Does the uncertainty principle apply at all to photons?

[Reallyfat]

I know this may sound strange, given that we cannot really work out where in space a photon is because it cannot be completely stopped. But here's a thought.
Let us assume that a photon has been emitted in vacuum going in a straight line. At any given moment in time, this photon will have traveled a distance of

c * t

from the source, where c is the speed of light and t the time since its emission. So technically, our Δx is in fact equal to zero.
Theoretically speaking, it does not matter now what Δp is, because 0 * n will always equal zero, and not a value greater than

hbar / 2

But we might even bring Δp down to 0.
Assume that the photon has been emitted from a monochromatic laser. Such a photon will have a known frequency and wavelength. Given that for a photon,

h / λ

gives us its momentum, we can know its momentum with an uncertainty of Δp=0.
Does the uncertainty principle apply at all to such a photon?

 

[ChrisVer]

0*∞=?
Once you know the position of a particle, it's wavefunction becomes $\delta(x-x_{0})$ where $x_{0}$ the position you measured with extreme certainty.
The wavefunction in the momentum space, is then the Fourier transform of the delta function- that's a wavefunction over all momenta space.

So your problem if you have extreme precision in measurements states that once you know have uncertainty in position 0 you'll have ∞ uncertainty of momentum .So when you measure the photon, you don't have anymore "details" about its wavelength.
That is also partially appearing in normal E/M waves.

http://en.wikipedia.org/wiki/Photon

 

[Reallyfat]

So your problem if you have extreme precision in measurements states that once you know have uncertainty in position 0 you'll have ∞ uncertainty of momentum .So when you measure the photon, you don't have anymore "details" about its wavelength.

Yes, but we already know the wavelength, do we not? I understand that if we tried to fit any parameters into our given equation (Δx*Δp = hbar/2) we would find that either value goes to infinity as the other goes to zero. But then we fall into the same trap again - what's ∞*0? 0 or ∞?
On the other hand, if we do what I did - bringing the parameters manually down to zero and then executing the equation, we would find our answer to be 0*0=0.

The laser is in a vacuum. Photons are emitted in a straight line. Thus dimensions x and z will stay at 0. Only the dimension y will change. This is the distance of the photon from the source. In a vacuum (and we know this precisely) the speed of light is c=299,792,458 m/s. Multiply this value by an arbitrary value for a time t, and the value will be a precise indication of its y-distance from the laser. For 1 second, the position of said photon will be exactly (0,299792458,0). Correct?
Now the laser is monochromatic. Its frequency must be a known, single value. Given the value of c, we can derive a precise wavelength. Let us say this is a UHF laser, with the photon at a wavelength of λ=1 meter. Now to calculate our momentum we can say h/λ. We now know a precise value for momentum too.

For this particular set of parameters I have mentioned, we know:
x=(0,299792458,0) with an uncertainty of Δx=0
p=6.626*10^-34 with an uncertainty of Δp=0

0*0=0, and not a value equal to hbar/2.
We know both values precisely, do we not? How then does the Uncertainty Principle apply?

 

[Cthugha]

Multiply this value by an arbitrary value for a time t, and the value will be a precise indication of its y-distance from the laser. For 1 second, the position of said photon will be exactly (0,299792458,0). Correct?

No. A laser is not a gun. The light field from a laser has some finite coherence time. You cannot nail down a photon "in transit" better than that. It is given by the Fourier transform of the spectrum. So if the energy is well defined, the coherence time gets large. Although it is a common naive picture, considering laser light as an army of tiny bullets is not an adequate model for light fields.

 

[Reallyfat]

I don't quite see how that would apply for a single photon. If my laser only releases one single photon, do we not know the speed and direction in which it travels to infinite precision? What my experiment relies on is calculating position, not interfering with the photon in any way. So we aren't really "seeing" the photon, we are simply calculating where it is.
Perhaps a laser is a bad source for the emission of the photon. But for one photon, I would argue that we can know its position precisely. Consider the photon at any point in space. After any given amount of time, we can know exactly where it will be down to extreme accuracy (provided that our photon travels in a straight line).

