Free particle probability distribution

In summary: No, I'm not saying that. If you want to know the answer to that question, you will need to do some more research. I am just providing a summary of the experiment.
  • #1
zekise
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Abstract:
If a laser shoots photons at a pinhole with a screen behind it, we get a circular non-interference pattern on the screen.

Is this distribution Guassian, and if not, what would its wave function be?

=====================

Assume a double-slit like experiment, but instead of double slits we have a single pinhole. If you draw a straight line from the laser to the pinhole to the screen, you get a point, which will be the origin x = 0 on the screen.

If we fire a large number of photons at the pinhole, we will get a (non-interference) pattern on the screen. My guts tell me it is a Gaussian distribution, with mean at x = 0.

If this is incorrect can you please give me the one dimensional (in space, or two dimensional) wave function for the particle?

If it is a Gaussian distribution, should not the screen be circular or spherical with center at the pinhole?

[This post has been edited by a mentor to remove some off-topic speculation]
Zekise
 
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  • #2
What exactly is your question?
It's much easier if you ask one question at a time.
 
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  • #3
I have only one question. I will make that clearer.

If we shoot photons at a pinhole with a screen behind it, we get a circular non-interference pattern on the screen.

Is this distribution Guassian, and if not, what would its wave function be?

Thanks
 
  • #4
zekise said:
I have only one question. I will make that clearer.

If we shoot photons at a pinhole with a screen behind it, we get a circular non-interference pattern on the screen.

Is this distribution Guassian, and if not, what would its wave function be?

Thanks
You get a pinhole diffraction pattern:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cirapp2.html
 
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  • #5
PeroK said:
You get a pinhole diffraction pattern:
Well yes, but you get a lot more than that. If that is all you get, then the double slit experiment would have no interference pattern. But each slit or pinhole produces a wave that interfere together creating a wide interference pattern. So if there is one slit or one pinhole, you still get the wave, and the particle will still get realized somewhere on the screen, and not just at the center.

Assume your slit is made of ultra thin material thinner than the wavelength. There would be no diffraction pattern, and if there was, it would be almost invisible. So are you saying you will only get a sharp dot on the screen? If so, if I punch a second pinhole close by, I should get a second sharp dot on the screen. But in reality we get a pretty wide interference pattern.
 
  • #6
zekise said:
Abstract:
If a laser shoots photons at a pinhole with a screen behind it, we get a circular non-interference pattern on the screen.
Why do you think it's a "non-interference pattern"? With a laser you should have a pretty coherent state (never single photons!), and thus get an interference pattern.

In Fraunhofer diffraction the diffraction pattern is given by the Fourier transform of the opening, and that's a Bessel function in case of a circular opening:

https://en.wikipedia.org/wiki/Airy_disk#Mathematical_formulation
 
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  • #7
zekise said:
So are you saying ...
I'm not saying anything other than to refer you to a reputable physics website.

What's not clear is where you are getting your theoretical or experimental physics.

You could search for double pinhole diffraction and see what you can find.
 
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  • #8
vanhees71 said:
Why do you think it's a "non-interference pattern"? With a laser you should have a pretty coherent state (never single photons!), and thus get an interference pattern.

In Fraunhofer diffraction the diffraction pattern is given by the Fourier transform of the opening, and that's a Bessel function in case of a circular opening:

https://en.wikipedia.org/wiki/Airy_disk#Mathematical_formulation
I am referring to a double slit experiment. If a laser is unsuitable for that, then please feel free to replace the laser with a single photon emitting source. That is not the issue and we are going on a tangent.

I am not speaking of diffraction patterns or the shape of the slit, another tangent. Are you saying that if we can somehow eliminate diffraction in the double slit experiment (such as by using nano thin materials), there will be no interference pattern? Yes or no?
 
  • #9
PeroK said:
I'm not saying anything other than to refer you to a reputable physics website.

What's not clear is where you are getting your theoretical or experimental physics.

