Does Heisenberg apply to a collapsed wavefunction?

[T'Pau]

Hi,

I'd like to argue Heisenberg doesn't apply to a collapsed wave...
I always interpretted the Heisenberg Uncertainty Principle (HUP) as follows:

1. HUP is *not* about measurement problems, it is fundamental
2. When (f.i.) an electron is a wave: HUP applies. The electron really *has* uncertain impulse/position
3. When the wave collapses, f.i. in a collission: the electron behaves like a particle and *does* have impulse/position.
4. *Immediately* after the collapse: the electron is a wave again, with uncertain impulse/position.

Point number 3, I learned, is controversial. I'd like to argue for it using an experiment.

I learned: HUP applies to photons as well (is that correct?).
Say, you perform a single-slit-experiment with a laser, say λ = 500 nm. (λ is well defined)
I use a very very narrow slit, λ << slit-width (no interference pattern behind the slit, only diffraction / bending of the laser light).

When you perform this experiment, on the screen you notice the same lightcolor as the laser emmited: is agreeable that behind the slit λ is still ≈ 500 nm? → Δλ < 1 nm?

I calculate the impulse-difference between λ = 500 nm and λ = 501 nm, → Δp ≈ 2.6 * 10^-30.

Δx Δp ≥ h/4π → Δx ≥ 20 μm.

But my slit-width is much smaller than that: it is < 0.5 μm...

So when you perform this experiment, for instance one photon at a time, everytime you see the screen light up, you know that at a very short instance just before that the photon:
a. *had* Δp ≈ 2.6 * 10^-30
b. *had* Δx < 0.5 μm.

This is better than what HUP allows for...

Notice: I wasn't able to predict where the screen would light up. My measurement doesn't allow me to make any predictions. But what I do know, is that for a very short moment in the past HUP has been broken...

I'm concerned with 'reality'. When something is a wave, it really *has* uncertain impulse/position.
But when this wave collapses, is that still true?

To put it otherwise: how can two electrons collide, when either position or impulse (or both) *is* uncertain?

I hope someone can help me out.

Paul

 

[bhobba]

I'm concerned with 'reality'. When something is a wave, it really *has* uncertain impulse/position.

QM does not ascribe any property until its measured to have that property.

It is never a wave - its a quantum particle: Simply ask yourself - waves of what?
https://www.physicsforums.com/showthread.php?t=511178
'In our ordinary world, “wave” and “particle” behavior are two different and opposite characteristics. It is difficult for us to think that they can be one of the same. Is light a particle or a wave? The simple, naïve answer to that is “both” or “neither”.

Light, or photon, was never defined as a “particle” the way we normally define a particle. Light is not defined to have a definite boundary in space like a ping-pong ball, or a grain of sand. Instead, light is defined as having quanta of energy. So the discreteness is not defined as discrete object in space, but rather in the energy it can carry. Already, this is not your regular “particle”, and should not be confused as such.

Secondly, in quantum mechanics, the description and properties of light has only ONE, single, consistent formulation, not two. This formulation (be it via the ordinary Schrodinger equation, or the more complex Quantum Electrodynamics or QED), describes ALL characteristics of light – both the wave-like behavior and the particle-like behavior. Unlike classical physics, quantum mechanics does not need to switch gears to describe the wave-like and particle-like observations. This is all accomplished by one consistent theory.

So there is no duality – at least not within quantum mechanics. We still use the “duality” description of light when we try to describe light to laymen because wave and particle are behavior most people are familiar with. However, it doesn’t mean that in physics, or in the working of physicists, such a duality has any significance.'

My gut tells me it might be wise for you to 'unlearn' the usual treatments found in beginning texts and popularizations and start with its actual conceptual core:
http://www.scottaaronson.com/democritus/lec9.html

If you are interested in the foundations of QM and seeing it developed in a proper axiomatic fashion while not skirting the interpretational issues THE book to get is Ballentine - Quantum Mechanics - A Modern Development:
https://www.amazon.com/dp/9810241054/?tag=pfamazon01-20

Thanks
Bill

 

[WannabeNewton]

The HUP is a constraint on the variances of observables. It comes right out of the inability of certain observables to commute. As such it applies to any and all wavefunctions when you're taking the averages of observables for a given wavefunction.