 

[Cthugha]

I don't quite see how that would apply for a single photon. If my laser only releases one single photon, do we not know the speed and direction in which it travels to infinite precision?

Lasers never emit single photons. Really creating single photons is one of the most complicated tasks you can do in an optics lab. With reducing laser intensity you can create states which have one photon per unit time ON AVERAGE. But you will always get a photon number distribution. The fields are pretty important. You get photon detection events due to interference effects between the fields from several emitters inside the laser, so you cannot in principle attribute "the photon" to one emitter or the other. Considering photons as ballistically traveling bullets will go wrong.

Perhaps a laser is a bad source for the emission of the photon. But for one photon, I would argue that we can know its position precisely. Consider the photon at any point in space. After any given amount of time, we can know exactly where it will be down to extreme accuracy (provided that our photon travels in a straight line).

Well, you can create light sources which emit single photons pretty well. You can also create single photon states which are pretty localized. However, these are necessarily very broad spectrally. There is no possibility, not even in principle, which allows light sources that create arbitrarily well localized and monochromatic light. You must sacrifice one or the other.

 

[Reallyfat]

Well, you can create light sources which emit single photons pretty well.

Okay then. I have this photon. This photon carries energy of an arbitrary value. Dividing the energy value by Planck's constant gives me a a frequency.
My single frequency has a single wavelength, and a single speed (it is in a vacuum). We can already gauge its momentum easily, dividing the energy by its wavelength λ. That's a precise value for magnitude. As for direction, we assume the photon is going along the y-axis of my three-dimensional space. We thus have a precise value for momentum.
Now if we know where it is in space for example, at (0,0,0) at the point of emission, we can know where it will be after any given amount of time down to a precise value.
Where's the catch?

 

[Cthugha]

Okay then. I have this photon. This photon carries energy of an arbitrary value. Dividing the energy value by Planck's constant gives me a a frequency.

"A photon" just means that you have a state of particle number 1. This photon may be monochromatic or it may be spectrally broad. A single photon does NOT need to have a single wavelength.

If it has a single wavelength, it is necessarily spread out in time. If it is well localized, it is necessarily spectrally broad. This is a simple Fourier transform relationship.

Where's the catch?

Your idea of what a photon is, is a common layman assumption, but it is wrong and leads to wrong conclusions.

 

[Reallyfat]

Fine, what if the photons emitted by the laser were absorbed, and only one allowed to continue?

 

[Nugatory]

I don't quite see how that would apply for a single photon. If my laser only releases one single photon, do we not know the speed and direction in which it travels to infinite precision?

We do not. The word "particle" may be giving you the impression that a photon behaves as if it were a bullet or a grain of sand, except smaller - it doesn't.

With quantum mechanics there's really no substitute for doing the math, but if you want a simple picture that you can visualize, try this:

Light spreads out just like a classical electromagnetic wave until it hits something. Only when it does hit something do you see a quantum mechanical effect: no matter how spread out the light is, its energy and momentum are delivered to a single point on whatever it hits, and then we say that a photon has arrived at ("has been detected at" would be more accurate) that point. It is pretty much random where that point is, except that it's more likely to be where the light wave is stronger than weaker. The frequency-momentum relationship you're using just tells us how much energy and momentum will be delivered at a single point.

This picture isn't right either (as I said above, there's no substitute for the math) but it does make it clear that the photon isn't something that moves from the source to the target like a little bullet. Indeed, it doesn't really make sense to talk about where the photon is while it's in flight - it's not anywhere until the light interacts with the target.

 

[jtbell]

Fine, what if the photons emitted by the laser were absorbed, and only one allowed to continue?

From post #6:

With reducing laser intensity you can create states which have one photon per unit time ON AVERAGE. But you will always get a photon number distribution.

Absorption is a probabilistic process. I would be very surprised if there were a way to absorb "all but one" photon.

 

[ZapperZ]

Fine, what if the photons emitted by the laser were absorbed, and only one allowed to continue?

