Position Eigenstates: Measurement & Imperfect Performance

[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

$$\rho:\mathbb{R}^2\to [0,\infty[$$

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

$$f(x_1,x_2)=|\psi(t,x_1,x_2,R)|^2$$

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

$$\int\limits_{\mathbb{R}^2}f(x_1,x_2)dx = 1$$

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

$$f(x_1,x_2)=|\psi(t,x_1,x_2,R)|^2$$

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

$$\int\limits_{\mathbb{R}^2}f(x_1,x_2)dx = 1$$

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:

$$\psi(t,x_1,x_2,x_3)$$

for all $x_3 < R$, and

$$\phi(t,x_1,x_2,R)$$

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

$$i\hbar\partial_t \psi = -\frac{\hbar^2}{2m}\nabla_x^2 \psi$$

and

$$\partial_t\phi = \frac{\hbar}{m}\textrm{Im}(\psi^*\partial_3\psi)$$

Or to be more precise:

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

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

$$\int\limits_{\Omega} |\psi(t,x)|^2 dx$$

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

$$\int\limits_{\Omega} \phi(t,x)dx$$

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

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

remains a constant under the time evolution. In other words

$$\partial_t P(t) = 0$$

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 $x_3<R$ 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 $\partial_t \phi < 0$ 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

 

[atyy]

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.

I attempted to interpret what you are suggesting below. See if I got it right?

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 $x_3<R$ 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 $\partial_t \phi < 0$ happens, it ruins the absorption interpretation.

But if one is simply going to write down phenomenological equations without using an explicit model of the interaction between the particle and the screen, then why not do what tom.stoer and Nugatory have been suggesting, ie: ρ(x₁,x₂) = c |ψ(x₁,x₂,x₃=R,t=Rm/p₃)|², where m is the mass of the particle, and c is a normalization constant so that ∫ρdx₁dx₂=1 ? The approximation is valid if the motion along x₃ is "classical", which means that the de Broglie wavelength should be much smaller than the distance between the slit and the screen.

One example of an experiment where they get interference patterns based on detection times which can be calculated using classical equations of motion along the slit-screen axis is http://www.atomwave.org/rmparticle/ao%20refs/aifm%20refs%20sorted%20by%20topic/ungrouped%20papers/wigner%20function/PFK97.pdf (see especially section 4 "Time-resolved diffraction patterns").

 

[bhobba]

then why not do what tom.stoer and Nugatory have been suggesting, ie: ρ(x₁,x₂) = c |ψ(x₁,x₂,x₃=R,t=Rm/p₃)|², where m is the mass of the particle, and c is a normalization constant so that ∫ρdx₁dx₂=1 ? The approximation is valid if the motion along x₃ is "classical", which means that the de Broglie wavelength should be much smaller than the distance between the slit and the screen.

Why worry about it at all.

The only observable is the position on the two dimensional screen - simply expand the state in that.

No need to make your job harder than necessary and have confusing issues like how a 3 dimensional function relates to a 2 dimensional one.

Even better - eschew it entirely and analyse it in terms of scattering angle as in the paper I linked to. That's one of its key ideas - by means of a little physical insight we see when it leaves whatever source you have, slit, or whatever, it was at a particular location so it's momentum is unknown. But its kinetic energy is fixed, hence its velocity in magnitude is fixed, so the only thing that can be uncertain is direction - and we can view the screen as an observation of that direction. Hence it makes sense to analyse the situation not in terms of screen position, but 'scattering' angles.

Its what we do in physics all the time to simplify problems.

Remember the freedom to represent a state how we like is built right into the foundations of QM - its why when Dirac came up with it it was called the transformation theory.

Thanks
Bill

 

[atyy]

Why worry about it at all.

The only observable is the position on the two dimensional screen - simply expand the state in that.

No need to make your job harder than necessary and have confusing issues like how a 3 dimensional function relates to a 2 dimensional one.

Even better - eschew it entirely and analyse it it terms of scattering angle as in the paper I linked to.

Remember the freedom to represent a state how we like is built right into the foundations of QM - its why when Dirac came up with it it was called the transformation theory.

Thanks
Bill

The scattering angle calculation is only an approximation, so the question is when is the approximation valid?

 

[bhobba]

The scattering angle calculation is only an approximation, so the question is when is the approximation valid?

Agreed.

But what are we trying to do here?

Are we trying to understand principles or are we trying to calculate the outcomes of increasingly sophisticated models?

If the latter there is another paper about that has issues with the analysis in my linked paper - I will dig it up anon (found it):
http://arxiv.org/pdf/1009.2408.pdf

But seriously - this is what goes on in physics all the time. You drill down on an issue, give a 'lay explanation', then see its actually more subtle than that so do that deeper look, then realize its not quite correct either.

When dealing with foundational issues like this, where you have someone questioning pretty basic stuff, you are simply making a rod to break your own back IMHO.

Still if its your won't be my quest.

Added Later:

Here is the concusion of the paper:
In sum, Marcella does make the valid point that quantum interference should be treated as a quantum phenomenon and quantum texts ought not immediately redirect the discussion to classical wave optics. But a more reasonable way to do this would be to simply show that the Schrodinger equation reduces to the Helmholtz equation, thus reducing the problem to one of classical scalar scattering with its concomitant approximations. This would also provide the opportunity of discussing relevant boundary conditions and to point out the difficulty of specifying them precisely in both the quantum and electromagnetic cases. As it stands, while Marcella’s procedure is useful in giving students practice with the Dirac formalism, it has introduced no quantum physics into the problem other than setting p = ~k, and has implicitly made all the assumptions that show this is indeed a problem of classical optics. That his result is the same as the one obtained by the simplest Huygens construction is merely a reflection of the fact that he has implicitly made the lowest-order approximations, where all methods converge to the same result.

