# Concervation of Momentum and Non-Classical Paths

1. Apr 9, 2010

### JDługosz

Consider, to give a concrete example, an Airy Ring pattern appearing on a phosphor screen when electrons are sent through a single hole.

Send one electron through.

If it passed through the hole while maintaining a straight path that it must have followed from the emitter (another small hole, perhaps) to the screen, then all is well.

If the electron landed elsewhere in the pattern—that is, it diffracted—then the momentum vector of the electron is different, as the straight line from its final position to the hole points in a different direction than the line from the hole back to the source.

How is momentum conserved?

―John

2. Apr 9, 2010

### ansgar

you are mixing the classical and quantum mechanical description of momentum and position....

3. Apr 10, 2010

### genneth

Why should momentum be conserved? There is no problem in there being momentum transfer between the electron and whatever surrounds the hole.

4. Apr 10, 2010

### ansgar

but the diffraction formalism has no source of scattering built in

5. Apr 10, 2010

### genneth

I don't know about you, but to me if something diffracts, that *is* scattering. Scattering of waves is nothing more or less than diffraction around obstacles.

6. Apr 10, 2010

### JDługosz

Why should momentum be conserved? Because that is fundamental. Space is the same everywhere.

The material containing the hole recoils? That would be my guess. But how does that happen? Re-reading Feynman's "QED", the classical path is said to emerge from the sum-of-possibilities, but no mention is made of momentum transfer with whatever causes one point along the path to become known. So, I'm asking if someone would elaborate on that.

Please keep going with that explanation.

7. Apr 11, 2010

### genneth

Yes, translational symmetry of a dynamic system leads to momentum conservation. However, the mathematical set up being described here is that the hole is not considered dynamic, and part of the background. Its presence explicitly breaks the symmetry.

Including the hole explicitly yields that the hole recoils.

8. Apr 17, 2010

### JDługosz

How does it do that?

9. Apr 18, 2010

### jeblack3

Consider the probability amplitude for the electron to go through the hole and hit a particular point on the screen. The phase of that probability amplitude depends on where the hole is. So the wavefunction of the object with the hole in it gets multiplied by a phase that depends on its position. That means it gets momentum.

10. Apr 18, 2010

### genneth

Consider a pendulum. Its momentum keeps changing! Why is momentum not being conserved?

11. Apr 19, 2010

### JDługosz

genneth: The pendulum and the earth are coupled with a classic force.

That doesn't explain how it happens in the Feynman phase explanation.

jeblack3: I understand from this thread so far that simply postulating a universe where the electron must pass through a particular region is not one in which momentum is conserved. A real hole must do more than is shown in these popular works.

Feynman's internal clocks, or the muddy tire, or the phase tape, etc. all explain the vector addition of a measure of each possible path length. Can someone show how the real hole does something that the abstract hole does not? "The hole has a position, that means it gets momentum" is leaving something out.

12. Apr 19, 2010

### genneth

I'm not seeing why you are happy with a classical force but not with a quantum one? Electrons interact mostly with photons. These photons then interact with the electrons (mostly) in the material making up the hole. When the electron you shoot through deflects, it imparts some momentum to some electron in the material (suitably simplified). Just like a pendulum, where the Earth oscillates in response, here, the material in the hole moves, i.e. the hole moves.

13. Apr 20, 2010

### JDługosz

Feynman's explanation is that simply knowing that the electron passes through this small region is enough to cause diffraction. The phase is a property of the length of each path, not something that is altered due to some interaction with the material.

Suppose you had no material forming a hole at all. Perhaps a ring of electric field strength meters, and when they all showed the same value, the screen was set to record the electron's final position. It passed an equal distance from all the sensors, was a considerable distance away from them, but still you know it passed through a small region.

I'm just trying to improve my understanding of what it says in "QED". Something is missing.

14. Apr 21, 2010

### genneth

If there's an electric field then the electron most certainly feels a force?

15. Apr 23, 2010

### JDługosz

Equal in all directions. The meters note that the field is the same on each one. But this is getting side tracked. Is there another place I might discuss this? Physics Forum had been helpful in the past, so I came here with my question.

16. Apr 23, 2010

### Antiphon

There is momentum exchanged with the screen by the diffracting particle.

This is true in both the classical and quantum cases.

17. Apr 24, 2010

### eaglelake

First, we must accept that quantum mechanics (QM) is not classical mechanics and quantum events cannot be explained using classical concepts. That is why we invented QM.

Quantum particles are scattered from a slit(s) without the benefit of a scattering force. QM does not require a momentum exchange with the slit. Nevertheless, the direction, but not the magnitude, of the momentum vector is changed. Actual diffraction experiments do not change the energy $$E = \hbar \omega$$ or the momentum $$p = \hbar k$$ of the scattered particle.

The QM explanation is given in arXiv:quant-ph/0703126. At the slit we know the particles position y, say. Hence, the particle has an uncertainty in the y-component of momentum and it can then be scattered at any angle.

There is no answer to your question, which begs a classical answer. There is only the QM explanation.
Best wishes

18. Apr 28, 2010

### JDługosz

Thanks.

Perhaps I can ask a better-formed question. Given that the screen bearing the slit recoils (I'm told by others on this thread that it does), how does that happen? The QM, as given in the article you pointed out, only covers that you somehow put the particle in a position state. There is nothing about the slit having walls or what they know about the eventual measured momentum state.

There is simply no concept of "re-action" or even a force. But somehow the books must balance, right?

19. Apr 28, 2010

### eaglelake

As far as we know, there is no classical-like interaction between the scattered (diffracted) particle and the slit. Classically, a particle can change direction only when an external force acts on it, which is what you want to happen. But, there is no force here! Further, we cannot explain the diffraction of particles by assuming conservation of momentum between the particle and the screen containing the slit. You are looking for a causal, i.e. a classical, explanation where there is none. You ask, "how does this happen"? We don't know! There is no classical explanation.

The "eventual measured value" of the momentum is a random value generated at the instant the particle is detected. Quantum mechanics does not "explain" how this occurs. It only gives us the statistical distribution of all possible momentum values. Neither the diffraction experiment nor quantum theory tells us anything about any underlying mechanism involving forces or conservation of momentum that might give rise to this randomness.

The closest we can get to conservation of momentum is this: The statistical average (called the expectation value) of all possible momentum values is $$\left\langle {p_y } \right\rangle = 0$$. If momentum is conserved, then there is no exchange of momentum and, on average, the wall does not recoil.
Best wishes

20. Apr 28, 2010

### LukeD

To make things more clear about the definition of momentum

In classical physics, momentum is a vector proportional to the velocity. It is defined as $$\vec{p}=m \vec{v}$$

In quantum mechanics, momentum is a vector proportional to the wave number. It is defined as $$\vec{p} = \hbar \vec{k}$$ The wave number tells you how quickly the phase of the quantum wave changes in space.

There is a very large difference in the contexts of these definitions. The classical momentum is defined along the path of a particle, but the quantum mechanical definition is only defined for the wave function, which is a sum of all paths. This is reflected in the fact that the particle, in a sense, goes through both slits at once.

And there is no momentum transfer with the slits. If there were, then this would be similar to their being a detection of which slit the particle went through as we could measure whether a wall got hotter or not. This would destroy some of the interference pattern.

Many -non-classical- paths do not obey momentum conservation in the -classical- sense, but this is ok because momentum does not exist in the classical sense, and non-classical paths do not have any direct physical meaning. We can only talk about momentum in connection to the wave function.