What experiment demonstrates Heisenberg's uncertainty principle?

[vanhees71]

Great paper. I'll have a close look at it. I must say, I can understand now, why the diffraction is not treated in standard textbooks. I couldn't solve the appropriate Helmholtz equation with the exact boundary conditions exactly, even not for a circular apperture. The best, I can come up with is the standard Sommerfeld-corrected approximate Kirchhoff solution, which should be a good approximation in the limit, where the de Broglie wavelength is small compared to the dimensions of the apperture, using the Green's function for the infinite plane. At the moment I'm at the process to get some numerical calculations done :-).

Of course the free-particle wave packet can be described pretty well in a classical approximation. Since there's no force, the leading-order classical approximations give the classical uniform motion: Ehrenfest's theorem together with the fact that the equations of motion in the Heisenberg picture are linear in position and momentum operators and are of the classical form; taking the averages gives then the classical equation of motion also for the expectation values.

 

[atyy]

Here are some other related papers, but I think they all make use of the approximation of the longitudinal motion being classical, which is hard to justify completely from purely quantum postulates.

http://arxiv.org/abs/quant-ph/0407245
http://arxiv.org/abs/0804.3006
http://arxiv.org/abs/0902.0234

 

[TeethWhitener]

I'm having a hard time understanding why the use of a plane wave is a big deal. Since the Schrodinger equation is linear, any superposition of plane waves will be a solution. And you can make a Gaussian wavepacket from a superposition of plane waves (this is in fact the definition of a Fourier transform). So I'm not seeing how the time-dependent behavior of a wavepacket would be qualitatively different from that of a plane wave.

 

[TrickyDicky]

Before the screen the wave-function is approximated as being in an eigenstate of momentum and energy - it really can't be because its wave-function would extend through all space - but that's what you do in modelling. After the screen the thin slits are modeled by having the position as exact and using a Dirac delta function - of course it's not exactly like that - but modelling wise its what's done to get a mathematical grip on the situation.

I have no problem with approximate models, that's physics. But one must know when a certain approximation is licit and when it is not. Zapper mentioned the use of plane waves in quantum tunnelling, but in that case the time-independent SE allows one not to think about any before or after the barrier, that acts as a spatial boundary condition, one doesn't need to have a localized electron at any given time. The HUP allows you to have a state with acertain probability in a classically forbidden zone.
The Marcella paper is different, it is a time-dependent model of an electron with no uncertainty about momentum(plane wave) interacting with the slit at some point in time modeled as you say with exact position(Dirac delta). When your approximation demands to model a quantum phenomenon throwing away the HUP I believe something's not right.

I'm having a hard time understanding why the use of a plane wave is a big deal. Since the Schrodinger equation is linear, any superposition of plane waves will be a solution. And you can make a Gaussian wavepacket from a superposition of plane waves (this is in fact the definition of a Fourier transform). So I'm not seeing how the time-dependent behavior of a wavepacket would be qualitatively different from that of a plane wave.

I think it is not granted in this particular case, see above. But I suspect it can't be done with wave packets either, there must be a reason why nobody tries to do this rigorously after all. There are theorems(i.e.Revisiting the First Postulate of Quantum Mechanics:Invariance and Physical Reality by Ronde, Massri) that point to problems for the dynamics of QM for quantum systems with Hilbert spaces of more than 2 dimensions.
Maybe some QFT scattering approach can do it.

 

[bhobba]

The Marcella paper is different, it is a time-dependent model of an electron with no uncertainty about momentum(plane wave) interacting with the slit at some point in time modeled as you say with exact position(Dirac delta). When your approximation demands to model a quantum phenomenon throwing away the HUP I believe something's not right.

Throwing away the HUP? That a position measurement as done by a slit has an unknown momentum after is the HUP.

It simple. Let's consider electrons - photons are more problematical. It leaves the source with a definite momentum and energy (NOT exact of course since that would mean a wave-function over all space - but is obviously true to a very good approximation). It passes through the slit or gets absorbed. If it passes through the slit then its position is known, so from the HUP its momentum is unknown. But the energy is the same which means the magnitude of the momentum is unchanged hence its direction is unknown and its scattered.