 

[ZapperZ]

Hi,

I'd like to argue Heisenberg doesn't apply to a collapsed wave...

I measure L_z. Do you think the values for L_x and L_y are also determined?

Zz.

 

[Nugatory]

So when you perform this experiment, for instance one photon at a time, everytime you see the screen light up, you know that at a very short instance just before that the photon:
a. *had* Δp ≈ 2.6 * 10^-30
b. *had* Δx < 0.5 μm.

You know the exact moment that the screen flashed, but that does not tell you an exact moment that the photon passed through the slit, nor does it tell you how long the photon took to pass through the slit. Thus, you still haven't localized the photon.
(Note also that I can correctly say "X does not tell you Y" even if Y is something completely meaningless - I am not suggesting that "the moment the photon passed through the slit" is a meaningful concept when there was no detector there).

You may want to approach the uncertainty principle not as Heisenberg originally described it, but as the more modern statistical interpretations approach it (Bhobba recommends Ballentine, I second it). The traditional approach invites all sorts of confusion.

You'll also want to understand what's behind ZapperZ's terse but spot-on rhetorical question about the components of angular momentum. That's essential to understanding what is and is not "definite" after a "collapse ofthe wave function".

 

[atyy]

Yes, the Heisenberg uncertainty principle applies to any wave function. For a wave to have a "wavelength", it must have many periods stretched out in space. If the wave is concentrated at a spot, then it doesn't have "a" wavelength, rather it is a superposition of waves of many wavelengths.

 

[Jilang]

I hope someone can help me out.

Paul

Hi Paul, it looks to me like you might confusing uncertainty in the wavelength (which is measured along the direction of motion) with the uncertainty in the position or width of the slit (which is at right angles to it). If you consider the components separately perhaps that will help.

Also I think you meant to say that for diffraction the wavelength is >> than the slit width.

 

[naima]

Hi Paul, it looks to me like you might confusing uncertainty in the wavelength (which is measured along the direction of motion) with the uncertainty in the position or width of the slit (which is at right angles to it).

I agree.
if Δx = 0.5 μm the wave function is a rectangular function rectf(x) which is null outside the slit
The momentum wave function will give you Δp:
the Fourier transform of rectf(x) gives you the cardinal sine sinc(p) and you can take Δp as the distance between the first zeros of this function.
You will see that you does get what you wrote.

 

[T'Pau]

Do I get it now?

Thanks everyone for helping me out.

I should have remembered: HUP is not about Δx times Δp_y, it's about Δx times Δp_x. So my experiment did not prove HUP wrong at all.

If I understand it correctly now, one of my miconceptions was that I thougt that a collapsed wave is not a wave. Is this true? → a collapsed wave is also a wave?

I'd like to check if I got it right this time.

Say, there is a totally free electron with zero uncertainty in impulse. The wavefunction of this electron is stretched out from minus infinity to plus infinity (it is as long as the universe).

Now, I trie to measure the position of this electron by shooting photons at it. Which is of course practically impossible: what are my chances of finding this electron if it's wave is that long? But say, after thousands of years, I get lucky: one of my photons strikes this electron.

The moment the photon strikes, the wave of the electron collapses. It becomes a new wave, say a rectf-function with Δx = 1 μm (I don't know the cross section of photons, is 1 μm agreeable?). This new wave does not have definite impuls anymore.

Just before I struck it: there was wave 1. Wave 1 has definite impulse, zero certainty in position.
I strike it: it becomes a new wave, wave 2. Wave 2 has uncertainty in impulse and uncertainty in position. The uncertainty in position is (say) Δx = 1 μm.
Just after I struck it: it becomes wave 3, or does it stay wave 2? Does wave 3 'keep' uncertainty in position of Δx = 1 μm? Or is there another wave 3, which just satisfies HUP?

(I know: I'm writing everywhere: 'it *is* a wave', I should write: 'I describe it as a wavefunction'.)

Thanks for all the help!

Paul

 

[bhobba]

one of my miconceptions was that I thougt that a collapsed wave is not a wave. Is this true? → a collapsed wave is also a wave?