The single-slit diffraction, which applies equally to photons, is one of the clearest example of the HUP at work.

https://www.physicsforums.com/blog.php?b=4364

Zz.

 

[Reallyfat]

Hm. Alright then. Perhaps I should consider something else. Why not a single electron? We know electron mass. Can we not choose a velocity with which to emit it? That should give us a precise momentum, as well as a position determined by multiplying the speed of the electron by the time since emission. I suppose I'm wrong here too, but maybe it could be possible

 

[ZapperZ]

Hm. Alright then. Perhaps I should consider something else. Why not a single electron? We know electron mass. Can we not choose a velocity with which to emit it? That should give us a precise momentum, as well as a position determined by multiplying the speed of the electron by the time since emission. I suppose I'm wrong here too, but maybe it could be possible

This thread is beginning to feel as if you're making things up as you go along.

Please note that we know a lot about "single electrons", considering that we have single-electron detectors. Many of the same quantum effects that govern photons are also applicable to electrons, single or not.

Zz.

 

[atyy]

The single-slit diffraction, which applies equally to photons, is one of the clearest example of the HUP at work.

https://www.physicsforums.com/blog.php?b=4364

Zz.

As I have said before, I find the statement "I have shown that there's nothing to prevent anyone from knowing both the position and momentum of a particle in a single mesurement with arbitrary accuracy that is limited only by our technology." dubious, given the measurement process proposed.

It is possible in special cases for conjugate variables to be jointly measured accurately. For example, if the particle is in an eigenstate of a particular observable, then measuring that observable will not disturb the state, leaving the same state available for an accurate measurement of the conjugate observable. However, for position and momentum, the eigenstates are not physical states, so this does not apply.

If the position were to be measured more accurately, the momentum distribution would be changed. So although the momentum measurement would remain "accurate", it would be an accurate measurement of a different state.

Also, the claim seems to violate the joint measurement inequality derived by Branciard in http://arxiv.org/abs/1304.2071. As shown by di Lorenzo in http://arxiv.org/abs/1212.2815 there are conditions under which Branciard's inequality does not hold, but these do not seem to be the conditions described in ZapperZ's link.

 

[ZapperZ]

As I have said before, I find the statement "I have shown that there's nothing to prevent anyone from knowing both the position and momentum of a particle in a single mesurement with arbitrary accuracy that is limited only by our technology." dubious, given the measurement process proposed.

It is possible in special cases for conjugate variables to be jointly measured accurately. For example, if the particle is in an eigenstate of a particular observable, then measuring that observable will not disturb the state, leaving the same state available for an accurate measurement of the conjugate observable. However, for position and momentum, the eigenstates are not physical states, so this does not apply.

If the position were to be measured more accurately, the momentum distribution would be changed. So although the momentum measurement would remain "accurate", it would be an accurate measurement of a different state.

Also, the claim seems to violate the joint measurement inequality in http://arxiv.org/abs/1304.2071 .

(1) I'm not going to debate this here in this thread because it will derail the topic.

(2) After the previous thread on this topic, I've shown my scenario to 3 other theorists, who had zero problem with the scenario.

(3) You have completely misunderstood that statement. It is more of the detection techniques that is involved here, not a joint measurement. How well I make my CCD to increase the density of the pixels and to reduce cross-talk, so that I can pinpoint the location where the photon hits the screen has nothing to do with the joint probability measurement of those two. I've measured the location where a photon hits the screen, and I've seen improvement in the accuracy of such determination by improving the detector. This changed nothing about the HUP relationship, it is a detector/technological issue!

I didn't reply and participate in that thread because I can see the difficulty in trying to explain this in this format, and without needing to make multiple sketches.

Zz.

 

[Nugatory]

Hm. Alright then. Perhaps I should consider something else. Why not a single electron? We know electron mass. Can we not choose a velocity with which to emit it? That should give us a precise momentum, as well as a position determined by multiplying the speed of the electron by the time since emission. I suppose I'm wrong here too, but maybe it could be possible

The same principle that I described in post #10 above applies to electrons as well. The big difference is that electrons have an electric charge so they're much more likely to interact with whatever is in their general neighborhood and deliver their energy and momentum to a single point (and then we say "that's where the electron is"). It takes fairly sophisticated lab equipment to avoid this effect and get quantum effects like interference and diffraction out of an electron; that's why we can more often get away with thinking of electrons as if they were like little tiny grains of sand moving around.

