For all these types of experiments, how do they “aim” the particle so it hits its target from far away? It would seem that the experimenters would know pretty much where the particle is when it shoots out of the gun (or whatever), so would not the velocity be all over the place? In the post on the Heisenberg principle, there was an example of letting the sun shine through a hole in a piece of paper, which caused the photons to spread pretty widely, pretty quickly.
Does the polarization vector change as the photon moves along? It seems to be very similar to a photon’s “main” wave function, as it can be represented as a complex number (and is even displayed as an arrow, like Feynman uses). But I know those Feynman arrows spin according to the photon’s wavelength.
Finally—and this is really tripping me up—why can we put in the minus sign in the equation that you say “we will need” later, instead of a + sign? If you have two blobs of amplitude, you need to add them to get the wave function, yes? If that is not the case, I have SEVERELY misunderstood the most basic posts of this sequence.
For all these types of experiments, how do they “aim” the particle so it hits its target from far away? It would seem that the experimenters would know pretty much where the particle is when it shoots out of the gun (or whatever), so would not the velocity be all over the place?
Only if they make the departing aperture small. A wider aperture allows the departing wave to be tight.
Does the polarization vector change as the photon moves along?
It depends which basis you look at it in. It is conventional to consider a photon’s ‘polarization’ to be ploarization subspace that contains all of its time dependence. The phase then indicates the rest of its state. However, you can look at it other ways. A circularly polarized photon moving +z can be considered as a rapid shift between various orientations of +x and +y polarization… but it’s simpler to just let it be in a circular polarization state and let the phase vary. A photon’s state in this sense IS its ‘main’ wavefunction as you call it. There is no distinction. People usually shorthand think of a photon to have perfectly-defined momentum, but of course that would mean the photon extends through all of space. Real photons have multiple momentum components, and form a wavepacket or a static state. In particular, and very relevantly, you can construct electromagnetic field states (photons) that are inverse square laws—the static electrical field from a charge—and these have a very broad momentum distribution.
why can we put in the minus sign in the eqation that you say “we will need” later, instead of a + sign?
I can’t find any minus signs in this post, but to take a stab in the dark at whatever it is you’re referring to, subtraction is the special case of addition after one of a particular set of phase shifts.
So this is sorta off-topic for this thread, but I cannot see where one can start a new one. I posted the following questions at http://lesswrong.com/lw/q2/spooky_action_at_a_distance_the_nocommunication/, as I cannot find the “rerun” version of it. Anyway, here goes. FWIW, the topic was about EPR experiments.
For all these types of experiments, how do they “aim” the particle so it hits its target from far away? It would seem that the experimenters would know pretty much where the particle is when it shoots out of the gun (or whatever), so would not the velocity be all over the place? In the post on the Heisenberg principle, there was an example of letting the sun shine through a hole in a piece of paper, which caused the photons to spread pretty widely, pretty quickly.
Does the polarization vector change as the photon moves along? It seems to be very similar to a photon’s “main” wave function, as it can be represented as a complex number (and is even displayed as an arrow, like Feynman uses). But I know those Feynman arrows spin according to the photon’s wavelength.
Finally—and this is really tripping me up—why can we put in the minus sign in the equation that you say “we will need” later, instead of a + sign? If you have two blobs of amplitude, you need to add them to get the wave function, yes? If that is not the case, I have SEVERELY misunderstood the most basic posts of this sequence.
Only if they make the departing aperture small. A wider aperture allows the departing wave to be tight.
It depends which basis you look at it in. It is conventional to consider a photon’s ‘polarization’ to be ploarization subspace that contains all of its time dependence. The phase then indicates the rest of its state. However, you can look at it other ways. A circularly polarized photon moving +z can be considered as a rapid shift between various orientations of +x and +y polarization… but it’s simpler to just let it be in a circular polarization state and let the phase vary. A photon’s state in this sense IS its ‘main’ wavefunction as you call it. There is no distinction. People usually shorthand think of a photon to have perfectly-defined momentum, but of course that would mean the photon extends through all of space. Real photons have multiple momentum components, and form a wavepacket or a static state. In particular, and very relevantly, you can construct electromagnetic field states (photons) that are inverse square laws—the static electrical field from a charge—and these have a very broad momentum distribution.
I can’t find any minus signs in this post, but to take a stab in the dark at whatever it is you’re referring to, subtraction is the special case of addition after one of a particular set of phase shifts.