Precisely why do you think waves are involved at all?

I know: I'm writing everywhere: 'it *is* a wave', I should write: 'I describe it as a wavefunction'.)

A wavefunction is a state expanded in terms of position eigenfunctions. Sometimes it's mathematically like waves in other areas of physics - but it's NOT a wave.

Thanks
Bill

 

[naima]

Just before I struck it: there was wave 1. Wave 1 has definite impulse, zero certainty in position.
I strike it: it becomes a new wave, wave 2. Wave 2 has uncertainty in impulse and uncertainty in position. The uncertainty in position is (say) Δx = 1 μm.
Just after I struck it: it becomes wave 3, or does it stay wave 2? Does wave 3 'keep' uncertainty in position of Δx = 1 μm? Or is there another wave 3, which just satisfies HUP?

Before the photon strikes the atom, the uncertainty of the atom's momentum is described py a probability function P1(k). We have not to think that a photon is something with a well known momentum it has also a probability function P2(k).
After the interaction the atom has a new probability function P3(k)
How can we get P3?
let us find the probability P3(k0).
if the photon had a k1 momentum we would get k0 if the atom had k0 - k1 as momentum
this has a P1(k0 - k1) P(k1) probability.
this is also true for other values of the atom's momentum
We must sum all these probabilities to find P3(k0)
$$P3(k0) = \int P2(k0 -k) P1(k) dk$$
the result is the convolution of P1 and P2

 

[ZapperZ]

Thanks everyone for helping me out.

I should have remembered: HUP is not about Δx times Δp_y, it's about Δx times Δp_x. So my experiment did not prove HUP wrong at all.

If I understand it correctly now, one of my miconceptions was that I thougt that a collapsed wave is not a wave. Is this true? → a collapsed wave is also a wave?

Actually, your biggest misconception is MORE fundamental than that. It appears that you do not understand at all the concept of commuting and non-commuting observables, and you do not understand the significance of it in terms of what can be known in a single measurement. It is why I think you didn't get the significance of my response earlier, and why I think you might not have been able to answer the question I posed.

In other words, you are focused on the shadow of the animal, without realizing what the actual animal looks like. The HUP and this so-called "collapse" are all consequences of the QM formulation. You need to go back to your QM lessons and trace the SOURCE of all this!

Zz.

 

[atyy]

I should have remembered: HUP is not about Δx times Δp_y, it's about Δx times Δp_x. So my experiment did not prove HUP wrong at all.

Yes, it is important to remember the uncertainty primciple applies to "canonically conjugate observables". Ballentine, author of a quantum mechanics text, makes a severe mistake by similarly forgetting this in his 1970 review. (Just in case you are thinking I'm criticising your question - I'm not - asking it on a forum and trying to learn from the discussion is different from publishing the claim in a top journal.)

If I understand it correctly now, one of my miconceptions was that I thougt that a collapsed wave is not a wave. Is this true? → a collapsed wave is also a wave?

Yes the collapsed wave function is a wave. It is a wave before collapse and after collapse. Before a measurement and after a measurement, the evolution of the wave function is governed by Schroedinger's equation. When a measurement is performed, the evolution is not governed by Schroedinger's equation, but collapses probabilistically according to the Born Rule.

The tricky part here is 'What is a measurement?", and quantum mechanics does not have a strict answer on this, but relies on our common sense ability to understand that a "measurement" is a "macroscopically irreversible reading".

But that aside, the wave function is a wave, before measurement and after measurement, ie. before and after collapse.

 

[bhobba]

But that aside, the wave function is a wave, before measurement and after measurement, ie. before and after collapse.

Is a Dirac Delta function a wave?

What I am about to say is not directed at Atyy - I know he knows this stuff.

What its directed at is sloppy use of terminology that leads to confusion

Its never a wave - ever. Its a state. Sometimes when expanded in terms of a certain basis it mathematically looks like wave equations found in other areas of physics - but it's not a wave - its a state.

A quantum system is described by a state at all times. Sometimes, as a result of an observation, it changes state. This is actually the same as a state preparation procedure and is generally how its viewed these days. But often what's being observed is destroyed by the observation so always looking at it that way is not correct. The correct way to look at it is via the formalism of QM - not through the prism of simplifications found in popularization's and beginning level texts.

