# Does Heisenberg apply to a collapsed wavefunction?

1. Feb 14, 2014

### 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

2. Feb 14, 2014

### bhobba

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?
'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/Quantum-Mechanics-A-Modern-Development/dp/9810241054

Thanks
Bill

Last edited by a moderator: May 6, 2017
3. Feb 14, 2014

### 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.

4. Feb 14, 2014

### ZapperZ

Staff Emeritus
I measure Lz. Do you think the values for Lx and Ly are also determined?

Zz.

5. Feb 14, 2014

### Staff: Mentor

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".

6. Feb 14, 2014

### 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.

7. Feb 14, 2014

### Jilang

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.

8. Feb 14, 2014

### naima

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.

9. Feb 15, 2014

### 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

10. Feb 15, 2014

### bhobba

Precisely why do you think waves are involved at all?

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

11. Feb 15, 2014

### naima

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

12. Feb 15, 2014

### ZapperZ

Staff Emeritus
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.

13. Feb 15, 2014

### atyy

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.)

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.

Last edited: Feb 15, 2014
14. Feb 15, 2014

### bhobba

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 whats 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/Quantum-Mechanics-A-Modern-Development/dp/9810241054

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

Last edited by a moderator: May 6, 2017
15. Feb 15, 2014

### atyy

Why isn't it a wave?

Also, is the Dirac delta a physical wave function?

16. Feb 15, 2014

### 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.

Last edited: Feb 15, 2014
17. Feb 15, 2014

### 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.

Last edited: Feb 15, 2014
18. Feb 15, 2014

### WannabeNewton

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.

19. Feb 16, 2014

### bhobba

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

Last edited: Feb 16, 2014
20. Feb 16, 2014

### 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).