# Does a photon's complex 'amplitude' rotate along its journey?

1. Oct 7, 2011

Hi, I was watching http://video.google.com/videoplay?docid=-2622437302869951111" [Broken] the other day and he described photons by an 'amplitude' which actually means a complex number, which rotates over time. This rotation speed gives the photon's colour/energy/frequency.

Then later in the lecture he says that the amplitude (complex number) doesn't actually rotate with time, he says that it is simply that the amplitude at emission rotates with time, and once that photon has been emitted with a certain amplitude, that amplitude doesn't change other than to decrease with the inverse square of distance.
Am I right so far? The lecture is also old, from 1979!

However, due to the uncertainty over when exactly the photon is emitted, the probability cloud of possible such photons could perhaps be considered as a small packet containing an oscillation of the amplitude, over perhaps a few oscillations, due to the probablistic time period over which it was emitted. But this actual little wave packet wouldn't be rotating those amplitudes over time. Does this make sense?

But, whether you wish to call photons the individual point, or the probability cloud, either way, at the receiving end you only get a single amplitude on detecting a photon right? You don't 'see' the whole probability cloud, just a single result. But this would mean that the receiving electron or whatever would not know the frequency of the photon, which it needs to know in order to jump to certain energy states.

Did I interpret Richard Feynman incorrectly? Is the frequency a property of a single photon, or a property of how it is emitted?
If it is the latter, then what is making the emitter (e.g. an electron) rotate the amplitude over time? It is based on the drop in electron energy levels right? So is it to do with the frequency at which the electron orbits?

Thanks, and sorry for the long question (BTW I'm a newby to QM if that wasn't obvious).

Last edited by a moderator: May 5, 2017
2. Oct 7, 2011

### Zarqon

Not entirely sure from your post, but it sounded like maybe you were mixing two things up.

1) the complex amplitude of the photon, a.k.a. it's phase. In classical EM theory this would correspond to the size and sign of the E-field.

2) the probability amplitude to find the photon at a particular point. This probability is given by the total size of the E-field (regardless of phase) and is thus calculated by taking the absolute value squared of the complex amplitude, which causes the complex oscillations to not enter here.

If you only measure on one photon, usually it's the second thing you measure, i.e. the probability distribution, but if you perform an experiment where you let multiple photons interfere coherently, then you will see that the photons add up using the complex amplitude first (including phases), and that the whole sum is then squared for the total probability amplitude to be measured.

Hope that helped.

3. Oct 7, 2011

Yes I think that helps.
Richard Feynman talked about a vector, which you rotate in proportion to how far the photon moves. This speed of rotation is the frequency. I was wondering whether this vector rotates as it moves along its journey, or whether it is only 'rotating at the emitter' as it were, and fixed between its source and destination (other than reducing in magnitude)?

He calls the vector an 'amplitude', and the probability is the 'square of the amplitude', using his wording.

4. Oct 7, 2011

### yoron

wow, you'll have to link that one :)

If we're speaking of frequencies of a photon, then I think of that as observer dependent. Photons can be described as red and blue shifted. That belongs to the energy they are expected to have relative the observer. They are also described as being invariant light quanta, never changing their energy intrinsically.

And both seems true :) They will be of one energy 'intrinsically', but they will also gain and lose energy relative the detector, like your eye. If you're standing on Earth looking up the photons that hit your retina will be blue shifted, as they 'accelerate'. In this case we know that all photons ever hitting your retina only can have one 'speed', (lights invariant speed in a vacuum, the same from any frame of reference you can think up, locally) so me using the word 'accelerate' here translate into getting a different (higher blue shifted) energy level relative you observing. And the 'energy' it express will be true.

Think of it as a car hitting another car, going the opposite direction, Gravity is in Einsteins definition a equivalence to accelerations, and so our Earth also 'accelerates' at a constant one G, uniformly and gravitationally. So 'the car' Earth with you on meets 'the car' the photon in the annihilation of it at your retina, sort of.

And frequency is a wave equivalence to energy. But most physicists today move from one to the other without problems when discussing their 'photon' experiments :) And in a way? I don't know. Photons are weird.

Last edited: Oct 7, 2011
5. Oct 8, 2011