For that you need to study the proper formalism that is found in texts like Ballentine - QM - A Modern Development:
https://www.amazon.com/dp/9810241054/?tag=pfamazon01-20

Here you will find QM developed from just 2 axioms with nothing mentioned about waves etc etc. It is regrettable that most beginning level texts speak about the wave particle duality and such - but strictly speaking its WRONG and leads to misunderstandings.

To illustrate it I will state the two axioms so you see nothing is mentioned at all about a wave or waves.

The first axiom is associated with any observation is a hermitian operator, O, whose eigenvalues give the possible outcomes of the observation.

The second axiom (also called Born's Rule) is there exists a positive operator of unit trace P such that the expected outcome of the observation E(O) = Trace (PO). By definition P is called the state of the system.

Immediately we see it has nothing to do with waves. A state, just like probabilities, is simply an aid in calculating the expected outcomes of observation. Saying a state is 'waves' is nonsense in a general sense. First, generally it's an operator, and secondly its not 'real' like a wave - its simply an aid in calculating the outcomes of observations. Sometimes, with so called pure states, the operator can be associated with elements of a vector space, and sometimes it makes sense to expand that element is terms of position eigenfunctions, and sometimes it mathematically looks like a wave - but that's it - that's all - only if those sometimes are satisfied does it even make sense to use the term wave-function - but its never a wave.

To answer the OP's original question the HUP is simply a theorem about non-commuting observable's. Its got nothing to do with waves, collapse or anything like that.

Thanks
Bill

 

[atyy]

Is a Dirac Delta function a wave?

Why isn't it a wave?

Also, is the Dirac delta a physical wave function?

 

[WannabeNewton]

"Wavefunction" generally refers to the position-space representation of a given state vector, $\langle x | \psi \rangle$ and the terminology is set in place for good reason. Even though Maxwell's equations in vacuum lead to equations that are formally wave equations i.e. second order in both time and space, Schrodinger's equation, while not formally a wave equation, also contains wave-like solutions i.e. solutions of the form $f(\vec{x} \pm \vec{v}t)$ which will of course just be Fourier decomposed into plane waves. Schrodinger's equation, just like Maxwell's equations, is linear so there's no problem having distributional solutions to the equation such as Dirac delta functions which can themselves be decomposed into Fourier modes. Distributional solutions are a problem for non-linear equations like Einstein's equation. So yes there certainly are solutions to Schrodinger's equation which are wave-like hence the term "wavefunction".

For example the general solution to Schrodinger's equation for a free particle is quite literally just a continuous superposition of Fourier modes which are individually of course sinusoidal waves acting as "stationary states" of the free particle. Whether or not the individual modes qualify as "physical" (hence the scare quotes in "stationary states") depends on whether you would be willing to call the non-normailzable generalized wavefunctions as "physical". Obviously the behavior of the localized solutions to the vacuum Maxwell equations will differ significantly from localized solutions to Schrodinger's equation for a free particle as the two are not in the same class of equations. For example an initial localized free Gaussian wave packet fed into Schrodinger's equation will tend to disperse whereas a vacuum EM wave packet will not disperse.

 

[atyy]

Could I get some commentary on this: the part that is perhaps misleading about thinking about the uncertainty principle as due to the wave-nature of the wave function is that the analogy to sound waves really comes from signal processing, which is the wave at a fixed position, and so may not have to do with a "wave" at all, but just a "signal". However, the part that I don't think is misleading is that in signal processing there is the conceptually analogous time-frequency uncertainty, and things like Cohen's class distributions, which are related to the Wigner function of quantum mechanics (known as the Wigner-Ville distribution in signal processing). Certainly one needs more than the time-frequency uncertainty, one needs that these correspond to the canonically conjugate observables of position and momentum. Also, once one goes to two particles, it becomes clear that the wave function is a wave in Hilbert space, not ordinary space. But basically, I think there is an uncertainty principle in a domain we are all familiar with - ordinary sound impinging on our ears - and it is related.

 

[WannabeNewton]

Also, once one goes to two particles, it becomes clear that the wave function is a wave in Hilbert space, not ordinary space.

