# Double slit experiment whith one photon

by alvaros
Tags: double, experiment, photon, slit, whith
 P: 166 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.
P: 1,746
 Quote by alvaros 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 ? "
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.
P: 166
meopemuk:
 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.
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.

HW Helper
P: 8,953
Double slit experiment whith one photon

 Quote by alvaros I dont know what is the rate appropiate to call "one photon at a time"
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.
P: 1,746
 Quote by 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.
Excellent answer! I couldn't say it better.

Eugene.
P: 166
all quotes from mgb_phys:

 There isn't really an answer to this, it's all a question of probabilites
.
I understand probabilities, lets suppose that the source emits a photon exactly each certain amount of time, like a clock.

 If ( ... ) there is only 1 photon/second emitted and the experiment is only 1foot long - the photon travels through the experiment in 1ns
( 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 ?

 If not reduce the brightness so there is only one photon/hour or 1/day.
It doesnt seem very scientific.
 Are those odds good enough to convince you there is only one photon?
Im convinced that there is just a photon at a time, this is not the question.
P: 1,746
 Quote by alvaros ( 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 ?
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
 Emeritus Sci Advisor PF Gold P: 11,155 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.
P: 166
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:
 The fact that photons never decay or disappear ...
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 ) ?
P: 1,746
 Quote by alvaros 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 )
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".

 Quote by alvaros meopemuk: The fact that photons never decay or disappear ... 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 ) ?
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.
 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.
P: 166
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
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:
 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.
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:
 Energy of a photon E = h. f Energy of an electromagnetic wave E = 1/2 E . H ( per volume ) Impedance of the empty space E / H = a given number From the tree equations you can derive the volume of a photon
What am I missing ?
P: 1,746
 Quote by alvaros 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.
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.
HW Helper
P: 8,953
 Quote by JeffKoch you can look up it's bandwidth and calculate it's coherence length. This gives you the temporal "duration" of a photon
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.
P: 400
 Quote by mgb_phys You cannot talk about the coherence length of a single photon.
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.
 Sci Advisor HW Helper P: 8,953 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.
P: 400
 Quote by mgb_phys The 'average' of a single photon (smirk)?.
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.
P: 166
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:
 Alvaros, can you clarify what - exactly - the homework question asks? Your reference is to an article, not a homework problem.
its not homework, just curiosity.

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