# Wave particle duality

1. Jan 10, 2016

### God Plays Dice

In the single slit experiment, individual photons may be diffracted. The electric field of the em wave should accelerate charged particles in the screen (if it were a capacitor for example) but only one packet of energy exists. The photon may be absorbed after many wavelengths of light have passed over the charges particles. So, how can this work out? How do the charged particles know not to accelerate or do anything due to the wave until it is decided that a photon is to be absorbed?

2. Jan 10, 2016

### Simon Bridge

Individual photons are not diffracted. Diffraction is what we call the emergent effect of the statistics of the apparatus on a large number of photons.
Charged particles know to accelerate when a photon is absorbed in exactly the same way that a pool ball knows to sit still until the cue ball hits it. The interaction with the photon causes the acceleration. In the photon description there is no such thing as a light wave.

3. Jan 11, 2016

### BvU

Yes, individual photons are diffracted, solely due to their wave character. So they do generate an interference pattern. There is NO electromagnetic interaction with the screen (which is realistically assumed to be an ideal blocking potential, because atomic dimensions << wavelength) in the experiment. See Huijgens principle.

4. Jan 11, 2016

### sophiecentaur

That is a difficult idea to justify. A convential diffraction pattern assumes a continuous wave - i.e. consisting of just one wavelength and exists for all time. If you want to assign a wavelength to a photon then you are assuming that the photon 'exists' for infinite time. It would have to have infinite extent (a zero bandwidth). How could that be when a photon is generated at its source and interacts with some matter when it reaches it and this all happens in a well defined time. If you modify your model of a photon to give it a certain extent (in time / space) then it will have a frequency spectrum (non-zero bandwidth). That spectrum would have to be dependent on the characteristics of the source. But all photons of a certain energy are considered to be identical and you are implying that they would all be different. It's very easy to make assumptions about photons by drawing a little 'squiggle' to represent one (as in the Feynman diagram, for instance) but that squiggle has no real basis, except to imply that there is a wave associated with it. As Simon says, talk photons and you have to talk statistics, talk waves and you can talk diffraction.

5. Jan 11, 2016

### BvU

I would be interested to know at what intensity the diffraction pattern would cease to exist in a simple experiment:

With a simple HeNe laser and a single narrow slit we see a clear diffraction pattern, even though the light isn't completely monochromatic (even HeNe lasers have a nonzero bandwidth and -- e.g. with an interferometer -- several modes can clearly be seen to contribute - never mind). With neutral density filters we can reduce the number of photons/unit of time coming in and with photomultipliers or some other devices we can still observe the diffraction pattern - albeit at the cost of counting time. I see no reason to expect the pattern to change by the time the intensity of the beam is so low that one can speak of single photons passing the slit.​

6. Jan 11, 2016

### sophiecentaur

And neither would I. The result of the statistics of large numbers will follow the wave prediction. One particle arriving does not constitute a diffraction pattern - is is just an 'event'. There is a real difference here.

7. Jan 11, 2016

### BvU

Thanks, I find that reassuring. I posted because of my interpretation of Simon's first sentence and now I am looking for a way to unify the statements somehow.

Single photons end up on the screen at locations with a probability that corresponds to the diffraction pattern

8. Jan 11, 2016

### sophiecentaur

That's fair enough and it avoids the necessity of individual particles being 'diffracted'. It's all too easy to write things in a way that can imply that - which can confuse people.

9. Jan 11, 2016

### Simon Bridge

Well that's the issue behind the different interpretations to QM.

The "diffraction pattern" is, mathematically, the probability distribution of photons arriving at the screen in particular positions.
How the source distribution turns into the screen distribution is what everyone is arguing about - with no resolution in sight.

Certainly - the diffraction pattern is in the probability distribution ... the visible pattern you see in, say, the standard HeNe Laser class demo, is the result of many individual photons. The relationship between the distribution and the actual events is like the relationship between statistics and the individual faces that show after throwing a die. A single photon passing through the apparatus only ends up in one place on the screen.

http://vega.org.uk/video/subseries/8
... Feynman covering the topic basics in this lecture series. His position is basically "shut up and do the math".

http://arxiv.org/pdf/quant-ph/0703126.pdf
Marcella 2002: Quantum interference at slits (a freshman approximation attempt to tackle the statistics)

http://scholarship.haverford.edu/cgi/viewcontent.cgi?article=1315&context=astronomy_facpubs
Rothman and Boughn 2010: Quantum interference with slits revisited (discussion of limitations of Marcella paper)

10. Jan 12, 2016

### sophiecentaur

Very wise advice. It carries the admission that we don't actually 'understand' very much of this stuff and that the Maths is a model that works. In fact, it's the only reliable way to predict things.

11. Jan 13, 2016

### God Plays Dice

As far as I know, individual photons do diffract. However the question was not do they diffract, but how does the wave cope with running into say, a free electron. It's electric field would accelerate the electron but that would take energy. So obviously the photon isn't sub divided. I can't see a redshift in the wave that wouldn't happen. So what happens?

12. Jan 13, 2016

### BvU

It's called Compton scattering

13. Jan 13, 2016

### sophiecentaur

And a free electron can have any energy change because it is not bound to an atom, with the constraints of discrete energy levels.

14. Jan 14, 2016

### God Plays Dice

It would have to be Compton/ Thomson scattering, however this is a different process entirely to a charge in an electric field and I don't think you can just neglect the latter

15. Jan 14, 2016

### God Plays Dice

Possible that the em wave divides into a range of frequencies, some lower, and these get used for acceleration?

16. Jan 14, 2016

### sophiecentaur

How? You would have to suggest a (linear) process that could achieve this and keep contiuity of fields everywhere. Why not accept the standard ideas? Are they wrong? Read about Thompson scattering. Wiki and other sources discussit at various levels.

Last edited: Jan 14, 2016