2
$\begingroup$

My admittedly limited understanding of the Schrodinger equation / uncertainty principle implies that as we narrow the range of possible values for the momentum of a photon, we necessarily increase the range of possible positions of the photon. Specifically, if we know the momentum of a photon with certainty, then its position is essentially spread all over space (see here http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html).

However, if we know the location of the source of a photon, and we know its momentum with certainty, then the only possible locations for the photon are along a straight line from the source, pointing in the known direction of its momentum. Further, because the photon must have a velocity of $c$, its location can't even be spread about that line, since we can simply calculate how far its traveled, based on how long it's been in flight.

This seems to suggest that if we know the location of the source of a photon, and we know its momentum, then we know its position at all times. Common sense agrees. However, the uncertainty principle suggests that the possible locations of the photon should in this case be spread all over space.

How do we reconcile this?

$\endgroup$
3
$\begingroup$

The Heisenberg uncertainty principle says that as per QM, it is any of a variety of mathematical inequalities, giving a fundamental limit to the precision with which certain pairs of physical properties of a particle can be known.

In your case, the particle is the photon, and the properties are position and momentum.

Now when the detector detects the photon's position, you can say that at that moment, the position is known. You are asking whether we can know the momentum at that moment too. The Heisenberg uncertainty principle specifically says that we cannot. The moment when the photon's position is known is when the detector shows it.

At that moment, what happens, is that the photon interacted with the detector screen's atom. That atom absorbed the photon, and the photon's energy transformed into the kinetic energy of the absorbing electron, that moved to a higher energy level as per QM.

Now at that moment, when the photon's energy was transformed into the kinetic energy of the absorbing electron, the photon seizes to exist. Its momentum cannot be interpreted anymore. What you know is that in the past, the photon's position, at the moment of absorption, was known.

But you cannot know it's momentum at the same moment. The momentum is a vector quantity, and since the photon seizes to exist in the moment of absorption, there is no momentum to measure anymore.

If you would interpret the position of absorption as the last known position of the photon, then you could try to find out its frequency at that moment (in the past). What you could check, would be the frequency of the photon maybe. To do that, you would need to check whether the absorbing electron moved to a certain energy level from its ground level, check the difference between the two energy levels, and that could be the past frequency of the photon. You could interpret that as the last known frequency of the photon.

But even that (the frequency) would not be momentum. Momentum is a vector quantity, and frequency is not. And even if you could figure out the energy difference of the absorbing electron's levels, that might not be with certainty the frequency of the original photon. There might be multiphoton absorption when a photon is absorbed by multiple electrons, or the electron might relax in multiple steps, and that would make it harder to check the original photon's frequency.

But the Heisenberg uncertainty principle would still work since you cannot know the position and the momentum at the same time.

Now that is for the time of the absorption.

You could say that you would like to find out the position and the momentum of the photon in flight. That is not possible either, because the photon is accepted to propagate as a wave. How would you measure the position of that wave? It is not possible. How would you measure the momentum of that wave? The only way to measure the photon's properties is to interact with it.

Now you could use inelastic scattering to detect a photon. It is very good to learn about the double slit experiment when there is a detector filter on one of the slits. They can use the detector to check whether the photon went through a certain slit. That creates an inelastic scattering, where the photon gets inelastically scattered off the detector's atom. But even then, that detector will not give you a certain position for the photon. And you cannot know it's momentum at all. That is the problem with inelastic scattering, it changes the energy and phase of the photon, and the photon changes angle too (momentum is a vector so with the change of angle the momentum changes too).

Now the only way to measure the position of the photon with certainty is to absorb it. But absorption will seize the photon's existence too, so it's momentum is not interpretable.