You could search for double pinhole diffraction and see what you can find.
But that website is irrelevant in this case. Why refer to it?

I asked a question. You are now asking the questions. Diffraction pattern is a red herring and a distraction/tangent.

Are you saying that if we can somehow eliminate diffraction in the double slit experiment (such as by using nano thin materials), there will be no interference pattern? Yes or no?
 
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  • #10
Above you said "pinhole", but no problem. You get the famous double-slit interference pattern when using a laser. A laser can never be a single-photon-emitting source. It always produces coherent states, and those have indetermined photon numbers, i.e., they are never single-photon sources.

To know, which pattern you get using true single-photon sources of course depends on the specific state you prepare the single photons in. If you prepare pretty well single-frequency modes you'll also get a diffraction pattern when collecting many such single photons on the screen.

Obviously I don't understand your question, which is not very clearly formulated.
 
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  • #11
vanhees71 said:
Above you said "pinhole", but no problem. You get the famous double-slit interference pattern when using a laser. A laser can never be a single-photon-emitting source. It always produces coherent states, and those have indetermined photon numbers, i.e., they are never single-photon sources.

To know, which pattern you get using true single-photon sources of course depends on the specific state you prepare the single photons in. If you prepare pretty well single-frequency modes you'll also get a diffraction pattern when collecting many such single photons on the screen.

Obviously I don't understand your question, which is not very clearly formulated.
Fair enough. I will try to explain. If a laser is the wrong source, then feel free to substitute that with a single photon source. My question has nothing to do with this laser issue or diffraction issue.

May I ask you what do you mean by a "coherent state"? Thanks.

Please assume we have a perfect apparatus with the proper photon source and which eliminates diffraction to the point that it is barely visible or invisible. And yes, it is a single frequency mode, or whatever, just like a proper double-slit experiment, which is what I declared at the very start.

So I am sending photons to a single pinhole. Just like in a two slit/pinhole model, the wave function would emit from the slit (pinhole - shape does not matter), and diverge, just like a wave diverges. There is no second pinhole to create an interfering wave. At some point, the photon gets realized on the screen. I conclude that the other two folks are saying the photons will all be concentrated in a tiny spot on the screen, and the wave function will not diverge, because that is due to diffraction. I don't thing this is correct, but you can correct me.

So what does the pattern on the screen now look like? It has to be smooth. It has to be fairly wide and dimming as we go to the edge of the circular pattern on the screen.

Question #1- Is this pattern on the screen a Gaussian? If we draw a line going through the center, will we find the number of dots on the line following a bell-shaped Gaussian?
Question #2 - If not Gaussian, then what would be the wave function?

If the question is still not well formed, please let me know.

Thanks.
 
  • #12
zekise said:
Fair enough. I will try to explain. If a laser is the wrong source, then feel free to substitute that with a single photon source. My question has nothing to do with this laser issue or diffraction issue.
What IS the question?
zekise said:
May I ask you what do you mean by a "coherent state"? Thanks.
A coherent state is a state, that it most close to the classical limit. Formally it describes the electromagnetic field from a classical charge-current distribution. Formally, for a single freqency mode, it's the eigenstate of the annihilation operator of the electromagnetic field.
zekise said:
Please assume we have a perfect apparatus with the proper photon source and which eliminates diffraction to the point that it is barely visible or invisible. And yes, it is a single frequency mode, or whatever, just like a proper double-slit experiment, which is what I declared at the very start.
I don't understand what you mean. If you have a single-frequency mode of a single photon, there'll always be diffraction.
zekise said:
So I am sending photons to a single pinhole. Just like in a two slit/pinhole model, the wave function would emit from the slit (pinhole - shape does not matter), and diverge, just like a wave diverges. There is no second pinhole to create an interfering wave. At some point, the photon gets realized on the screen. I conclude that the other two folks are saying the photons will all be concentrated in a tiny spot on the screen, and the wave function will not diverge, because that is due to diffraction. I don't thing this is correct, but you can correct me.
There's always diffraction, even with a single slit. There's no diffraction with incoherent photons. They are described by a mixed state (e.g., by the thermal state, i.e., black-body radiation).
zekise said:
So what does the pattern on the screen now look like? It has to be smooth. It has to be fairly wide and dimming as we go to the edge of the circular pattern on the screen.
That depends on the specific state you prepare your photons in.
zekise said:
Question #1- Is this pattern on the screen a Gaussian? If we draw a line going through the center, will we find the number of dots on the line following a bell-shaped Gaussian?
Question #2 - If not Gaussian, then what would be the wave function?