Exactly. So the antiquated wave-mechanics picture that strongly associates the wave-function with the physical space of a particle and describes a particle quite literally by a spatially extended wave in physical space will conceptually break down if one starts considering systems of more than one particle or non-coherent states in contrast with the coherent states allowed by, for example, the quantum harmonic oscillator. This is why one has to be careful when using physical space wavefunctions to provide intuition for the uncertainty relations associated with observables and quantum states. Furthermore QM allows for uncertainty relations between any pair of non-commuting observables such as the components of the spin operator so the intuitive descriptions making use of the antiquated wave-mechanics picture have limited scope.

 

[bhobba]

Why isn't it a wave? Also, is the Dirac delta a physical wave function?

Waves generally have peaks and troughs.

Its a wave-function by the definition of wave-function which is the state expanded in eigenstates of position. Thinking in terms of wave-functions is generally not a good idea because you are singling out position eigenvectors as somehow special - they aren't.

Thanks
Bill

 

[Jilang]

Exactly so. If you measure the position is becomes a delta function for the position (not so wavy), but in the momentum basis it is still is expressed a superposition of many eigenvectors with different momenta (very wavy).

 

[ZapperZ]

Exactly so. If you measure the position is becomes a delta function for the position (not so wavy), but in the momentum basis it is still is expressed a superposition of many eigenvectors with different momenta (very wavy).

Which was why I stressed that the OP appears to not understand the principle involved in having commuting and non-commuting observables. It is why, for example, we can detect the presence of superposition by NOT measuring the observable that is in a superposition of state, but by measuring something else that it non-commuting with. It is why we can detect a superposition of the direction of a supercurrent by not detecting the direction of the current itself, but by detecting the presence of the coherence energy gap (such as in the Delft/Stony Brook experiments).

When an observable is measured, it only "collapses" the value for THAT particular observable, and all other observables that it commutes with (in a non-degenerate system). All other observables that it doesn't commute with are STILL UNDETERMINED and still in a superposition of states.

Zz.

 

[atyy]

Waves generally have peaks and troughs.

http://www.acoustics.salford.ac.uk/feschools/waves/standing_waves.php is an example of how in plain speech, an impulse is a wave - a sound wave. In any case, if calling the wave function a "wave" is misleading, then saying that sound is a wave is also misleading.

Its a wave-function by the definition of wave-function which is the state expanded in eigenstates of position. Thinking in terms of wave-functions is generally not a good idea because you are singling out position eigenvectors as somehow special - they aren't.

The idea that sound is a wave or light is a wave already has this idea that one thing can be represented in many ways. When we say white light contains all colours, we are using the Fourier representation, and it is a simple example of how one thing can be described in many representations. Similarly, when we say white noise is all frequencies with randomized phase, we are using the Fourier representation.

The uncertainty principle for a single particle is a particular case of the general principles of quantum mechanics, so the question is one of taking care when generalizing from this specific example.

 

[bhobba]

The idea that sound is a wave or light is a wave already has this idea that one thing can be represented in many ways. When we say white light contains all colours, we are using the Fourier representation, and it is a simple example of how one thing can be described in many representations. Similarly, when we say white noise is all frequencies with randomized phase, we are using the Fourier representation.

Its a fact Jack that virtually any function can be decomposed into Fourier components. In that sense any function is a wave - even for example a parabola. But would people say a parabola is a wave - highly doubtful. Even distribution's like the Dirac delta function can be so decomposed (provided they are what is called 'well tempered') and wavelike in any usual sense they most definitely are not - even a function in the usual sense they aren't.

Thanks
Bill

 

[T'Pau]

Thanks again, everyone, for making me stop and think,

Mayby I better explain why I asked: I'm involved in writing a QM-chapter for a high school book (in Europe). It has to be introductory level, and in my opinion, it has to have (some) popularization. What it does not need, are my misconceptions. Better to be slaughtered here :-), then teaching 17 year-olds rubbish.
I cannot (and will not) use words like 'Hilbert', 'operator', 'Born Rule', 'non-linear equations' (sorry)...