$\endgroup$
  • $\begingroup$ Thanks! I understand that the photon propagates as a wave, and during that time, it is impossible to say what its momentum is, without interacting with it. However, once the photon hits the detector, interacting with the detector, I believe it has a definite momentum with which it strikes the atoms within the detector, which is evidenced by Compton Scattering. The magnitude of that momentum, as you note, can be easily calculated once we know its energy. How could the direction of the momentum (upon collision) be anything other than the straight line that connects its source to the detector? $\endgroup$ – Feynmanfan85 Sep 3 '18 at 17:37
  • 1
    $\begingroup$ You are correct that whenever it hits the screen, that is the detector in this case, a pixel will light up, but that in this case is not scattering but absorption. Only in the case of absorption could we tell the exact energy of the past photon with its last known position. The problem is that the momentum is a vector. Now as the photon travels in air. It interacts with air molecules, and that is elastic (Rayleigh) scattering. $\endgroup$ – Árpád Szendrei Sep 4 '18 at 17:19
  • 1
    $\begingroup$ The photon keeps its energy, phase, but changes angle. The photon usually does not take a straight path, but a zigzag random path as it scatters from air molecule to air molecule. The overall path looks like a straight path on the bigger scale because the wavefront keeps its phase. So between the emission (laser), and absorption (screen) you could look at it as a straight path. But when it is absorbed by an atom in the screen, the last scattering with the last air molecule might have come from a different angle, making the momentum vector different then the straight path (of the wavefront). $\endgroup$ – Árpád Szendrei Sep 4 '18 at 17:22
  • 1
    $\begingroup$ You could do this experiment in vacuum and then see if the last known momentum vector could be calculated but I do not know of such an experiment. $\endgroup$ – Árpád Szendrei Sep 4 '18 at 17:23
  • $\begingroup$ "the photon seizes to exist" seizes -> ceases $\endgroup$ – Acccumulation Apr 22 at 17:10
1
$\begingroup$

Since we know the position of the source of the photon, we can determine its trajectory from the source to the detector, once we know its final position on the detector.

Photons don't have trajectories. It's perfectly possible to consider a situation where you know the starting point of the photon to a high accuracy (say, you know it was produced at a molecule which is tightly bound at a specific location) and that you know the position at which it was detected (you know which pixel clicked), but that knowledge does not tell you anything about what happened between those two times. If you are not actively performing a projective measurement on position (resp. momentum) at a given time $t$, then QM is explicitly restrained from saying anything about that position (resp. momentum) at that time.

I see nothing that would prevent, as a practical matter, such a detector from also giving us the energy of the photon to arbitrary precision, subject only to practical, technological constraints.

Since the speed of light is always $c$, and the formula for the momentum of a photon is simply $h/\lambda$, it follows that we can calculate the position, and momentum of the photon, without any apparent constraints that relate these two values.

It's perfectly reasonable to add a spectrometer behind your detector pixel so that you also do a projective measurement on frequency, and this does give you a definite value for the photon momentum.

However, that procedure doesn't tell you anything about the direction in which that photon is headed, so you don't have any real knowledge of the photon momentum, which is a vector quantity.

Moreover, the fact that you're post-selecting on the photon frequency means that the initial pretense that the photon frequency was undetermined can be stripped away in this context, and your configuration is equivalent to one in which you knew perfectly well (say, because of the way you've tuned your molecule's emission lines) what the photon frequency was. In that picture, you have a molecule emitting a narrow-band single-photon wavepacket, which immediately proceeds to fill up all of the available space in whatever Helmholtz mode is driven by the transition, evolving into a complicated spatial wavefunction with a large spatial and momentum support, which covers a large fraction of your detector, after which one of the detector pixels clicks.

And once you frame things like that, it should be clearer that the information offered by that detector click is rather limited, since the next time you re-run the experiment you'll get a click from a different pixel.

