
#1
Jul2807, 05:09 PM

P: 167

1. The problem statement, all variables and given/known data
From "http://physicsweb.org/articles/world/15/9/1": In 1909 Geoffrey Ingram (G I) Taylor conducted an experiment in which he showed that even the feeblest light source  equivalent to "a candle burning at a distance slightly exceeding a mile"  could lead to interference fringes. This led to Dirac's famous statement that "each photon then interferes only with itself". To be sure that just one photon is producing interference you need to calculate how long the photon last ( t ). Since s = c . t ( space, velocity of light and time ), this will give us a length of the photon. "length is this experiment" an operative definition. How can be calculated the power of ligth that must reach the double slit to be shure that there is just a photon at a time ? 2. Relevant equations Energy of a photon E = h. nu Energy of an electromagnetic wave E = 1/2 E . H ( per volume ) Impedance of the empty space E / H = a given number 3. The attempt at a solution From the tree equations you can derive the volume of a photon, but: volume = surface . length and we dont know the surface of the photon. Besides this seems that is a nosense concept ( volume, length, surface ) when applied to a photon. So my question is "How was calculated the power of ligth that must reach the double slit to be shure that there is just a photon at a time ? " I dont understand QM, so I just want an answer to my question. Thanks in advance. 



#2
Jul2807, 06:07 PM

P: 1,746

You'll see that the lightbulb emits a huge number of photons per second: billions upon billions... In order to reduce the rate of photon emission you can apply filters or let the light to pass through a small aperture (thus blocking a large portion of photons). You'll need to repeat this filtering/blocking procedure several times until you get to the 1 photon/second (or whatever rate you think is appropriate to call "one photon at a time") light source. Eugene. 



#3
Jul2907, 03:27 PM

P: 167

meopemuk:
Thanks. 



#4
Jul2907, 05:10 PM

Sci Advisor
HW Helper
P: 8,961

double slit experiment whith one photonIf you reduce the power of the lamp so that ON AVERAGE there is only 1 photon/second emitted and the experiment is only 1foot long  the photon travels through the experiment in 1ns so there is only a 1 in 1billion chance of another photon being emitted in the same nanosecond and being in the experiment at the same time. Are those odds good enough to convince you there is only one photon? If not reduce the brightness so there is only one photon/hour or 1/day. 



#5
Jul2907, 06:45 PM

P: 1,746

Eugene. 



#6
Jul3007, 03:59 PM

P: 167

all quotes from mgb_phys:
I understand probabilities, lets suppose that the source emits a photon exactly each certain amount of time, like a clock. Can you be shure that a photon cant last 2 seconds or 10 minutes, why ? 



#7
Jul3007, 04:56 PM

P: 1,746

The fact that they are sizeless particles is supported (among other things) by the observation that each photon interacts with a single AgBr cluster of the photographic plate emulsion. The fact that photons never decay or disappear is supported (among other things) by observations of distant galaxies. However, the infinite lifetime of photons does not imply that their "length" is infinite as well. Eugene 



#8
Jul3107, 06:30 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

I imagine you need to do something like the following for the problem in the OP:
Assume you have an isotropic source of power P, some distance R from the slits. Let the slit area be A  that gives you a total power through the slit. Assume the light is monochromatic at frequency f to find the photon rate, and hence the time between photons arriving at the slits. Finally, assume a distance D from the slits to the screen, and ensure that there must never be more than one photon in this space at any time. 



#9
Aug107, 05:22 PM

P: 167

To all posters: I read carefully your posts and I wanted to discuss each sentence you wrote but I want to focus in the main idea.
Lets suppose you can control exactly the rate of the photons arriving at the double slit ( 1 photon/second ... 1 photon/ns ), which rate would you choose ? ( to be shure that each photon interferes whith itselt ) meopemuk: 



#10
Aug107, 05:58 PM

P: 1,746

some other atom absorbs it. The photons emitted by distant galaxies apparently haven't met any absorbers on their way to Earth. If the photons were unstable particles with a finite lifetime t, we would have trouble seeing anything at distances greater than ct from Earth. Eugene. 



#11
Aug107, 11:10 PM

P: 400

Taking the Heisenberg uncertainty point of view, the "duration" of a photon can be related to it's bandwidth. This is how I'd approach the problem  if you know that your source is a HeNe laser, say, then you can look up it's bandwidth and calculate it's coherence length. This gives you the temporal "duration" of a photon  if the average time separation between photon emissions is longer than this, then you can say that there is very probably only one photon at a time interacting with the slits.




#12
Aug207, 03:55 PM

P: 167

meopemuk:
If your argument is true you cant set up this experiment using the photons from stars because you would need years between photon and photon. JeffKoch: What does coherence length mean ? If the bandwith of the laser is 10 Hz, what is the coherence length ? But I want to further explain my posts. I want to know how is related the idea of a photon whith the electromagnetic waves. A photon is a piece of electromagnetic wave ( true ? ), so it must start and end, and occupy a certain volume. From my first post: 



#13
Aug207, 04:13 PM

P: 1,746

I think this is a wrong idea. In my opinion, it is more correct to imagine photon as a point particle, and the electromagnetic wave as a (sort of) wavefunction that describes the probability amplitudes of finding this particle in different regions of space. This situation is very analogous to the QM description of the electron. We consider electron as a point particle, and not as a piece of the De Broglie wavefunction. This is a classic example of the "waveparticle duality" in quantum mechanics. Eugene. 



#14
Aug207, 06:37 PM

Sci Advisor
HW Helper
P: 8,961





#15
Aug207, 09:16 PM

P: 400

I can't think of any other measure for the "length" of a photon, though I agree that perhaps this will just confuse the poster. Alvaros, can you clarify what  exactly  the homework question asks? Your reference is to an article, not a homework problem. 



#16
Aug207, 11:53 PM

Sci Advisor
HW Helper
P: 8,961

The 'average' of a single photon (smirk)?
I got asked how many constellations there are! Ok I'm an astronomer but my Phd was in building closure phase interferometers. 



#17
Aug307, 02:16 PM

P: 400

If you have a better measure for the "length" of a photon, I'm all ears  I can't think of one that doesn't invoke the ghost of Heisenberg. You could consider lifetime instead of length, but it's the same thing. 



#18
Aug307, 03:55 PM

P: 167

Concepts ( like: time, space, coherence length, sizeless particles, wavefunction ) provoke endless discussions. This forum proves it.
I just want to talk about this experiment, if you agree. 1 The experiment doesnt change with the distance sourcedouble slit ( True / False ) 2 To be shure that we are proving the interference of a single photon with itself we need to reduce the number ( /second ) of photons arriving at the double slit ( T/F ) How can we calculate the highest rate of photons arriving at the double slit in order to prove the interference photonitself ? JeffKoch: 


Register to reply 
Related Discussions  
The Photoelectric Effect, Photon Duality, and The DoubleSlit Experiment  Quantum Physics  5  
Problem: Plastic Covering One Slit in Double Slit Experiment  Introductory Physics Homework  5  
Double Slit Experiment  Quantum Physics  16  
Exact solution of the double slit experiment with a single photon  Quantum Physics  18  
Single Photon Double Slit Experiment  Quantum Physics  10 