I notice there is discussion about calling it a 'state' or calling it a 'wave-function'. I would really like to stick with 'wave-function'. In an early version:
I called a free electron/photon a traveling wave,
An electron in a one-dimensional infintelly deep square well I called a standing wave,
Electrons in an atom I also a standing wave...

Is this to far from the 'thruth'?

The tricky part here is 'What is a measurement?", and quantum mechanics does not have a strict answer on this, but relies on our common sense ability to understand that a "measurement" is a "macroscopically irreversible reading".QUOTE]

I fear I still have a misconception there. I don't understand why it needs the word 'macroscopical' (Wikipedia uses the word 'classical' in stead of 'macroscopical'
http://en.wikipedia.org/wiki/Uncertainty_principle. In their source I found examples of measurements that involved something classical but not necessarily macroscopical.)

To me, (probably naively) it seems, that all measuring involves collisions, it involves particle-like behaviour. To me, (probably naively), collisions and particle-like-behaviour are synonyms.
(I know: the double-slit experiment measures wave-like behaviour. But how do we see this → because a screen lights up: that lighting up is particle-like behaviour.)

Why can't I simply interpret 'measurement' as 'collision'. Why are not *all* collisions between two electrons a measurement? Why is it only a measurement when I perform something classical with one of the electrons, for instance: put it through an ampere-meter?
When two electrons collide when no one watches, do their wave-functions not collapse? Don't the collide?

To be more precise (putting what I guess is my misconception as plainly on the line as I can): to me, four things are synonyms (for now):
'collision', 'measurement', 'collapse of wave functions', 'particle-like behaviour'.

Thanks, again,

Paul

 

[bhobba]

Well if you are doing that you could do much worse than simply use Feynmans introduction which was designed for and pitched at exactly that level:
https://www.amazon.com/dp/0691125759/?tag=pfamazon01-20

Its cheap, and has associated video lectures:
http://vega.org.uk/video/subseries/8

Thanks
Bill

 

[bhobba]

Why can't I simply interpret 'measurement' as 'collision'. Why are not *all* collisions between two electrons a measurement? Why is it only a measurement when I perform something classical with one of the electrons, for instance: put it through an ampere-meter?
When two electrons collide when no one watches, do their wave-functions not collapse? Don't the collide?

What measurements are is a difficult issue that requires a detailed understanding of QM - especially a phenomena called decoherence.

Why you think it's collisions is beyond me - it most definitely is not that. In accelerators when particles collide its not 'measuring' anything - it generally scatters the particles and spews forth a whole heap of other particles - its those other particles those that conduct such experiments are interested in.

Something classical is a bit closer to the mark - it's basically when marks are left here in the commonsense classical world. That however is very imprecise and the difficulty is making it precise. That is one reason, but only one reason, its a difficult issue.

In fact its the heart of the real issue with QM - namely how can a theory that is about marks left in an assumed classical world explain such a world since it is composed of quantum objects.

This however is advanced stuff, most definitely NOT suitable for HS students. At that level its wise to simply gloss over it as most beginning books do.

Thanks
Bill

 

[bhobba]

I notice there is discussion about calling it a 'state' or calling it a 'wave-function'. I would really like to stick with 'wave-function'. In an early version:
I called a free electron/photon a traveling wave,
An electron in a one-dimensional infintelly deep square well I called a standing wave,
Electrons in an atom I also a standing wave...

Is this to far from the 'thruth'?

Well actually it is.

But beginning books often do something like this, telling the story of Plank, how he introduced a particle like aspect to light, how Einstein took it literally to explain the photoelectric effect, and De-Broglie said if light can be particles then maybe matter can be waves. Standing waves then explain electron orbits etc.

But if you go down that path make sure you tell the students this is just an introduction and isn't what really is going on. When De-Broglie proposed his matter waves within a few years it was completely outdated and seen as basically wrong by a series of fast paced developments - Schrodinger's wave mechanics, Heisenberg's Matrix Mechanics and its final culmination Dirac's transformation theory which bought all this together as one cohesive whole - and showed matter waves etc were basically a crock of the proverbial. But understanding that will need to wait for college.