 

[atyy]

(1) I'm not going to debate this here in this thread because it will derail the topic.

(2) After the previous thread on this topic, I've shown my scenario to 3 other theorists, who had zero problem with the scenario.

(3) You have completely misunderstood that statement. It is more of the detection techniques that is involved here, not a joint measurement. How well I make my CCD to increase the density of the pixels and to reduce cross-talk, so that I can pinpoint the location where the photon hits the screen has nothing to do with the joint probability measurement of those two. I've measured the location where a photon hits the screen, and I've seen improvement in the accuracy of such determination by improving the detector. This changed nothing about the HUP relationship, it is a detector/technological issue!

I didn't reply and participate in that thread because I can see the difficulty in trying to explain this in this format, and without needing to make multiple sketches.

Zz.

I agree the momentum measurement at large distance is accurate, even if your derivation of it is a little informal. I am mainly concerned that the statement seems to be about joint measurement, so the sentence as written doesn't seem to convey the meaning you intend. At any rate, it's good to get this cleared up that it's a matter of language, and not physics.

 

[Reallyfat]

This thread is beginning to feel as if you're making things up as you go along.

Well, I obviously am, aren't I? My purpose is to find a possible scenario that doesn't fit with this principle. So what do you think - If we had a mechanism that could emit an electron at a particular chosen velocity, could the uncertainty principle be violated? This takes place in a vacuum, under no external influences, keep in mind.

 

[Cthugha]

Well, I obviously am, aren't I? My purpose is to find a possible scenario that doesn't fit with this principle. So what do you think - If we had a mechanism that could emit an electron at a particular chosen velocity, could the uncertainty principle be violated? This takes place in a vacuum, under no external influences, keep in mind.

I appreciate your curiosity, but this kind of question boils down to just one question: If I assume an initial situation that violates the laws of physics, will I be able to violate the laws of physics?

You will be able to learn a lot about physics by finding out what the general principle is that renders your initial assumption non-physical. You will unfortunately not learn much by coming up with lots of "case studies".

 

[ZapperZ]

Well, I obviously am, aren't I? My purpose is to find a possible scenario that doesn't fit with this principle. So what do you think - If we had a mechanism that could emit an electron at a particular chosen velocity, could the uncertainty principle be violated? This takes place in a vacuum, under no external influences, keep in mind.

No, because you need to MEASURE that velocity to determine its value. When you do that, then the corresponding conjugate observable will obey the HUP relationship depending on how well you determined that velocity.

There is no escape.

The thing you are not understanding here is that the HUP is NOT the starting point, but rather the CONSEQUENCE of the quantum formalism. HUP is not fundamental. It is the result of the quantum description.

So unless you can supplant quantum mechanics, you have to live with the HUP.

Zz.

 

[DrChinese]

Okay then. I have this photon. This photon carries energy of an arbitrary value. Dividing the energy value by Planck's constant gives me a a frequency.
My single frequency has a single wavelength, and a single speed (it is in a vacuum). We can already gauge its momentum easily, dividing the energy by its wavelength λ. That's a precise value for magnitude. As for direction, we assume the photon is going along the y-axis of my three-dimensional space. We thus have a precise value for momentum.
Now if we know where it is in space for example, at (0,0,0) at the point of emission, we can know where it will be after any given amount of time down to a precise value.
Where's the catch?

OK, think of it this way. You emit a photon from a laser through a filter which tells us its momentum/wavelength to a high precision. That is allowed. Using a very fast shutter (at least in principle), you localize its position to a high precision. That too is allowed. Does it still have its original wavelength?

The answer is no. Place an identical filter after the shutter and nothing will come through (except perhaps the very rare photon). That's the HUP for you.

 

[Reallyfat]

You will unfortunately not learn much by coming up with lots of "case studies".

