Position Eigenstates: Measurement & Imperfect Performance

[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

$$\int\limits_{\mathbb{R}^2\times \{R\}} dx |x\rangle\langle x| \neq \textrm{id}$$

 

[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

$$\rho\big((x_n,y_n)_{1\leq n\leq N}\big) = \prod_{n=1}^N f(x_n,y_n)$$

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

$$\int\limits_{\mathbb{R}^2\times \{R\}} dx |x\rangle\langle x| \neq \textrm{id}$$

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:

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

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.

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

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

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

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

with

$$n=(0,0,-1)$$

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

$$\textrm{Im}(z) = -\frac{i}{2}(z - z^*)$$

we can write the same thing as

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

and this is equivalent to

$$\phi(t,x) = \phi(0,x) - \int\limits_0^t n\cdot j(t',x)dt'$$

with

$$j(t',x) = -\frac{i\hbar}{2m}\big(\psi^*(t',x)\nabla\psi(t',x) - \psi(t',x)\nabla\psi^*(t',x)\big)$$

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

$$f(x_1,x_2)$$

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?