Hi,
I'm so confused with what a photon is...
Are photons short waves? Can a photon have more than one color/hz ?

Hi,
I'm so confused with what a photon is...
Are photons short waves? Can a photon have more than one color/hz ?
A photon can be viewed as both a wave and a particle, depending on how an experimental setup interacts with it; this is true for all quantum objects.Are photons short waves?
No, a single photon has a welldefined wavelength, so it can't have more than one "colour" at a time. However, the wavelength can change, if that is what you mean.Can a photon have more than one color/hz ?
So, a photon is 1 cycle of a specific wavelength?
This is not true. A photon has a single wavelength or frequency if and only if it is precisely a momentum or energy eigenstate. For an example of a photon that does not have a single frequency, if a shortlived excited state emits a photon, the frequency of that photon will be broadened by the form of the Heisenberg uncertainty principle. This is a significant limitation in nuclear magnetic resonance (NMR) spectroscopy. Because the energy separation between the nuclear spin states is quite small, the photons are low (radio) frequency, and therefore are significantly broadened by the HUP, which results if the relaxation time is too short. But if the relaxation time is too long, then the signal will too readily saturate because the populations of the two energy levels are almost equal (at the temperature typically used), also because the difference in energy of the two levels is quite small. The solution to the problem is a more powerful magnetic field, which increases the separation between the two levels, both reducing the significance of the HUP, and improving the population statistics.
That's true, but I think the OP had a much simpler scenario in mind  a single photon with a well defined energy in free space. At least that is how I understood the original question. I am aware that once we consider the intricacies of emission mechanisms and atomic spectra, then things become a lot more complicated...
It's not just that. The OP asked what a photon looks like, and a photon looks like whatever shape it is given. And given its shape, its frequency and wavenumber will be the corresponding Fourier transformation. Thus, if a photon has compact support in any dimension, the corresponding wavenumber will not have compact support (e.g. a photon of single frequency cannot be a singlecycle pulse). In the doubleslit experiment (in the case of an interference pattern being produced), the photon is actually two separate pulses, which is what is meant by the photon passing through both slits.
I felt the need to stress that a photon does not have a single frequency (or wavelength).
One remark I need to make and it's not clear to me how it affects what I said above, is that photon number is not necessarily a definite quantity. Electromagnetic radiation with a welldefined electromagnetic field (in particular, the phase), e.g. coherent radiation, has an indefinite photon number. Conversely, a quantum state with a definite photon number has an indefinite electromagnetic field (and is incoherent). Thus, a photon doesn't have a wavefunction. I suspect that this differs from other particles because photons do not possess a charge of any type, and is something I do not properly understand.
See, I still don't know what a photon is, how it can possibly behave like a particle, how it can possibly contain more "information" than a single cycle of a single wavelength... Aren't photons the minimum possible quantity of light?
My caveman brain ain't helping.
KJW is very true in his statements. However photon or EM field does have wavefunction (we usualy call it state of field).
The question what single photon looks like is actually very good question because we encounter problems of formal definition of photon. What quantum optics mean by one photon (or state with definite umber of photons) is usualy this
where is creation operator that puts one photon of precise frequency into state it operate on. In this case it operates on vacuum and resulting state is so called single photon state. This is often very good approximation. But the problem is that since frequency = energy is precisely defined this photon has to be infinite in both time and space (since k vector is precisely defined too). Also if you try to measure say electric field in this state you will get zero because
where a is annihilation operator (hermitian conjugate to creation operator).
therefore
because states with precisely defined photon number are orthogonal and a acts on state by substracting photon of defined frequency and also .
What this means is that these states that we call single photons are not real physical states. From properties of Fourier transform it is clear that if we want photon that is localized in space we need to have some distribution of frequencies/wavevectors. Eg. lorenzian probability of localizing photon in space will result in lorenzian probability of measuring frequency.
Now if we want to have something that we want to call photon in sense of real "particle" that is coming to us from say supernova we have to have some distribution in frequency. We define creation operator for bundle with one photon.
where is some normalized distribution in frequencies.
Problem is that if we measure E field te same way as before we will still get zero.
On our pathway to real photon state we have to define one more thing. That is coherent state. Without any math I will just write here that coherent state satisfies
and
where is complex number and therefore it immediatelly imply that
These are first real physical states that we come across. This for example can describe continuous laser radiation.