If the question is still not well formed, please let me know.

Thanks.
 
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  • #13
zekise said:
Question #1- Is this pattern on the screen a Gaussian? If we draw a line going through the center, will we find the number of dots on the line following a bell-shaped Gaussian?
Good, but then it is a question about wave optics, not about quantum mechanics. So you want to have a very narrow hole, such that the wave going through should look somewhat like the (amplitude of the) wave emitted by a spherical point source: ##\exp(ikr)/r##.

The pattern you will get on the screen will not be a Gaussian.
zekise said:
Question #2 - If not Gaussian, then what would be the wave function?
The wave function itself will depend on details like whether you look at a time-harmonic Maxwell equation, or a Helmholtz equation (a "time-harmonic wave equation"), and on the details of the boundary conditions imposed near the pin-hole. But let us assume that the wave function was ##\exp(ikr)/r##.

Then the intensity distribution on the screen will be determined by two effects:
  1. The geometric ##\frac{1}{r^2}## factor describing the energy spread into the volume.
  2. A factor ##\cos(\theta')## where ##\theta'## is the angle between the surface normal and the connection line from the surface point to the pinhole, counteracting the possibility to increase the surface area (of a patch) by letting it wiggle or otherwise increase its area compared to the solid angle covered by the patch.
Hope this answers your question, which seems to be concerned with the global shape of the amplitude/intensity in case of a very small pinhole. For "some boundary conditions," the amplitude of a very small pinhole could also be something like ##\cos(\theta)\exp(ikr)/r##, where ##\theta## is the angle of the connection line from the "current" point to the pinhole, with the surface normal of the screen (that has the pinhole). But I guess such details are currently not your concern.
 
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  • #14
zekise said:
Well yes, but you get a lot more than that. If that is all you get, then the double slit experiment would have no interference pattern. But each slit or pinhole produces a wave that interfere together creating a wide interference pattern. So if there is one slit or one pinhole, you still get the wave, and the particle will still get realized somewhere on the screen, and not just at the center.
You may have been confused by the terminology here. It is traditional to call the single-opening pattern a "diffraction" pattern and the multi-opening patterns "interference" patterns, but they are all interference patterns and they are all caused the same way: by interference between the various paths the wave follows. Even when there is only one opening, the special case that is called "diffraction", there are multiple interfering paths because of the non-zero extent of the opening - the wave passing through one side of the opening interferes with the wave passing through the other side.
 
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  • #15
Hi Nugatory -- Thanks. So let us assume the pinhole is tiny enough and the slit material is thin enough. In a double slit experiment, we do not usually bother with the perfection of the slits and the thickness or composition of the material. Because that is not the point and they have little bearing on the experiment. I have never seen these issues raised in this context, especially in a thought experiment. We see the interference fringes on the screen and nobody raises the objection that there are 2nd or 3rd order phenomena taking place. So I am not sure why the other replies repeatedly bring this matter up? If we have a single pinhole vs. a double pinhole is there a reason to bring them up? I am under the impression that diffraction refers to photons that bounce off the side walls of the slit, but apparently I am mistaken.

So let us set aside these imperfections. Now let's say we have a double slit. We have detectors that discover which slit the electron passed. Then we lose the interference pattern. So what shall we call this? It is neither diffraction nor interference.

