of that observable.

The observable associated with the screen is O = ∑ x(ij) |x(ij)><x(ij)| where ij represents the 'disreetised' position on the screen to avoid irrelevant issues associated with having a continuum and dealing with Dirac Delta functions etc.

You expand the state in that basis to get |u> = ∑ c(ij) |x(ij)> or simply c(ij) if you simply take it as a representation like we do with a wave function.

Now you apply the Born Rule, which I explained in a previous post, in this case reduces to simply squaring the absolute value, but will run though the gory detail.

First we note the observable for getting 1 if the outcome is x(ij) and zero otherwise is simply |x(ij)><x(ij)|. So probability getting (x(ij)) = <u||x(ij)><x(ij)|u> = |c(ij)|^2.

The problem you had is you are analysing the problem incorrectly, and not understanding the difference between a state and its representation.

I also want to say the way I analysed it above is not really the best either, the following link, I gave previously, is much better IMHO:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

There it is analysed in terms not of positions on the screen, but of scattering angles.

Thanks
Bill

 

[bhobba]

Sounds good. I'm sure that in the end the entagelment with the measurement device and the decoherence is the way to the right solution. Also, for years I have thought that somekind of "effective collapse" and perhaps even an "effective collapsing density with respect to time" could be derived from the decoherence interpretation, but I never managed to figure out any details, so to me it has remained only as a vague idea.

If you want to go down that path then THE book to get is Decoherence and the Quantum-to-Classical Transition by Schlosshauer:
https://www.amazon.com/dp/3540357734/?tag=pfamazon01-20

The following also gives a good overview of the issues:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

But from my reading of it, the issue of exactly how the screen carries out the observation wasn't what seemed to concern you.

For me it was choosing an inappropriate representation to analyse the problem.

Thanks
Bill

 

[bhobba]

I don't think that this is the problem jostpuur has.

I don't think so either.

I think it was understanding the wavefunction is an entirely arbitrary representation of the state, freely chosen as an aid to solving problems. A 3d expansion for a 2d observation is simply making the analysis harder than it needs to be, and introducing irrelevant side issues.

Thanks
Bill