If we simply take this form and instead of standard creation operator we put in "creation operator for bundle with one photon" we get
which is coherent wave bundle that among other things can describe real single photons.
Last edited by Gere; October 7th, 2013 at 05:04 PM.
Well photon is just some object that has harmonic electric and magnetic field, is somewhat localized in space, posseses momentum and energy an also has angular momentum either or . It doesn`t contain information it IS information. In same way that Morse code was used in naval communication by blinking light photons are used by humans in same way. Also yes. Single photon is single cycle of wavelenght and yes they are minimum possible quantity of light. It behave like a particle because it is somewhat localized and it possesses momentum, nothing more nothing less.
So an orange photon can be 600nm long?
Well orange isn`t a pure spectrum color but say red photon will be 600+ nm long.
I read through the interchange between Markus and KJW, then came to the detailed exposition by Gere, totally bewildered by it all and in a general state of awe that the gentlemen (or gents/ladies) involved understand this as thoroughly as they do and can express it with mathematical precision. And I smiled, rather glad I had chosen the field of geology where the position of the rocks is clearly defined because the damn things move so very slowly and you can check out whether they are made of particles or a waves by hitting them with a hammer. So, nandoanalog, I recommend basalts and blueschist facies and bellerephon ammonites as a field of interest. Your 'caveman brain' will feel at home.
I have no idea what you just saidI recommend basalts and blueschist facies and bellerephon ammonites
Yes. Continuous beam is continuous flow of a lot of photons. What I meant by single photon being single cycle of wavelenght is that single photon is single period which is propagating in space if that makes sense. For example you can`t have attosecond pulse combined of 600nm photons because their wavelenght is just too long.
But how could something which has a duration of a single period or the size of single wavelength be described as monochromatic  i.e. with a small spread of frequency components?
If one considers a radio wave with a wavelength of 1000m, are you proposing that such a wave is represented by photons which are a 1000m long? (or with a duration equal to the corresponding period.) If you are, how could such a wave be used to transmit a signal over a distance of, say, 200m?
Regarding the original question:
We can't see photons and it is unlikely that anyone ever will "see" one. No one can say what they look like. The question doesn't have much meaning. All we can do is list the properties of photons that are measurable or calculable and note how properties such as energy and momentum are related to the properties of corresponding electromagnetic waves.
Yes exactly. I don`t know how radio work exactly but say we have two molecules 200 meters next to each other, one excited one in ground state. Energy difference between state is proportional to say 1000 m. The excited molecule will deexcite and emit a photon in direction of ground state molecule. The ground state molecule will begin to detect changes in E field WHILE the excited molecule is still goes through deexcitation (ie. is emitting photon). After and only after the ground state molecule detects full period of E changing (ie. sees full photon) will it be excited to higher state.
This has to happen like this for various reasons but to be honest the exact dynamics of excitation and deexcitation and creation of light aren`t really clear to me too.
This is exactly the problem with photons and their formal definition. Every photon by nature has to have some distribution of probability of measuring it`s energy. Therefore that "spread" in energy leads to spread in frequency which leads to spread in wavelenght which leads to probability distribution of finding photon in time.
The precisely monochromatic photons are just components of planewave basis of Hilbert space from which real states of light are composed.
I agree with jonG.. you cannot see how photons look like.. here are the reasons:
*Suppose if you want to see how a train looks like, either you can go along the length of a train, or you can stand and let the train move across in front of you.. you could see how a train looks like, all along its length..
*Suppose imagine if a pulse of light is travelling across in front of you ( in a similar way as the train moves across in front you). In this case you cannot see the moving pulse of light for some reasons:
1)you see the moving train's length because light is reflected from it, in other words, you use light to see it.. but you cannot use 'light' to see a 'light' moving across.
2)you cannot see a pulse of light from anywhere you wanted..!! you could see it, only when it comes straight towards you. Suppose if it travels in positive direction of X axis, you could see it only in X axis, appearing as a point; that too only when it approaches you, but not when it is moving away from you.
