Position Eigenstates: Measurement & Imperfect Performance

[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

 

[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&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.

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?

<br /> \int\limits_{\mathbb{R}^2\times \{R\}} |x\rangle\langle x|\psi(t)\rangle dx<br />

This quantity contains the same information as the function

<br /> g(x_1,x_2) = \psi(t,x_1,x_2,R)<br />

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.

 

[jostpuur]

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

The idea that you can expand the state as a representation on the two dimensional screen doesn't work because of the fact

<br /> \int\limits_{\mathbb{R}^2\times \{R\}} dx |x\rangle\langle x| \neq \textrm{id}<br />

 

[jostpuur]

This eliminates the time variable.

It would be clearer to state that we wish to eliminate the time variable in the solution process, but how it is accomplished, is not trivial.

I'll clarify on the original problem: If the number of particles, which have hit the screen, is N and is very large, obviously we assume that the particles were all sent towards the screen with identical initial states. So if the hitting points are (x_1,y_1),(x_2,y_2)\ldots, (x_N,y_N), the probability density for this 2N dimensional random vector has the form

<br /> \rho\big((x_n,y_n)_{1\leq n\leq N}\big) = \prod_{n=1}^N f(x_n,y_n)<br />

with some function f. So it is sufficient to solve the function f for one particle, and then the remaining problem is this:

What we have in the beginning: An initial state of a wave packet \psi(0,x,y,z) at time t=0, the Schrödinger equation for time evolution, and the location of the screen z=R.

What we want in the end: A two dimensional probability density f(x,y) which does not involve a time parameter.

So in other words we want to eliminate the time parameter t when producing the solution, but in that process we will have to deal with it.

 

[tom.stoer]

So in other words we want to eliminate the time parameter t when producing the solution, but in that process we will have to deal with it.

Yes, this is what we want to achieve

The simplest idea is to provide a rule which
1) eliminates time
2) provides predictions in agreement with experiment

I think we provided a couple of ideas ;-)

But it seems that you are not satisfied and you want to understand in more detail how this rule can be justified. Then I think you have to do the following: model the screen as a many-body system or some "effective Hamiltonian"; and solve e.g. the equations of time-dependent scattering theory

 

[atyy]

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.

Actually the substitution t=Rm/p₃ is a wave packet sort of idea, with the idea that packet is localized enough that the time of detection is sharply peaked. So I think this approximation will fail if the longitudinal momentum along the slit-screen axis is small comapared to transverse momentum, and if the slit-screen distance is small compared to the de Broglie wavelength.

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.

Yes, the method of substituting t=Rm/p₃ is ad hoc and only works because in some of these experiments the time at which the position of the atom is registered on the screen is sharply peaked, the screen is far enough from the slit, and the longitudinal momentum along the slit-screen axis is much larger than the transverse momentum. And yes, these are all "physical intuition" considerations, so it is hard to give a quantitative analysis of how the approximation fails as the screen is moved nearer and nearer the slit.

Although the "physical intuition" is ad hoc, in some cases (not above), one can show in simpler conditons that it is a well-controlled approximation. For example, shows that the centroid of the Gaussian packet is well described by classical equations.

I do think your method can eventually be justified, but one should start with an effective interaction between the particle and the screen as tom.stoer and I have mentioned (a lot of work!)

 

[tom.stoer]

I do think your method can eventually be justified, but one should start with an effective interaction between the particle and the screen as tom.stoer and I have mentioned (a lot of work!)

Note that with time-dep. scattering theory based on wave pakets ψ(x) you would essentially use the solution of the stationary problem u(x) - with double slit but w/o any screen - to expand the scattering states; so through the backdoor you introduce the well-known wave functions u(x).

The next step would then be to introduce an interaction of the wave paket ψ(x) - i.e. the stationary states u(x) - with the screen via some potential V (which models the absorption and/or scattering within the screen). Neglecting the microstructure of the screen, i.e. assuming V to be xy-independent, the potential looks like V(x,y,z) = V χ(R,R+ε) where χ is the characteristic function of the (small) z-interval [R,R+ε].

From perturbation theory we know that we have to evaluate integrals like

\langle \psi | \hat{V} | \psi \rangle \to V \epsilon \int_{\mathbb{R}^2}dx\,dy\,|u(R)|^2

So it becomes clear the absorption of the particle by the screen involves something like the probability density of the stationary problem evaluated on the screen.

In a last step on may ask for the absorption of the particle in some area [X,X+a] * [Y,Y+b] of the screen. In that case we replace the above mentioned integral by

\int_{[X,X+a] \times [Y,Y+b] }dx\,dy\,|u(R)|^2

So I think even w/o a detailed microscopic model it becomes clear that the probability density derived from ψ and evaluated on the screen becomes important.

