How does uncertainty princ. apply to photons?

[JesseM]

For a particle moving at a small fraction of the speed of light, momentum is just rest mass times velocity, and since the rest mass doesn't vary, uncertainty in momentum is equivalent to uncertainty in velocity...but what would uncertainty in momentum mean for a photon? Would it be uncertainty about the frequency, or uncertainty about the distance the photon will have moved between two successive position measurements? Maybe you'd need quantum field theory to deal with this question, and I don't know if the uncertainty principle has the same form in quantum field theory as it does in nonrelativistic QM...

 

[Edgardo]

Hi,

I think uncertainty in momentum just means that the photon is diffracted.
For example if light is incident on a slit, the momentum in y-direction changes.


- Edgardo

 

[JesseM]

Hi,

I think uncertainty in momentum just means that the photon is diffracted.
For example if light is incident on a slit, the momentum in y-direction changes.

How would you express this in terms of an equation for the uncertainty relation between position and momentum (or between position and the three components of momentum $$p_x$$, $$p_y$$ and $$p_z$$)?

 

[Edgardo]

Ok, let me try ;)

You know the uncertainty relation (momentum space):
(Delta_x)*(Delta_p) > h
This is often cited in books and on websites.


For the photon incident on the slit:
The slit shall be Delta_y wide, so
you try to localize the photon in an interval (Delta_y).
According to the uncertainty relation you will change the photon's momentum
in y-direction, so (Delta_py).

Therefore the equation is:
(Delta_y)*(Delta_py) >h

:shy:

Sorry for the pseudo TeX formulae, but this is my first time I posted on the forums and I just registered a few minutes ago.

- Edgardo

 

[Edgardo]

Maybe to make things clearer:
the photon flies in x-direction to the right (horizontal) and y-direction
is vertical.


Before the slit:
Photon has only momentum in x-direction and NONE in y-direction.

After the slit:
Photon still has momentum in x-direction BUT a p_y component for the
momentum in y-direction.

-Edgardo

 

[JesseM]

Thanks. But if the magnitude of the total momentum can be determined to arbitrary precision (assuming there is no uncertainty relation involving the frequency), then that would seem to place an upper bound on the uncertainty in the momentum along each axis (for example, if the total momentum is known to be exactly p, then $$p_x$$ must be between p and -p, so the uncertainty in $$p_x$$ could be no greater than 2p). Would this mean there is also a lower bound on the uncertainty in the position? In nonrelativistic QM you normally assume position can be determined with as much precision as you like, it's just that the uncertainty in momentum will get larger and larger the smaller the uncertainty in position.

 

[Edgardo]

First of all, I think I understand now, why you consider the frequency.
p = hbar k = hbar (2Pi / lambda)
And if p changes, does lambda also change? (Did you have the same thought?)
Seems strange to me, now that I think about it...I get an headache .
Because it would be somehow odd if the photon changed its wavelength after passing the slit. This would violate energy conservation.

I also ask myself now if my explanation with the slit is correct
because the photon gets a momentum 'kick' due to the position
measurement. But where does this kick come from (hmm...momentum conservation). I think it's rather the DIRECTION of the momentum vector
that is changed but not its magnitude. Therefore the x and y component of p
are only distributed in a another way. (My second reply to your post
doesnt seem correct then, because px is not changed in my explanation)

2 things that I also don't understand:

Firstly: The change of distribution in x and y means, there's not only
a Delta_py for position measurement Delta_y, but also
a Delta_px. Quite strange for me, because it's not contained
in the uncertainty equation, that Delta_px occurs too.

Secondly: it also follows that Delta_y is indeed bounded, like you've said.
I also don't understand this because the uncertainty equation says
Delta_y * Delta_py > hbar/2. So if I made the slit smaller and smaller,
then Delta_py would go to infinity...




Hmm...I hope someone else can clear things up, lol.
Nevertheless, thanks for this question, it's very inspiring.


-Edgardo

 

[JesseM]

First of all, I think I understand now, why you consider the frequency.
p = hbar k = hbar (2Pi / lambda)
And if p changes, does lambda also change? (Did you have the same thought?)