These reasons limit us that we cannot see the waves/particles of light. But you can make an assumption that they behave as waves or particles.. By the term 'light moves across', it means that, 'energy is moving across'. You introduce some situations, such as twoslit, and say it undergoes interference, which means they behave as waves.. Sometimes you introduce other situations like black body radiation, and say that it behaves as particles.. And one has to understand that, particles in this sense, doesnt mean classical particles, but that the energy is quantized, so that it could only be absorbed or emitted in packets> particle nature.
Another thing i would like to stress is that, consider uncertainty principle. when you try to measure the position of a particle more accurately, you must use light of smaller wavelength, which has higher energy that it disturbs the particle in a way that you cannot predict the original momentum of it.. but that doesnt mean that a particle cannot have position & momentum at the same time; it only says that we cannot measure its original values simultaneously.. So, the way you try to measure something, interacts with that, and gives you a conclusion which is actually due to such interaction.. By this way, you could only CONSIDER it to be particles or waves, only under such "situations"( that you introduce).
A photon is not a single cycle of a specific wavelength, as some have discussed above!!.. In a given time, a light of low frequency and a light of high frequency both cover the same distance. But, the low frequency light has cast out few waves(or cycles of wavelength) along the distance, while the high frequency light has cast out more number of cycles of wavelength through the same distance. Then, if one photon is one cycle, it means that in a low freq light, few photons have reached the distance, and in high freq light, many photons have reached it..!! So, this is not true.. What actually means is, a photon(or a quantum) represents the amount of energy travelled in a specific period of time. By this, it means that, in a given time, in a low freq light and in a high freq light, same number of photons have crossed the distance, but the photon of high freq light has many cycles in it (compared to a photon of low freq light).. And hence, in a given time, in high freq light, more energy has crossed the distance(=more number of cycles), but equal number of photons, compared to that of low freq light.. So, a photon is not a single cycle of a wavelength, but an integer number of cycles!
There are two things wrong with this:
(1): The number of photons depends on the amplitude of the electromagnetic field (proportional to the square of the field).
(2): All photons are relativistically identical in the sense that a photon of one frequency (or wavelength) is a photon of a different frequency (or wavelength) in a different frame of reference. Thus, if a photon is N cycles in one frame of reference, it is N cycles in all frames of reference, and therefore it is N cycles regardless of the frequency (or wavelength).
Last edited by KJW; October 11th, 2013 at 04:47 AM.
*I specified: For whatever the frequency of light, in a given time, the number of photons is constant(=number of photons is independent of freq)
*frequency= number of cycles of wavelength in unit time. so, a photon of high freq light has more cycles(high freq, more energy)
*A photon of one frequency in one frame of reference, is of different freq in other frame,, ofcourse, this is due to doppler effect.. and had i mentioned freq of a photon as constant?
* my conclusion is: a photon is not necessarily a single cycle of wavelength, but an integer number of cycles of wavelength( which is same as i stated above)
*To KJW: i dont know where is the mistake seriously..
Guys, don't we have sensors that can detect single photons? Can't frogs also see single photons?
I am not a scientist. But I think some responders are over confusing the OP.
I can not answer the wave form question... but:... is an invalid question, if you do not add the specifics of the observer.Can a photon have more than one color/hz ?
You need to add all the specifics for the situation. If the observer is moving, perhaps irradicly, then the light observed but the observer, provided the light seen and measured, with the same consistancy, will shift in color. Red and Blue shift.
However,... as lightspeed is still the biggest speed in our observable universe, the observer would need to move faster than the photon in order to observe it continuesly. At the moment the photon is observed though,... you as the observer become part of the observation and will effect the photon. As I understand it photons are funny things. I lack the knowledge to go into what happens at the moment of observation, as this indeed is one of the fundamental questions.
But to answer the question, can a photon have more then one color/hz.... I would say... this depends on the observer. If the state of the observer is different at the moment the photon is observed,... then yes the photon will have a different color,... because besides the single photon, the observer needs to be included in the observation calculations too.
But... I might be way off. My scientific knowledge does not go that far.
I would like to add though... Does one tree make a forrest?
KJW actually accommodated relative motion a couple of posts ago.
Why is continuous observation necessary? What does it even mean, in your view?However,... as lightspeed is still the biggest speed in our observable universe, the observer would need to move faster than the photon in order to observe it continuesly.