 

[atyy]

@tom.stoer, I agree. What's a good potential? I thought to simplify to a 2D problem, and a 1D screen. So the question would be how to get ρ(x) from ψ(x,y,t). At first I thought to just put a negative step at one end of an square well with infinitely high walls, but I think it will only localize the particle along one direction, not both.

 

[tom.stoer]

I think you must not localize the particle, but you need to absorb it. So you need a constant potential with an imaginary term. Unfortunately I am not familiar with these models and I do not know which formulas from scattering theory remain valid.

Another idea would be a square well with a second particle (or an ensemble of non-interacting particles) having exactly the correct energy levels to absorb the incoming wave paket.

Anyway, this is a new idea and it will take some time to work out the details.

 

[vanhees71]

I'd use the probability current. For a Schroedinger wave fct. it's
\vec{j}=\frac{\hbar}{2m\mathrm{i}}(\psi^*\vec{\nabla}\psi-\psi \vec{\nabla}\psi^*).
The pattern on the screen is then
P=\int_{\mathbb{R}} \mathrm{d} t \vec{n} \cdot \vec{j}.
Here \vec{n} is the unit-normal vector of the surface.

 

[Jano L.]

I'd use the probability current. For a Schroedinger wave fct. it's
\vec{j}=\frac{\hbar}{2m\mathrm{i}}(\psi^*\vec{\nabla}\psi-\psi \vec{\nabla}\psi^*).
The pattern on the screen is then
P=\int_{\mathbb{R}} \mathrm{d} t \vec{n} \cdot \vec{j}.
Here \vec{n} is the unit-normal vector of the surface.

According to the Born interpretation, we should use $|\psi|^2$ to get probability. This gives non-zero probability density even for cases when $\mathbf j = 0$ and $\psi \neq 0$ as it happens, for example, for the eigenfunction of the Hamiltonian in the hydrogen atom model.

Using $\mathbf j$ to get probabilities does not seem to be justified by the Born interpretation.

 

[jostpuur]

Jano L., vanhees71 is talking about an answer to a problem. You cannot understand the answer, if you don't understand the problem first. In order to learn what the problem is, you must go through the thread. I explained the problem in posts #20, #25 and #72. Then I also proposed a solution in post #57. The solution is essentially the same what vanhees71 is talking about.

 

[bhobba]

The idea that you can expand the state as a representation on the two dimensional screen doesn't work because of the fact

<br /> \int\limits_{\mathbb{R}^2\times \{R\}} dx |x\rangle\langle x| \neq \textrm{id}<br />

Come again.

That makes zero sense.

That bit of math needs fleshing out.

Its a fundamental property that you can expand the state in any compete set of eigenvectors which, in this problem, since it is eventually observed by the screen, is the eigenvectors of the observable of the screen.

You seem to be missing the point.

You have done nothing wrong - you simply have made a poor choice in analysing the problem that is making it more difficult than necessary.

I suspect that's because you do not understand that the wave-function is an arbitrary representation and you are free to choose any you like.

Vanhees has posted a possible way of doing it via your method - but why you want to make your life hard beats me.

Thanks
Bill

 

[atyy]

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.

I'd use the probability current. For a Schroedinger wave fct. it's
\vec{j}=\frac{\hbar}{2m\mathrm{i}}(\psi^*\vec{\nabla}\psi-\psi \vec{\nabla}\psi^*).
The pattern on the screen is then
P=\int_{\mathbb{R}} \mathrm{d} t \vec{n} \cdot \vec{j}.
Here \vec{n} is the unit-normal vector of the surface.

So these are identical if we take jostpuur's $\partial_t\phi$ the same as vanhees71's $\vec{n} \cdot \vec{j}$ ? Seeing it written in vanhees71's form I understand jostpuur's solution better, and yes it seems correct, complete, and a very nice method.

 

[jostpuur]

Well, guys, how do you accomplish the same with relativistic particles, which don't have known probability currents?

 

[vanhees71]

I still don't know, what \phi might be. There is no such thing as two state representations as in posting #54.

Due to unitarity of time evolution, what's for sure is the continuity equation. Defining the "probability density" (i.e., the probability distribution) by
\rho(t,\vec{x})=|\psi(t,\vec{x})|^2
and
\vec{j}(t,\vec{x})=\frac{\hbar}{2 m \mathrm{i}} (\psi^* \vec{\nabla} \psi - \psi \vec{\nabla} \psi),
the continuity equation holds
\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0,
which ensures that the time evolution conserves the total probability,
\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho(t,\vec{x})=1=\text{const},
as it must be.