More of a thought experiment, really.

OK, think of it this way. You emit a photon from a laser through a filter which tells us its momentum/wavelength to a high precision. That is allowed. Using a very fast shutter (at least in principle), you localize its position to a high precision. That too is allowed. Does it still have its original wavelength?

The answer is no. Place an identical filter after the shutter and nothing will come through (except perhaps the very rare photon). That's the HUP for you.

Say we remove the filter.
Correct me if I am wrong here:

Using my extremely-high-precision-shutter, I can localize my photon with an uncertainty of 0, right? Now, would that mean that my uncertainty in momentum has actually gone up to infinity? Being a vector quantity, the magnitude could not have changed (it carries the same energy), so direction must have changed. With an uncertainty of infinity? Really?
The thing about my experiment is that we are not interfering with the photon. We are simply calculating. How, then can the uncertainty principle apply?
If I worded it as such:
A photon is emitted from a laser in a straight line along the y-axis of a three-dimensional vacuum from the origin (0,0,0). The laser emits photons with a frequency of 300 MHz. After exactly one second, the photon collides with a screen. Calculate its [the photon's] position and momentum at the time of collision.
How can you NOT calculate them with absolutely no (or arbitrarily small) uncertainty?

 

[QuasiParticle]

I think it would perhaps be better to forget about lasers for a moment and to just consider the case of spontaneous emission. That is, radiative transition of an electron in an atom from a higher energy level to a lower. In the textbooks they usually just note that a photon is created with definite energy hf = E2-E1. For some reason spontaneous emission still bugs me, even after having taken all kinds of QM courses. There is something in it that I'm not fully getting. (the photon cannot have a definite energy, what is the state of the electron before & after emission, etc.) I know ordinary QM is not completely capable of handling the situation, but QED is required. However, I'm not sure if even in the "ordinary QM" there is something I don't quite understand.

 

[TrickyDicky]

... the HUP is NOT the starting point, but rather the CONSEQUENCE of the quantum formalism. HUP is not fundamental. It is the result of the quantum description.

So unless you can supplant quantum mechanics, you have to live with the HUP.

Well put.
The HUP is unavoidable as long as one considers objects with particle properties as fundamental. Of course it is not written in stone that you must, you could try "supplanting" QM with a different model that reproduces the experimental results. QFT supposedly would try to do that(it does change certain key things in the QM formalism) if it truly was a theory of fields(of force), but the fact is it isn't really, it is a theory of "particle fields" which is not the same thing, so the HUP is still with us.

Then again we haven't had a true theory of force fields since Faraday's.

 

[ChrisVer]

I will repeat myself...
Once you measure the position with infinite precision , you have absolutely no clue what the momentum of the measured photon is...
If you tell me that the photon starts with some momentum $p$ and I measure it at point $x_{0}$ then you'll probably never know its momentum again- not until you measure it. It can still be $p$ of course, but it can as well be $p+Δp$... In fact the momentum is totally undetermined, it spans the whole p-space.
Why does this happen?
Because the position and momentum quantum operators don't commute. And also because of that, if you measure the $x$ component of the photon's position, you will lose information about the $p_{x}$ component of momentum, not necessarily $p_{y,z}$ since they indeed commute with the operator you measure. Of course that's nonsence, because what you measure is a position vector, so you lose momentum's vector.
As I said this problem corresponds to finding the delta function's Fourier Transform.

Also before the photon collides on your measuring machine, you have absolutely no idea what its position is... In fact the $Δx=L$ where L is the distance between your source and your measuring machine (where the photon scatters or is absorbed to be measured). That's the natural way of choosing the x uncertainty in that thought experiment. What about p? If p is fully determined (that means you can somehow measure it before you measure the photon's position), then the photon's propagation is space is going to be like a wave - you won't know anything about the x.

 

[DrChinese]

...With an uncertainty of infinity? Really?

Sure, as the other approaches zero. That's the experiment! In other words, the HUP is confirmed in practical situation, including observations of entangled particles.

As others say, the HUP itself is not fundamental, it is a consequence of other assumptions.