Photons aren't just funny, they're hilarious (especially after a few pints). Perhaps you are thinking of the "observer effect" already alluded to in an earlier post, in connection with the HUP? It is true that any practical measurement method will perturb the thing under measurement, but that shouldn't prevent answering the question of whether a photon can have more than one color.At the moment the photon is observed though,... you as the observer become part of the observation and will effect the photon. As I understand it photons are funny things. I lack the knowledge to go into what happens at the moment of observation, as this indeed is one of the fundamental questions.
I added the continuation of the observation... to explain that the colour of photon in observation is irrelevant,... if you do not add the specifics for the observer. If the conditions for the observer changes, so will the observation for that specific photon. ... I could not think of a better way of explaning myself so quickly.
And I do think it is relevant, the observer. Is the observation not one of the problems?
I did not want to mention it trying to keep it simple... most likely not being a scientist and lacking a fundament to explain it better then I can understand it myself...
... but I actually had the problem of quantum mechanics in mind, known to effect this:
Doubleslit experiment  Wikipedia, the free encyclopedia
And as we particularly are talking about a single photon here (atleast that was my assumption), and not a stream of photons from the same source and conditions, observed at a particular time and place by a single observer,... I tried to explain it from a changing observer perspective, trying to observe that one single photon various times.
Because,.. what happens to that photon,... once it has been observed by the observer?
Heck,... Im probably making it more difficult... aint I? ... and I tried to avoid that. Well I did say I am not a scientist. But I honoustly try my best too, too understand it (all).
The very concept of framedependence already embeds within it a concept of measurement. What one sees (observes) is framedependent. Thus, KJW's comment about it already subsumes the relevant observer dependence, and without the unnecessary complications of "the observer effect," with which too much quantum woo is associated. Observation is not one of the problems; it is a fundamental part of the experiment, and one which has already been accommodated from the start.
The doubleslit experiment has little or nothing to do with the OP's question (at least, not as far as I am able to discern). I think you have probably been misled by some popsci article or TV programme that took a quantumwoo approach to the question of what an observer does.I did not want to mention it trying to keep it simple... most likely not being a scientist and lacking a fundament to explain it better then I can understand it myself...
... but I actually had the problem of quantum mechanics in mind, known to effect this:
Doubleslit experiment  Wikipedia, the free encyclopedia
And that is precisely the set of conditions I was assuming as well. And again, as KJW noted, what one sees is frame dependent. You will get, e.g., a wavelength shift. If whatever you observe is emitting multiple wavelengths, you will still observe multiple wavelengths, even in a different frame. That is the key point you are missing. Thus, the "observer effect" has no effect on the qualitative character of whatever solution one ultimately arrives at. If it starts off as a single wavelength, that's what you'll see in another (inertial) frame. If it starts off as a multitude of wavelengths, that's again what you'll see.And as we particularly are talking about a single photon here (atleast that was my assumption), and not a stream of photons from the same source and conditions, observed at a particular time and place by a single observer,... I tried to explain it from a changing observer perspective, trying to observe that one single photon various times.
It can be wholly or partially absorbed or reflected. It could end up shifted in wavelength.Because,.. what happens to that photon,... once it has been observed by the observer?
Yes, but for a good reason  you are eager to apply things you've heard about, and to use that knowledge to help others. But before offering advice on how to explain something, it's best to have a somewhat deeper understanding first. You didn't realize that KJW had already raised the issue you raised. And even with further explanation, you still didn't quite realize it. But I have faith that you will get there.Heck,... Im probably making it more difficult... aint I? ... and I tried to avoid that.
I would like to expand upon this point, for the massless nature of the photon makes it quite special in comparison to massive particles. In the case of a massive particle, observation from different frames of reference are distinguishable simply by the particle having different speeds. Not so in the case of photons which have the same speed in all frames of reference. Thus, suppose there are two observers, A and B, in different frames of reference, and there are two photons of frequency and in A's frame of reference. Suppose that B observes A's photon as having frequency . Then there is nothing that distinguishes A's photon from B's photon. They are identical in every respect (even having the same speed). But B's photon is A's photon, and thus we can conclude that there is a simple relationship between photons of different frequency, noting that and are arbitrary (they can be as low or high frequency as we choose).
Now I got my answer! All photons look the same!
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