Really - did that illuminate any foundational issue or make things clearer?

Thanks
Bill

 

[bhobba]

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.

Its in Schlosshauer.

Its a general consideration related to the fact the interaction Hamiltonian is 'radial'.

He gives a general argument in that case decoherence must single out position as the observable.

In fact since most physical situations are like that is why position is usually the observable.

I can dig up the page if you like.

Thanks
Bill

 

[jostpuur]

This thread proves that when a human being doesn't know an answer to a question, he can unconsciously prefer to not understanding the question as well.

The only observable is the position on the two dimensional screen - simply expand the state in that.

No need to make your job harder than necessary and have confusing issues like how a 3 dimensional function relates to a 2 dimensional one.

What $t$ do you use for "the" state? By "expanding" you mean something like this?

$$\int\limits_{\mathbb{R}^2\times \{R\}} |x\rangle\langle x|\psi(t)\rangle dx$$

This quantity contains the same information as the function

$$g(x_1,x_2) = \psi(t,x_1,x_2,R)$$

That doesn't work because the left side doesn't have $t$ and right side does. How many times do I need to point out the same thing? You cannot solve this problem by choosing a new notation for the right side.

But if one is simply going to write down phenomenological equations without using an explicit model of the interaction between the particle and the screen, then why not do what tom.stoer and Nugatory have been suggesting, ie: ρ(x₁,x₂) = c |ψ(x₁,x₂,x₃=R,t=Rm/p₃)|², where m is the mass of the particle, and c is a normalization constant so that ∫ρdx₁dx₂=1 ? The approximation is valid if the motion along x₃ is "classical", which means that the de Broglie wavelength should be much smaller than the distance between the slit and the screen.

Oh! You substituted $t=\frac{Rm}{p_3}$. Well it is true that with this additional information the idea starts to mean something, and now it can be applied. Clearly this is not equivalent to my ad hoc idea. In my formulas the probability current would smoothly get absorbed to the plate during a long time interval, which looks qualitatively different from simply restricting the wave function on the plate at some singular average hitting time. In other words my idea is less ad hoc, while this "averate hitting time" is more ad hoc. Also, it is not impossible that a better justification could be found for my formulas eventually, reducing the ad hocness. The probability current in three dimensional space is unique after all, and I don't believe that you can device anything else than what I just proposed, that would conserve the total probability similarly.

 

[bhobba]

What $t$ do you use for "the" state? By "expanding" you means something like this?

Since t is irrelevant to the problem it's ignored - we are only interested in what happens at the screen - not when it happens.

Back off a bit.

To understand what the state is see the link I posted previously - see post 137:
https://www.physicsforums.com/showthr...=763139&page=8

Its simply something that helps us calculate outcome probabilities.

Now for pure states they can be mapped to a vector space.

I am sure you have studied linear algebra and know vectors can be expanded in terms of many different basis, and its quite arbitrary. The only reason for choosing one over the other is utility - they are all equally valid.

What you did is expand the state in terms of 3d position eigenstates - by definition that is called a wave-function.

But since the observation is a 2d screen so that's a very poor choice - and why you run into problems.

Expand it in terms of the actual observation - the position on the screen.

So my question to you is why did you use a 3d representation of the state?

I suspect you think the wavefunction is the state - it isn't.

How many times do I need to point out the same thing? You cannot solve this problem by choosing a new notation for the right side.

You introduce irrelevancies like a time dependence in the state, when such is not required, you expand the state out in positions over all space when we have a two dimensional screen and when your choice proves poor in analysing the situation you claim people do not understand what going on.

Mate - I think you need to understand QM a lot better.

Thanks
Bill

 

[jostpuur]

We are talking about a process where a wave packet first flyes in three dimensional space freely, and then hits a detector plate. The ordinary three dimensional spatial representation is the most suitable one to describe the flying wave packet, and the time parameter is very much needed.

 

[bhobba]

We are talking about a process where a wave packet first flyes in three dimensional space freely, and then hits a detector plate. The ordinary three dimensional spatial representation is the most suitable one to describe the flying wave packet, and the time parameter is very much needed.

That's not what we are talking about.

We are talking about a quantum object emitted in a certain state and observed at the screen. What happens at the screen is the only relevancy.

How we model it is arbitrary. You chose a bad way in the form of 3d wave packets.

I chose a way that is much simpler and with less baggage.

You can probably analyse it your way, but it is much more complicated for zero gain eg you would need to model the screen as some kind of two dimensional Dirac Delta function so that you actually get a non zero probability.

Thanks
Bill

 

[tom.stoer]

I agree with Bill, even so I propose a different method.

In order to understand the pattern on the screen you have to look at many wave pakets over a very long time, therefore you are not observing a single wave paket and the probability where and when it will hit the screen, but your are observing wave pakets which have been absorbed with certainty within a very long time; under this condition you then evaluate the probability where the wave pakets hit the screen. This eliminates the time variable.