Thanks
Bill

 

[atyy]

Mayby I better explain why I asked: I'm involved in writing a QM-chapter for a high school book (in Europe). It has to be introductory level, and in my opinion, it has to have (some) popularization. What it does not need, are my misconceptions. Better to be slaughtered here :-), then teaching 17 year-olds rubbish.
I cannot (and will not) use words like 'Hilbert', 'operator', 'Born Rule', 'non-linear equations' (sorry)...

I notice there is discussion about calling it a 'state' or calling it a 'wave-function'. I would really like to stick with 'wave-function'. In an early version:
I called a free electron/photon a traveling wave,
An electron in a one-dimensional infintelly deep square well I called a standing wave,
Electrons in an atom I also a standing wave...

In my opinion, it ok for an elementary exposition. The uncertainty principle is seen even for sound. A sound that is a well-localised impulse is a superposition of sinusoidal wave with well-defined wavelength. http://www.acoustics.salford.ac.uk/feschools/waves/standing_waves.php

A sinusoid has a well-defined (sinusoidal) periodicity, while a localized impulse clearly does not because it is a superposition of waves of many wavelengths. So there is for any wave a localization-wavelength uncertainty principle.

In the quantum mechanics of a single particle, our naive concept of a localized wave is basically still ok. The strange thing is that we now associate "momentum" with wavelength (inverse wavelength) - this is the de Broglie relationship. Why? Well, that is a quantum mechanical concept that experiment has verified. There are several reasons we can use to justify calling the inverse wavelength "momentum". One is that measured over many experiments on a single particle set up in the same way, the average momentum over all experiments obeys the same equation as the classical momentum - this is Ehrenfest's theorem.

The main reason this analogy is misleading is that the wave is a wave in "Hilbert space", and this can be clearly seen when we extend the formalism to two particles. So more properly, the wave is not a wave in the usual 3D space of classical physics, but a wave in an abstract space, which can be used to calculate the probabilities of things we observe like position or momentum.

I fear I still have a misconception there. I don't understand why it needs the word 'macroscopical' (Wikipedia uses the word 'classical' in stead of 'macroscopical'
http://en.wikipedia.org/wiki/Uncertainty_principle. In their source I found examples of measurements that involved something classical but not necessarily macroscopical.)

To me, (probably naively) it seems, that all measuring involves collisions, it involves particle-like behaviour. To me, (probably naively), collisions and particle-like-behaviour are synonyms.
(I know: the double-slit experiment measures wave-like behaviour. But how do we see this → because a screen lights up: that lighting up is particle-like behaviour.)

Why can't I simply interpret 'measurement' as 'collision'. Why are not *all* collisions between two electrons a measurement? Why is it only a measurement when I perform something classical with one of the electrons, for instance: put it through an ampere-meter?
When two electrons collide when no one watches, do their wave-functions not collapse? Don't the collide?

To be more precise (putting what I guess is my misconception as plainly on the line as I can): to me, four things are synonyms (for now):
'collision', 'measurement', 'collapse of wave functions', 'particle-like behaviour'.

Let us use the "shut-up-and calculate" interpretation of quantum mechanics, in colloquial physics speak, this is a "Copenhagen"-like interpretation. In this view, to do quantum mechanics we must always divide the universe into a macroscopic realm and a quantum realm. So there is no such thing as a "wave function of the universe". In this view, quantum mechanics is just a way to calculate the probabilties of observations we get when we use macroscopic measuring instruments to probe quantum systems.

The unsatisfying thing about this is that we know quantum mechanics applies even on the largest scales - for example for describing the anisotropies in the cosmic microwave background. So how can the whole universe not be quantum? The answer in this view, is that even when we observe a large part of the universe, we still consider ourselves in the readings on our measinring instruments to be macroscopic, because they are not in a superposition of readings, they produce a definite reading.

That is the common sense or shut-up-and-calculate view. In this view, quantum mechanics, although powerful and far reaching, is perhaps incomplete. There are other views such as the many-worlds approach to quantum mechanics in which the wave function of the universe exists, and quantum mechanics is complete. In the many-worlds view, each measurement does not produce one outcome, instead all outcomes happen.