Yeah, that's what I was thinking of...in nonrelativistic QM the idea of uncertainty in wavelength shouldn't even make sense, because a particle's wavelength really is determined by how the amplitude of the wavefunction is varying over different position eigenvalues, and there shouldn't be any uncertainty in the state of the wavefunction itself. So it seems like the formula $$p = h/\lambda$$ should allow you to find the total momentum exactly, so if the uncertainty was only in the components of the momentum along different spatial axes as you suggested, you'd get that problem of a lower bound on the uncertainty in position. But my guess is that we're just getting nonsensical answers because we're trying to use nonrelativistic QM to understand a photon, which probably isn't kosher...so, anyone here know enough about quantum field theory to explain how the uncertainty principle applies to particles moving at light speed?

edit: hmm, on second thought maybe the answer has nothing to do with quantum field theory, I just remembered that the equation $$\lambda = h/p$$ is just the De Broglie wavelength which is supposed to apply to all particles, including electrons and others that may be moving at nonrelativistic speeds.

 

[JesseM]

in nonrelativistic QM the idea of uncertainty in wavelength shouldn't even make sense, because a particle's wavelength really is determined by how the amplitude of the wavefunction is varying over different position eigenvalues, and there shouldn't be any uncertainty in the state of the wavefunction itself. So it seems like the formula $$p = h/\lambda$$ should allow you to find the total momentum exactly

OK, after googling around a little I came across http://physics-qa.com/html/kqm09.htm which made me realize the flaw in my argument above...the problem is that even if you know the way the amplitude of the wavefunction is varying in space, that doesn't give you an exact wavelength unless the amplitude varies like a perfect sine wave--if instead it looks more like a wavepacket localized in space, then by Fourier analysis this can be decomposed into a sum of pure sine waves with different wavelengths, which means there is no single $$\lambda$$ you can plug into $$p = h/\lambda$$.

But I'm still not sure about my original question--what does uncertainty in momentum mean experimentally for a photon? Does it only mean that the photon has been localized into a wavepacket and thus the wavelength is fuzzy as above, or does it also mean uncertainty in how fast the photon will get from one location to another? Say both the emission and detection of a photon are extremely well-localized in space and time--perhaps I turn on a tiny light for an almost instantaneous time-interval, and then detect photons emitted by this light at another location. If I calculate the distance/time between these two localized events, will the answer always be as close to c as my measurements can resolve, or does the large uncertainty in momentum due to the precise position-measurements mean that distance/time may sometimes depart significantly from c?

 

[vanesch]

But I'm still not sure about my original question--what does uncertainty in momentum mean experimentally for a photon?

Exactly that: uncertainty in momentum ! The amount of momentum that will be transferred upon absorption. And because for a photon, E = p c, also, uncertainty in energy.

A very short light pulse must be polychromatic. BTW, that's a trick to make very short laser pulses: you use mode locking, so that you use light of many different frequencies to make a short pulse.
Of course the geometrical speed of the light pulse in vacuum will always be equal to c. For photons, momentum has nothing to do with speed.

cheers,
Patrick.

 

[JesseM]

Exactly that: uncertainty in momentum ! The amount of momentum that will be transferred upon absorption. And because for a photon, E = p c, also, uncertainty in energy.

A very short light pulse must be polychromatic. BTW, that's a trick to make very short laser pulses: you use mode locking, so that you use light of many different frequencies to make a short pulse.
Of course the geometrical speed of the light pulse in vacuum will always be equal to c. For photons, momentum has nothing to do with speed.

cheers,
Patrick.

Thanks Patrick, the issue of speed was the main thing I was wondering about. So can this be understood as a limiting case? If I see two electrons, one previously measured to be traveling at 0.2c relative to me and another previously measured to be traveling at 0.99c relative to me, and then I measure each electron's position in a way that creates the same amount of uncertainty about each one's momentum, will I have any less uncertainty about the speed of the electron which was previously known to be moving at 0.99c?

 

[vanesch]

Thanks Patrick, the issue of speed was the main thing I was wondering about. So can this be understood as a limiting case? If I see two electrons, one previously measured to be traveling at 0.2c relative to me and another previously measured to be traveling at 0.99c relative to me, and then I measure each electron's position in a way that creates the same amount of uncertainty about each one's momentum, will I have any less uncertainty about the speed of the electron which was previously known to be moving at 0.99c?

Yes ! It is even the case in most high energy experiments that you assume all particles to have essentially light speed, no matter what their momentum. You measure the momentum of charged particles for instance by the curvature of their track in a magnetic field (but they all go essentially at 0.99c or so) ; or you measure the total energy they deposit in a "calorimeter".

cheers,
Patrick.