Back to the original question. I'll rephrase it. We have a double pinhole experiment and we see the interference pattern. Now we put a piece of tape on one of the two pinholes.

Question 0# - Do we see that the photons spread across the length and width of the screen or detector, fading as we go out the center (call this a circular fade pattern CFP), or do we see all the photons bunched up in a tiny bright center dot?

Question #1 - In case of the former, what is the probability distribution function for the pattern in one dimension? Is it a Gaussian?

For further clarity I'll rephrase this - In a double slit experiment with an interference pattern, we put a detector to discover the path and we lose the IP. Then what do we see on the screen? Two tiny bright dots or a CFP pattern? Yes I am aware that if it is a CFP, it is actually made of two overlapping CFP patterns with centers very close together.

Question #2 - If CFP is not a Gaussian, can you please give me the distribution on the screen or the wave function past the pinhole?

Thank you so much.
Zekise
 
  • #16
vanhees71 said:
A coherent state is a state, that it most close to the classical limit. Formally it describes the electromagnetic field from a classical charge-current distribution. Formally, for a single freqency mode, it's the eigenstate of the annihilation operator of the electromagnetic field.
In a double slit experiment, when we shoot the particle to the slits, the particle (photon, electron. buckyball) must have a condition that a non-measuring observer has lost the path information of the particle. Otherwise the interference pattern will not happen. What is this condition called? Is this coherence?

The particle may be an atom or molecule with no EM field.
 
  • #17
zekise said:
Hi Nugatory -- Thanks. So let us assume the pinhole is tiny enough and the slit material is thin enough. In a double slit experiment, we do not usually bother with the perfection of the slits and the thickness or composition of the material. Because that is not the point and they have little bearing on the experiment. I have never seen these issues raised in this context, especially in a thought experiment. We see the interference fringes on the screen and nobody raises the objection that there are 2nd or 3rd order phenomena taking place. So I am not sure why the other replies repeatedly bring this matter up? If we have a single pinhole vs. a double pinhole is there a reason to bring them up? I am under the impression that diffraction refers to photons that bounce off the side walls of the slit, but apparently I am mistaken.
Yes, you are mistaken. The Quantum Mechanics that describes, for example, electron diffraction and the double-slit experiment is not simply classical mechanics with random collisions against the walls of the slits.
zekise said:
So let us set aside these imperfections. Now let's say we have a double slit. We have detectors that discover which slit the electron passed. Then we lose the interference pattern. So what shall we call this? It is neither diffraction nor interference.
We get the sum of two single slit patterns. In fact, the simplest way of getting which way information is to close one slit. That the double-slit pattern is different from the sum of two single slit patterns is the key point, and represents a fundamental departure from classical mechanics.
zekise said:
Back to the original question. I'll rephrase it. We have a double pinhole experiment and we see the interference pattern. Now we put a piece of tape on one of the two pinholes.

zekise said:
Question 0# - Do we see that the photons spread across the length and width of the screen or detector, fading as we go out the center (call this a circular fade pattern CFP), or do we see all the photons bunched up in a tiny bright center dot?
We get a single pinhole pattern.
zekise said:
Question #1 - In case of the former, what is the probability distribution function for the pattern in one dimension? Is it a Gaussian?
No. It's as descibed in the relevant webpage.
zekise said:
For further clarity I'll rephrase this - In a double slit experiment with an interference pattern, we put a detector to discover the path and we lose the IP. Then what do we see on the screen?
It depends how you detect the electron. If you could practically detect it without distrubing it, you would get the sum of two single slit patterns. But, the detection itself will tend to disrupt the electron path. Unless, of course, you simply close one slit.

Feymans Messenger lecture on QM presents a detailed analysis of this.
zekise said:
Two tiny bright dots or a CFP pattern? Yes I am aware that if it is a CFP, it is actually made of two overlapping CFP patterns with centers very close together.

