# Questions on Feynman's view on Double slit experiment

1. Sep 5, 2008

### suomi

hi, i'm new to this forum and mainly because i have some questions regarding a book i am reading. Brian Greene's The Elegant Universe: Superstrings, Hidden Dimensions, and the Quest for the Ultimate Theory

mainly he explains the view that Feynman have on the double slit experiment

as shown on the picture Feynman is saying that every electrons use all the "roads" possible including going to Andromeda and coming back to go through slit A." the electrons taste every trajectories possible from its start until it gets to the final detector on the other side of the double slits planel. ok every body know that if we put a detector in one or two of the slits the interference disappear. my question is what if we put detectors around the experiment except in front of the experiment (the panel and the 2 slits) will they stop interefering too ? Brian Greene explains more on Feynman's view but i'm not gonna translate more since i'm pretty sure the ones answering here already know

Also i'm not finished yet with the book but he tries to defend the superstring theory in that book, he's saying that this experiment is the core of quantum mechanics but doesnt explain the results with the superstring theory...anyone ?

by the way i'm french canadian so english is only my second language.

File size:
47.1 KB
Views:
74
2. Sep 5, 2008

### peter0302

Yes, that is Feynman's view, but if you understand Feynman's view in detail, you'll see that any paths that deviate significantly from the center have such miniscule probabilities of being true that they don't contribute anything significant to the interference pattern, and so if you eliminate them (by placing detectors there) they are not missed.

Ironically Feynman's view is rather indistinguishable from the Many Worlds Interpretation, although I don't think he was an open supporter of it.

3. Sep 6, 2008

### lightarrow

The far you put the detectrors from the slits, the more interference will be strong and then visible.

4. Sep 6, 2008

### f95toli

I don't think it is quite correct to talk about "Feynman's view" in this case since it somehow implies that it is an interpretation.
The path integral formulation (which is what you are describing) is a generalization of the action principle (from classical mechanics); i.e. it is first and foremost a very powerful mathematical method which is also the basis for quantum field theory.
Now, the most "obvious" way of explaining this method is to imagine that a particle to take every possible path and then to sum over all histories. However, you do not HAVE to interpret it that way (in fact you don't have to interpret it at all).
Since we know that the path integral formulation of QM is correct (it correctly predicts the outcome of experiments) it follows that ALL interpretations of QM must be "compatible" with it and vice-versa.

5. Sep 6, 2008

### suomi

and what if the panel containing the two slits would be a detector ? i mean light or electrons are shot in this direction...so if i understand what lightarrow said the interference will be very low... but i mean this must take months of caculations. i'd love to see my experiment being done.

6. Sep 6, 2008

### haushofer

Is that true? Does one sum over all thinkable paths ( including going from Andromeda and back ), or only over physically allowed paths?

7. Sep 6, 2008

### peter0302

Yes, you sum over all physically possible paths. Including to and from Andromeda. As I said, those paths have such a miniscule contribution to the result, though, that you can disregard them.

8. Sep 7, 2008

### Hans de Vries

There a limited time in which the electrons travel through the experiment.
Paths which require a speed higher then c have no physical contribution at all.

Regards, Hans

9. Sep 8, 2008

### lightarrow

How would you evaluate this time?

10. Sep 8, 2008

### Hans de Vries

This is simply the time from electron emission to the impact on the detector screen. All
paths must be included but the total path length may not be longer as permitted by c.

Technically these paths are arbitrary, like zigzag paths for instance. They represent
how the field propagates. The field may be propagating forwards, it thereby pertubates
the field in the front. This pertubation is re-propagated in all directions, so also to the
back, and so on.

A significant part of the propagation, for any electron independent of its speed, is on
or near the light cone. These components however are interfered out by the finite
field which stays more or less localized over time. The field does spread out however
(when not in a bound state)

You can commonly hear statements like "the electron" as a point particle follows all these
zigzag paths, and does so at the same time. Clearly this invokes a *QM interpretation*
about what a field is. I personally prefer to stay interpretation agnostic and just talk
about field propagation.

Regards, Hans

11. Sep 8, 2008

### lightarrow

Exactly; this is my concern. How can I evaluate the travelling time t if I have to consider the lenght of the average path which depends on the lenght of all the possible paths? I Should exclude those longer than c*t but t is what I have to evaluate...

12. Sep 8, 2008

### peter0302

Are you sure you exclude paths that are longer than ct? I would think those paths would just contribute so little to the result that they are negligible. I also thought that, in theory, the Schrodinger Eq does permit a particle to appear somewhere as though it had travelled faster than light. If that's true, I would think you would *not* exclude FTL paths.

13. Sep 8, 2008

### Hans de Vries

If you have an initial condition of the field of a just emitted electron and you want to
calculate the probability that the electron hits the screen after a given time then you
have sufficient information to exclude the FTL paths.

In practice one uses the propagator which already contains all the parts. The paths
which go straight from A to B have the shortest arrival while the zigzag paths or the
paths that take a detour arrive later.

The propagator from A to B gives a value for every possible "flight" time t, where the
shortest time t is defined by c and the latest time is in principle at $t=\infty$

Regards, Hans

14. Sep 8, 2008

### Hans de Vries

Forget the Schrödinger equation in this respect. It's a non-relativistic theory so
it shouldn't surprice that violates SR.

For a relativistic theory there should not be any FTL contribution obviously...

There's are some technical issues for instance with the Feynman propagators which
can cause some confusion in this respect here. The propagator at t=0 has a small
non vanishing part "outside the light cone" in the order of the Compton radius.

This part decreases rapidly linear with time. So it's initially ~10-13m at t=0, but
after a propagation of 1 micrometer it has decreased to ~10-20m and diminishes
further with increasing propagation time.

The technical reason is that a point-source at t=0 is expected to contain only
positive or negative frequencies going in one or the other direction in time. Such
a Fourier decomposition does not exist but requires a complentary "Hilbert partner"
(Most phycisist will probably be more familiar with the term Kramers-Krönig relation)

This Hilbert partner is not a point but a slightly spread out function with a size of
roughly the Compton radius. This Hilbert partner is exactly the inititial part "outside
the light-cone". There isn't any propagation outside the light-cone because the
actual source can not be a point if it contains only positive or negative frequencies,
therefor the propagator can not be confined to a point either at t=0.

This is a rather technical subject and one of the reasons that the current version
of my chapter on Klein Gordon operators currently spans 33 pages with 10 illustrations
now. (it's unfortunately not on line yet, it needs some more work to get finalized)

Anyway, In QFT one always postulates that there are no space-like causal relations,
(no FTL paths) for the reason that the commutators vanish.

Regards, Hans

15. Sep 8, 2008

### suomi

still my 2nd question went unanswered...

we know that if we put detectors near the experiment it will reduce the interference and if ew put two detectors in the slits the interference will stop. what if the panel containing the two slits would be a detector ? light or electrons are shot in this direction...and if i understand what lightarrow said the interference will be very low because we are detecting electrons very near the experiment

In other words when an electron is shot it knows how many detectors there is!!! am i understanding it right ?

16. Sep 8, 2008

### peter0302

Thanks Hans, very helpful!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?