$\endgroup$
  • $\begingroup$ Thanks! I understand that the photon propagates as a wave, and during that time, it is impossible to say what its momentum is, without interacting with it. However, once the photon hits the detector, interacting with the detector, I believe it has a definite momentum with which it strikes the atoms within the detector, which is evidenced by Compton Scattering. The magnitude of that momentum, as you note, can be easily calculated once we know its energy. How could the direction of the momentum (upon collision) be anything other than the straight line that connects its source to the detector? $\endgroup$ – Feynmanfan85 Sep 4 '18 at 8:33
  • 2
    $\begingroup$ "I believe it has a definite momentum" - it doesn't, and no amount of "believing" will change that. The photon is not a particle (it is an excitation of the electromagnetic field) and it does not have a trajectory, which is why your "straight line that connects its source to the detector" intuition fails. In QM, if you are not actively performing a projective measurement of a quantity (and in your case, with momentum, you're not) then it doesn't have a well-defined value. $\endgroup$ – Emilio Pisanty Sep 4 '18 at 9:21
  • $\begingroup$ Thanks, so in the case of Compton Scattering, the electron will gain momentum from an incident photon. Are you saying that when we measure the change in frequency / direction of the incident photon, we won't get a definite value? That doesn't sound right. $\endgroup$ – Feynmanfan85 Sep 4 '18 at 11:27
  • 1
    $\begingroup$ I'm saying that you cannot mix-and-match the results of different experiments. It is possible to perform Compton-scattering experiments where the photons have a well-defined momentum (and, when you do them, you lose the ability to talk about their position). The experiment you described isn't one of those, so you don't get to use the conclusions that are specific to those experiments. $\endgroup$ – Emilio Pisanty Sep 4 '18 at 11:30
  • 1
    $\begingroup$ That's already been answered. Given the scant evidence that you're asking in good faith or that you're interested in any answer that doesn't agree with your existing misconceptions, I'm stepping out of this conversation. $\endgroup$ – Emilio Pisanty Sep 4 '18 at 13:06
1
$\begingroup$

A simple explanation:

To know where the photon is, we need to know exactly when it is released. Suppose $\Delta t$ is our uncertainty in this time. The uncertainty in position is thus $\Delta x = c\Delta t$. Meanwhile, our uncertainty in its energy is at best $\Delta E > \hbar/\Delta t$. Since for photons energy and momentum have a simple relation $E = cp$, we have $c \Delta p > \hbar/ \Delta t$, so $\Delta p > \hbar/c \Delta t = \hbar/ \Delta x$ and everything is consistent.

$\endgroup$
1
$\begingroup$

Photons are elementary particles and obey quantum mechanical constraints in production and detection. The Heisenberg uncertainty principle is a lower bound to the product of position and momentum.

$$dx \, dp>h/{2π}$$

If you substitute the numbers, you will see that any measurement done in the lab will obey the "larger than" of the inequality.

$h/{2π}$ is of order $10^{-15}\: \mathrm{eV \, s}$, a very small number

Detectors can give vertices of the order of microns, and if the energy of the photon is smaller than a few electron volts it will not interact with the atoms and molecules.

In addition, the original energy levels from which the photon is produced have a width, which introduces uncertainty in frequency.

Take the $\pi_0 \to \gamma \gamma$, the error in the vertex of the e+e- pairs adds to the indeterminacy of original vertex as well as the error in the momentum of the leptons.

pi0 pi02

If you take the trouble to do the calculations, you will see that the HUP is fulfilled automatically macroscopically. It is only on direct quantum interactions that it has a restrictive role.

$\endgroup$
1
$\begingroup$

This seems to suggest that if we know the location of the source of a photon, and we know its momentum, then we know its position at all times. Common sense agrees. However, the uncertainty principle suggests that the possible locations of the photon should in this case be spread all over space.

I'm not sure what your confusion is. If we know the location of the source with infinite precision, and we know its momentum with infinite precision, that would violate the Uncertainty Principle. So you're saying "If the Uncertainty Principle didn't hold, then we would have a situation that contradicts the Uncertainty Principle".

When a photon is emitted, the emitter has some width, so we have some uncertainty as to its original position. And there is some uncertainty as to its direction, as well. If you could somehow make an emitter of infinitely small width, then the photon's wave would spread out in all directions. You can see that in the single slit experiment: as the slit gets smaller, the photons get more and more spread out.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.