Question #2 - If CFP is not a Gaussian, can you please give me the distribution on the screen or the wave function past the pinhole?
This is given in the relevant webpage. Note that wven a single slit, if narrow enough, produces as interference pattern. Note also that as you narrow the slit you get a narrower band on screen until the UP (uncertainty principle) becomes relevant. Then, the narrower the slit, the wider the pattern on the screen. Walter Lewin demonstrates this in one of his online lectures.
zekise said:
Thank you so much.
Zekise
QM and the double-slit interference are not mystical phenomena. If you study QM it makes perfect sense.

I would recommend a more structured approach to learning QM than diving in with random questions about specific experiments.
 
  • #19
PS if there is no barrier with a slit or if the slit is sufficiently wide that there is little or no interaction, then the pattern on the screen may well be a Gaussian. Normally when we are talking about single slit, pinhole or double slit interference, the slits are assumed to be sufficiently narrow to disrupt that Gaussian distribution and produce a different pattern. Technically, the free Gaussian is decomposed into bands of dark and light - which get more pronounced the narrower the slit.

The double-slit pattern entails an added complexity on top of the complexity of two single slit patterns.
 
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  • #20
gentzen said:
Good, but then it is a question about wave optics, not about quantum mechanics. So you want to have a very narrow hole, such that the wave going through should look somewhat like the (amplitude of the) wave emitted by a spherical point source: ##\exp(ikr)/r##.
OK -- so even though the photon is in a superposition of states, it's wave function past the pinhole will be governed by wave optics.
 
  • #21
vanhees71 said:
A coherent state is a state, that it most close to the classical limit. Formally it describes the electromagnetic field from a classical charge-current distribution. Formally, for a single freqency mode, it's the eigenstate of the annihilation operator of the electromagnetic field.
Hi vanhees71 -- Thank you for your long post and advice. In the above reply to my question -- I am trying to find the right terminology for a particle that is in a superposition of states w.r.t. observer D (i.e. a detector screen which is embedded in E). As we know, the particle emitted by the source in a double-slit experiment is in a superposition of states before and after it passes through the pinhole.

Is this what is meant by a quantum coherent particle (w.r.t. D the detector)? I.e. a particle in a superposition of states (w.r.t. D)?

Wiki says
When the incident beam is represented by a quantum pure state, the split beams downstream of the two slits are represented as a superposition of the pure states representing each split beam.[17] The quantum description of imperfectly coherent paths is called a mixed state. A perfectly coherent state has a density matrix (also called the "statistical operator") that is a projection onto the pure coherent state and is equivalent to a wave function, while a mixed state is described by a classical probability distribution for the pure states that make up the mixture.
Why does it say "pure state", instead of just "state" or "mixed state"? If by "pure" it means that the beam is NOT in superposition, then the above is incorrect. The beam is in a superposition of states (w.r.t. D), i.e. in a mixed state, even before it reaches the slits, and as soon as it is emitted.

It refers to an "imperfectly coherent path". Is 'coherent' in this context refer to optical coherency? What is a "coherent path"?

Thanks a lot
Zekise
 
  • #22
zekise said:
Why does it say "pure state", instead of just "state" or "mixed state"? If by "pure" it means that the beam is NOT in superposition, then the above is incorrect. The beam is in a superposition of states (w.r.t. D), i.e. in a mixed state, even before it reaches the slits, and as soon as it is emitted.
No, none of what you say here is correct.
You’ve reached the point where you’re going to need a real QM textbook (and to unlearn some things you think you know) to make further progress.

A pure state (what you’ve been accustomed to calling a “wave function”) is a vector in an abstract mathematical vector space, while a mixed state is a more complicated mathematical object, the “statistical operator” or “density matrix”. All pure states can be described with a density operator, but there are also “mixed” or “impure” states that can only be described by a density operator.

All superpositions are pure states and whether a pure state is a superposition or not is just a matter of how we write it - pure states are abstract mathematical vectors and any vector can be written as a sum/superposition of other vectors if we so choose.
 
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  • #23
Yes, I guess I have reached my limits.
 

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