When both slits are open the state is obviously a superposition of the state with a single slit open - that is the principle of superposition. The symmetry of the situation leads to equation 9 in that paper - and - wonder of wonders - you get an interference pattern.

There really isn't much to it. Of course approximations are made - but that's done all the time in modelling. I have zero problem with people pointing that out - it must really be a single wave function etc etc. But it doesn't invalidate the analysis any more than for example modelling a ball rolling down an incline by a point invalidates that.

Thanks
Bill

 

[bhobba]

I'm having a hard time understanding why the use of a plane wave is a big deal. Since the Schrodinger equation is linear, any superposition of plane waves will be a solution. And you can make a Gaussian wavepacket from a superposition of plane waves (this is in fact the definition of a Fourier transform). So I'm not seeing how the time-dependent behavior of a wavepacket would be qualitatively different from that of a plane wave.

The issue is that plane waves are not square integrable hence can't be a valid solution to the Schrödinger equation. We have to make a few reasonableness assumptions to mathematically get a handle on it. While true its done all the time, often without even saying that's what's being done eg:
http://www.physics.ox.ac.uk/Users/smithb/website/coursenotes/qi/QILectureNotes3.pdf

See the free particle solution. It's not square integrable so its not valid - but no-one really worries about it.

Thanks
Bill

 

[TrickyDicky]

Throwing away the HUP? That a position measurement as done by a slit has an unknown momentum after is the HUP.

It simple. Let's consider electrons - photons are more problematical. It leaves the source with a definite momentum and energy (NOT exact of course since that would mean a wave-function over all space - but is obviously true to a very good approximation). It passes through the slit or gets absorbed. If it passes through the slit then its position is known, so from the HUP its momentum is unknown. But the energy is the same which means the magnitude of the momentum is unchanged hence its direction is unknown and its scattered.

When both slits are open the state is obviously a superposition of the state with a single slit open - that is the principle of superposition. The symmetry of the situation leads to equation 9 in that paper - and - wonder of wonders - you get an interference pattern.

There really isn't much to it. Of course approximations are made - but that's done all the time in modelling. I have zero problem with people pointing that out - it must really be a single wave function etc etc. But it doesn't invalidate the analysis any more than for example modelling a ball rolling down an incline by a point invalidates that.

Thanks
Bill

If you read the paper you linked in post #15, which I read after posting above, you might understand what's wrong with your analysis: Marcella's paper makes a purely classical approximation to a quantum problem, the "after" is the classical optics treatment, so it is quite reasonable to say there is no HUP in his treatment, as vanhees told you: "look at what he does, not what he says", the HUP doesn't show up magically just by invoking it because this is QM, right?, you must justify it with the math and the physics in a specific problem, not just imply it by using Dirac formalism.
So again approximations are fine, when they don't directly affect the essence of what you want to model, in this case the quantum nature of the interference phenomenon. This is so basic that I cannot imagine how it can scape anyone.

 

[TrickyDicky]

The issue is that plane waves are not square integrable hence can't be a valid solution to the Schrödinger equation. We have to make a few reasonableness assumptions to mathematically get a handle on it. While true its done all the time, often without even saying that's what's being done eg:
http://www.physics.ox.ac.uk/Users/smithb/website/coursenotes/qi/QILectureNotes3.pdf

See the free particle solution. It's not square integrable so its not valid - but no-one really worries about it.

Thanks
Bill

It is enphasized in that section about free particles!, you either use the time-independent SE like it's done in tunnelling etc, or if you are going to deal with a time-dependent problem and use the traveling plane wave like we are here you can never localize the electron with a slit, only in the well known classical macroscopic optics case you can.

 

[bhobba]

it is quite reasonable to say there is no HUP in his treatment,

It obviously does - as I explained in my post. Its why the particle is scattered by the slit.

Thanks
Bill

 

[TrickyDicky]

It obviously does - as I explained in my post. Its why the particle is scattered by the slit.