The proof is simple. For a single non-relativistic particle you have the Schrödinger equation
\mathrm{i} \hbar \partial_t \psi=-\frac{\hbar^2}{2m} \Delta \psi+V(\vec{x}) \psi.
Then you get
\vec{\nabla} \cdot \vec{j}=\frac{\hbar}{2 m \mathrm{i}} ( \psi^* \Delta \psi-\psi \Delta \psi^*).
Now from the Schrödinger equation and its conjugate complex you get
\frac{\hbar}{2m \mathrm{i}} \Delta \psi=-\mathrm{i} \partial_t \psi +\frac{V}{\hbar} \psi, \quad \frac{\hbar}{2m \mathrm{i}} \Delta \psi^*=+\mathrm{i} \partial_t \psi^* +\frac{V}{\hbar} \psi^*.
Plugging this into the above equation, you get
\vec{\nabla} \cdot \vec{j}=-(\psi^* \partial_t \psi+\psi \partial_t \psi^*)=-\partial_t (\psi^* \psi)=-\partial_t \rho,
and this is just the continuity equation.

So what I assumed is that the screen registers all particles running towards it. Thus the time integral of \vec{n} \cdot \vec{j} over time gives the probability distribution for particles hitting the screen.

This is the same in relativistic theory. The only difference is that all this only makes sense for asymptotic free particles, i.e., for particles far away from the screen. If this condition is not fulfilled, i.e., if the screen is to close to the slits you have quantum fields interacting with the slits, and then an interpretation of particle numbers, their densities, or particle-number currents is difficult (if not impossible)!

 

[tom.stoer]

Fine, makes sense!

This is by far the most simple restriction to the screen surface b/c it does not require a specific interaction with the screen.

 

[jostpuur]

I still don't know, what \phi might be.

Well watch carefully. I said that it is something that satisfies

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

However, I had also clearly stated that the detector screen is the plane x_3=R, and the particle is coming from the x_3&lt;R side. So obviously we can write the same thing as

<br /> \partial_t\phi(t,x) = -n\cdot \frac{\hbar}{m}\textrm{Im}\big(\psi^*(t,x)\nabla \psi(t,x)\big)<br />

with

<br /> n=(0,0,-1)<br />

and this is how the formula is generalized for other normal vectors n too. Then by using the formula

<br /> \textrm{Im}(z) = -\frac{i}{2}(z - z^*)<br />

we can write the same thing as

<br /> \partial_t \phi(t,x) = n\cdot\frac{i\hbar}{2m}\big(\psi^*(t,x)\nabla\psi(t,x) - \psi(t,x)\nabla\psi^*(t,x)\big)<br />

and this is equivalent to

<br /> \phi(t,x) = \phi(0,x) - \int\limits_0^t n\cdot j(t&#039;,x)dt&#039; <br />

with

<br /> j(t&#039;,x) = -\frac{i\hbar}{2m}\big(\psi^*(t&#039;,x)\nabla\psi(t&#039;,x) - \psi(t&#039;,x)\nabla\psi^*(t&#039;,x)\big)<br />

So that is what \phi is based on what I wrote in #57.

 

[vanhees71]

Ok, that was a bit unclear to me. So what you wrote is basically the same thing as I did. Your \phi is simply the probability distribution, i.e., |\psi(t,\vec{x})|^2. If the screen is in the xy plane at z=z_0 after the slits, then of course \vec{n}=\vec{e}_z, and the interference pattern is described by
P(x,y)=\int_{-\infty}^{\infty} \mathrm{d} t j_z(t,x,y,z_0).

 

[jostpuur]

This is the same in relativistic theory. The only difference is that...

The word "only" is what doesn't look right

The experimental set up where relativistic particles are shot at a screen does exist though, and the final answer, which is a two dimensional density

<br /> f(x_1,x_2)<br />

exists too. It makes sense to inquire how to compute this, and I believe nobody knows the answer at this point.

 

[vanhees71]

You calculate it as you calculate any scattering cross section, using the S-matrix. Where is the problem? In fact, you always observe asymptotic free states, if you can talk sensibly about particles. So there is no conceptional problem. The only thing that changes is the expression for the current. Usually what you measure is the energy density (photons) or the charge density (charged particles). Then you can use the corresponding four-currents that are conserved for free particles, and there is no problem.

 

[jostpuur]

Nobody knows how to describe probability currents for relativistic particles, and it may be that probability currents don't even exists for relativistic particles, so certainly that is one major problem. If you don't see any problems ahead, how about you show what is the time evolution of the quantity that you are going integrate with respect to time?