# Homework Help: Double slit experiment whith one photon

1. Jul 28, 2007

### alvaros

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.

2. Jul 28, 2007

### meopemuk

You are right, the notions of volume, length, and surface are not applicable to photons. However, you can easily calculate how many photons a light source emits each second. You need to know the light frequency $\nu$, the energy of each photon $E = h \nu$, and the energy emitted by your light source in one second. The latter quantity is called the power of the source. For example, a 100 W(att) lightbulb emits 100 Joules of light energy each second.

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. Jul 29, 2007

### alvaros

meopemuk:
I dont know what is the rate appropiate to call "one photon at a time" and this is the relevant answer I want to know.
Thanks.

4. Jul 29, 2007

### mgb_phys

There isn't really an answer to this, it's all a question of probabilites.
If 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. Jul 29, 2007

### meopemuk

Excellent answer! I couldn't say it better.

Eugene.

6. Jul 30, 2007

### alvaros

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.

( c = 1 feet/ns ) You are assuming that the length of the photon is 0, why ?
Can you be shure that a photon cant last 2 seconds or 10 minutes, why ?

It doesnt seem very scientific.
Im convinced that there is just a photon at a time, this is not the question.

7. Jul 30, 2007

### meopemuk

From everything we know about photons, they are point-size particles with infinite lifetime.

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. Jul 31, 2007

### Gokul43201

Staff Emeritus
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. Aug 1, 2007

### alvaros

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:
From this sentence I infer that you never can be shure that there is just a photon at a time. Is there any distribution of probabilities of the "life time" of the photons ? The "life time" depends of the source ( distant galaxies ) ?

10. Aug 1, 2007

### meopemuk

Suppose that the distance between your proton source and the photographic plate is 1 foot. For a photon it takes 1ns to travel this distance. So, if your source emits 1 photon per 1 ns (let's make it 1 photon per 2ns, to be certain), then you can be sure that when a photon is emitted, the previous photon has already traveled the full length (1 foot), interacted with the photographic plate, and left its mark there. So, the previous photon cannot have any effect on what the present photon will be doing. With this emission rate you can confidently say that each photon behaves on its own, there can be no any photon-photon interaction, and if an intereference pattern is formed in these conditions, then you must conclude that each photon "interferes with itself".

I am not sure I understand your question. I said that the "lifetime" of photons in infinite. This means that once a photon is created (e.g, in an atomic transition, or whatever) it lives forever. It travels with the speed of light until
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. Aug 1, 2007

### JeffKoch

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. Aug 2, 2007

### alvaros

meopemuk:
The experiment doesnt change whit the distance source-double slit, the interference happens when the photon ( or two photons ) arrive at the double slit.
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:
I dont understand.
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:
What am I missing ?

13. Aug 2, 2007

### meopemuk

alvaros,

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 "wave-particle duality" in quantum mechanics.

Eugene.

Last edited: Aug 2, 2007
14. Aug 2, 2007

### mgb_phys

You cannot talk about the coherence length of a single photon. The coherence length is a measure of how long the source gives 'equivalent' photons. Given that the poster is already confused about the size of a photon I don't think it is useful to imagine the coherence length as the length of a photon in an interference experiment.

15. Aug 2, 2007

### JeffKoch

Sure you can, in an average sense. An excellent working model of a photon (whatever that really is - in fact I was asked "What is a photon?" during my dissertation oral exam I don't think anyone can really answer this.) is a wave with a single oscillation frequency within an envelope function. A Fourier transform gives the bandwidth.

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. Aug 2, 2007

### mgb_phys

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. Aug 3, 2007

### JeffKoch

More like, "an average photon" from the source.

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. Aug 3, 2007

### alvaros

Concepts ( like: time, space, coherence length, sizeless particles, wavefunction ) provoke endless discussions. This forum proves it.

1 -The experiment doesnt change with the distance source-double 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 photon-itself ?

JeffKoch:
its not homework, just curiosity.

19. Aug 5, 2007

### alvaros

I think that a experiment from 100 years ago is easy.
And also its easy to give an answer.

I repeat my last post for your convenience:
Sorry, I forgot a possible answer: nosense. ( true / false / nosense )
Thanks.

20. Aug 5, 2007

### meopemuk

Hi alvaros,

It is well-established that free photons of visible light do not interact with each other (there is a photon-photon interaction predicted by QED, but it is much too weak to have any effect on interference experiments). So, for the interference picture, it doesn't matter what is the intensity of the light source (the number of photons per second). It could be a super-powerful laser, or it could be a source emitting one photon per year. The accumulated interference picture will be exactly the same. This fact has been proven by experiment.

Eugene.

21. Aug 5, 2007

### alvaros

meopemuk:
I assume this is the answer to my first question:
True

But
says nothing about how must be calculated the highest power to prove:
Yes, everybody would agree that one photon per year is enough slow rate, but this is not a satisfactory answer for a physicist.

I infer that there is no way of knowing where is the boundary between interference (photon-another photon) and (photon-itself) and I cant understand why nobody says it clearly.

22. Aug 5, 2007

### Staff: Mentor

But does such a source exist? I don't remember ever reading about one. Are there any photon sources (even idealized, gedankenexperiment-type sources) that are not probabilistic in nature?

23. Aug 5, 2007

### meopemuk

alvaros,

in physics we have statements of various degrees of plausibility: from "absolutely true" to "absolutely false" and everything in between. I think that the statement "a photon interferes with itself" belongs to the category "most likely to be true". Of course, one can doubt that and imagine (in the case of the source emitting 1 photon per year) that the photon absorbed by the photographic plate 1 year ago has some influence on the photon that we are shooting today. However this assumption is very weird, and not many people would take it seriously. You are welcome to pursue your line of reasoning, but I doubt that it will lead you to any better understanding of physics.

Eugene.

24. Aug 6, 2007

### alvaros

meopemuk:
I understood.

jtbell:
This is not the main question, but thanks for teaching me something.

25. Aug 6, 2007

### JeffKoch

Apparently not.

If the source is a point source, then it does change because the flux of photons incident on the slit scales as 1/R^2. The source/slit distance only matters to the extent that it affects the number of photons per unit time per unit area at the slit.

True.

As a reasonable first guess, consider a thermal line-emission source. Filter it so that you're almost certain that all the photons hitting your slit are from a single line transition. Measure the power per unit area at your slit plane, and calculate the number of photons incident in the area of the slit per unit time (treat them as photon bullets). Look up the spontaneous emission lifetime of this transition, which is the 1/e time constant of the exponential photon emission probability distribution. Then, attenuate the source emission until the calculated number of photons incident in the area of the slit is 1 per spontaneous emission lifetime. This is the boundary between "multiple hits" and "single hits" - for extra confidence, attenutate the source by another factor of 10 or 100. At that point you're pretty certain that only one photon at a time is incident on the slit.