Thanks
Bill

It is claimed in your post rather than explained, it is explained in the lecture notes you linked above: Plane waves are not proper solutions and are not valid approximations for position dynamical wave functions(you need wave packets there), they are perfectly valid approximations for energy solutions of the TISE, that's why no one worries about it in tunnelling kind of problems.

 

[bhobba]

It is claimed in your post rather than explained,

I carefully explained it - reread it if you don't get it. But to repeat it - particles that go through the slit have been subjected to a position measurement. By the HUP there momentum is now unknown. If you still don't get it - nothing much more I can do.

Thanks
Bill

 

[TrickyDicky]

particles that go through the slit have been subjected to a position measurement.

An exact position (Dirac delta point) measurement cannot be performed to a wave function.

By the HUP there momentum is now unknown.

By the HUP that you put by hand there is no state anymore either :), besides if momentum is unknown one cannot use the y component, momentum is either known or not, the rest is the classical optics treatment, which is what Marcella does disguised in Dirac formalism.

 

[vanhees71]

Yes, it's completely right what TrickyDicky wrote. Take a good standard textbook of quantum theory, where scattering is described correctly. Then you'll see that you need wave packets to define the cross section. In the generalized (!) momentum eigenstates, aka plane waves, you get
$$S_{fi}=\delta_{fi} - (2 \pi)^4 \mathrm{i} T_{fi} \delta^{(4)}(p_f-p_i).$$
The transition probability is the modulus squared of the amplitude, which naively is $S_{fi}$. Of course, this doesn't make sense, but you first have to fold it with the incoming wave packets take the square and then take the limit to vanishing momentum spread. The final result is that there's only one energy-momentum conserving $\delta^{(4)}$ distribution in the transition probability density, which can then be meaningfully integrated out.

Admittedly, here the whole issue is complicated also by the problem of how to define asymptotic free states, where you have to use an appropriate switching procedure. If I remember right, a good source for the discussion of this is Messiah's classical quantum mechanics text on this issue. For the relativistic case, it's nicely dealt with in Peskin&Schroeder.

Of course, there are shortcuts in the literature like in Landau/Lifshitz vol. 4, using a "box regularization", where the momenta become descrete, and there's no problem in squaring the S-matrix elements to get transition probabability rates. After deviding over the finite four-volume and then taking the limit of the four-volume to $\infty$ you get the same result as with the proper procedure with the wave packets. However, the latter is much more physical, although a bit more complicated mathematics wise. When discussing the physics, one should use the wave-packet approach.

 

[TeethWhitener]

The issue is that plane waves are not square integrable hence can't be a valid solution to the Schrödinger equation.

I think you meant to say that plane waves aren't a valid physical solution to the Schrodinger equation. But that's my point. You can sum them in a way that they are a valid physical solution, namely:$$\psi (x) = \int_{-\infty}^{\infty}f(k)e^{ikx}dk$$ with a proper weighting function $f(k)$ such that square integrability holds at all times for this solution. But since this solution is built up from plane waves, then solving the time-dependent behavior of a single plane wave going through a slit should give you an idea of how the time dependent behavior of a linear combination of plane waves will act.

 

[bhobba]

An exact position (Dirac delta point) measurement cannot be performed to a wave function.

That's the modelling bit. Its used all the time in applied math eg short electrical impulses are modeled by a Dirac delta function - that the slit is of very small width is modeled similarly.

Thanks
Bill

 

[TrickyDicky]

That's the modelling bit. Its used all the time in applied math eg short electrical impulses are modeled by a Dirac delta function - that the slit is of very small width is modeled similarly.

Thanks
Bill

That's fine for electrical impulses, now go ahead and try that modelling bit to approximate Heisenberg inequality for position and momentum, you are welcome to try Δx=0 or Δp=0 and see what you get. It's no big deal, just modelling...

 

[Quds Akbar]

Theoretically, they take every possible path connecting the two points and they can exist anywhere during the trip. It is all based on probability.