 

[jtbell]

consider the case of spontaneous emission. That is, radiative transition of an electron in an atom from a higher energy level to a lower. In the textbooks they usually just note that a photon is created with definite energy hf = E2-E1.

Textbooks generally assume that the atom is at rest initially, for simplicity. However, the atom cannot be definitely at rest, nor in any other exact state of motion. The "fuzziness" in the atom's initial state leads to "fuzziness" in the emitted photon's energy.

 

[Reallyfat]

But that doesn't make sense. You are trying to imply that if I know a source which emits photons in a certain direction and with a certain frequency, I can never know its momentum and position with infinite certainty even if I know when it was emitted! If I know its momentum with exact precision, given to me by the frequency of the source, and the direction of emission, will my very calculation will make me infinitely uncertain about its position? That's impossible. Will the calculation of momentum send the photon out into infinity?

 

[ChrisVer]

knowing the momentum p, the probability to find the particle is propagating as a normal wave with momentum p...

 

[DrChinese]

If I know its momentum with exact precision, given to me by the frequency of the source, and the direction of emission, will my very calculation will make me infinitely uncertain about its position? That's impossible. Will the calculation of momentum send the photon out into infinity?

You may be being a little bit dramatic by talking about the calculation thing. But the fact is, as you localize a quantum particle, its momentum does spread wider.

It still has an average value for momentum. But that average value becomes less and less likely to be observed in an experiment.

If you repeat the experiment many times, the predicted distribution will appear. If you repeated enough times, a few extreme values would result as well.

 

[Nugatory]

But that doesn't make sense. You are trying to imply that if I know a source which emits photons in a certain direction and with a certain frequency, I can never know its momentum and position with infinite certainty even if I know when it was emitted! If I know its momentum with exact precision, given to me by the frequency of the source, and the direction of emission, will my very calculation will make me infinitely uncertain about its position? That's impossible. Will the calculation of momentum send the photon out into infinity?

You're still making the same mistake... Photons are not what you think they are, and therefore your entire model of the thought experiment is misleading you.

Photons are always fuzzy and spread out through space so that you can no more specify an exact position for them than you could specify the edge of a cloud of smoke. That's the uncertainty in spatial position (and equivalently, time of emission/detection - when exactly does a cloud of smoke start to make your eyes water?). This is one of the ways that a photon does not behave like a little tiny grain of sand, despite our using the word "particle" to describe it.

The more tightly you constrain the spatial extent of the cloud, the greater the spread of momentum (and equivalently, frequency, energy and wavelength). That's an inherent property of a photon, and another of the ways that a photon does not behave the way the word "particle" suggests that it might.

The calculation itself has nothing to do with the uncertainty and the way that we can trade uncertainty in position against uncertainty in momentum. That's always there, and the calculation just tells us how that tradeoff is working for us in a particular setup.

 

[vanhees71]

Photons are indeed far from anything you associate with "particle". Already massive "particles" are different from miniature billard balls when it comes to situations where you have to use quantum theory to adequately describe them. At least massive particles always have a proper position observable, i.e., there exist self-adjoint operators $\hat{\vec{x}}$ that have the Heisenberg-Born commutation relations with momentum, i.e., fulfill
$$[\hat{x}_j,\hat{x}_k]=[\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk}.$$

This is not the case for massless particles with a spin $\geq 1$. There exists a well-defined momentum operator but not a position operator. For details, see

http://arnold-neumaier.at/physfaq/topics/position.html

 

[Reallyfat]

Not while it's in space, but upon collision with a screen, perhaps?

 

[QuasiParticle]

Textbooks generally assume that the atom is at rest initially, for simplicity. However, the atom cannot be definitely at rest, nor in any other exact state of motion. The "fuzziness" in the atom's initial state leads to "fuzziness" in the emitted photon's energy.

I like this answer. Often, though, the time-energy uncertainty is invoked in this context, which might be the source of my confusion. If I'm not mistaken, spontaneous emission is not allowed in the old quantum theory. But still it is claimed that some of the broadening of the spectrum is due to the time-energy uncertainty.