Now what has this got to do with whether a collision is a measurement or not? In the shut-up-and-calculate Copenhagenish view, we divide the universe into macroscopic and quantum realms using common sense. The measuring apparatus is on the classical side, and the reading is "irreversible" to the experimenter. But what is irreversible? The laws of quantum mechanics using Schroedinger's equation are reversible in time, only when a measurement is made and an irreversible mark is made does collapse happen - which is not time reversible. This is undoubted subjective, since there are things that look like collisions and decoherence, but for a skilled experimentalist with great experimental control, these are not irreversible, and collapse does not occur. This need for common sense, fuzziness is the sign within the Copenhagen interpretation of the "measurememt problem" of quantum mechanics.

A good introduction to the measurement problem is John Bell's http://www.tau.ac.il/~quantum/Vaidman/IQM/BellAM.pdf .

Here is an example of decoherence which is in some sense "undone" by the ability of the skilled experimenter to extract more information from the environment that caused the decoherence http://chapmanlabs.gatech.edu/papers/scattering_ifm_prl95.pdf .

 

[kith]

Why are not *all* collisions between two electrons a measurement?

"Measurement" is not a purely physical term but is related to how a person acquires knowledge. If you remove the pointer from your amperemeter, you don't have a measurement apparatus anymore although the physical interaction between your system of interest and the amperemeter doesn't change.

Likewise, you can't learn anything from the collision of two electrons until they leave a human-readable mark somewhere.

 

[WannabeNewton]

I called a free electron/photon a traveling wave,
An electron in a one-dimensional infintelly deep square well I called a standing wave,
Electrons in an atom I also a standing wave...

Is this to far from the 'thruth'?

Well for starters, only the stationary states of an electron in an infinite square well correspond to standing waves in the sense that each stationary state can be written as a product of a spatially varying sinusoidal function and a temporally varying sinusoidal function. A general state of the electron will be a discrete but infinite superposition of such standing waves of every normal mode allowed by the boundary conditions of the infinite square well. Each stationary state corresponding to a different allowed normal mode will have a different number of fixed nodes apart from those at the ends which is characteristic of standing waves of course, but for a general state the only fixed nodes are those at the endpoints of the infinite square well and these nodes exist only because of the boundary conditions. A free electron on the other hand has a general state given by a continuous superposition of Fourier modes-there are no boundary conditions so all possible modes are allowed.

Regardless it has already been mentioned previously in this thread that associating the wave-function of an electron with something that propagates in the physical space that the electron travels through, in an attempt to describe the electron as a spatially extended "smeared out" wave, is conceptually flawed. This antiquated picture of the wave-function was replaced by Born's rule and the statistical interpretation of the wave-function a long time ago so I don't know why the conceptually flawed picture lingers on.

 

[bhobba]

"Measurement" is not a purely physical term but is related to how a person acquires knowledge. If you remove the pointer from your amperemeter, you don't have a measurement apparatus anymore although the physical interaction between your system of interest and the amperemeter doesn't change.

This is part of the difficulty with the whole measurement paradigm of what QM is and why its a difficult issue.

Basically in modern times a measurement is replaced with dechoherence which is independent of pointers in measurement apparatus etc. But this is an advanced issue best glossed over for HS students.

Thanks
Bill

 

[T'Pau]

Hi,

And 'collisions' and 'particle-like-behaviour', are these more or less synonyms?

-> is there particle-like-behaviour that does not involve collisions?

Paul, who is grateful for al the responses.

 

[atyy]

Hi,

And 'collisions' and 'particle-like-behaviour', are these more or less synonyms?

-> is there particle-like-behaviour that does not involve collisions?

Paul, who is grateful for al the responses.

Tricky, tricky, tricky.

Yes, collisions are particle like behaviour.

However, in LHC type collisions, the rough idea is that one sends in free, non-interacting particles, the particles collide and interact, and then the outgoing particles from the collision are again free, non-interacting particles. The tricky part is that a free particle from a collision is the most unlocalized state, because it is a particle with a definite momentum. Since it has a definite momentum, it must have very undefinite position. But then what about all those particle tracks? Well, although the tracks looks small and localized to us, they are big from the point of view of the uncertainty principle.

http://books.google.com/books?id=CNCHDIobj0IC&vq=particle&source=gbs_navlinks_s
p117: To a particle the beam is the whole universe, and it is big!