3 easy QFT questions

Sterj

I have three questions about QFT. I know you can answer this questions, sure you can.

- How does a standing wave look like in 3 dimensions (x,y,z). I mean an electro magnetic oscillation between two walls or in free space. I can't really imagine that thing and looked for pictures (google) but coudn't find what I searched for.

- Let's say we have an electromagnetic oscillation with quantum number n=3. Then this has energy of 7/2*w*h(bar) and contains 3 particles (photons).
Here we say this particles make the field oscillating. And thus the oscillation of this particles can go from earth to moon. But if we describe this 3 photons as wave packets they don't have such an influence (from moon to earth). Something has to be false, but what?

- Consider two plates. Between that plates there is a harmonic oscillation with energy 7/2*w*h(bar). How can we describe this oscillation with formulas. I tried to solve this with 2 wave functions. One comes from left and one from the right side and they interfer. But my wave functions don't involve the energy, arghhhhh. Can someone give me a guess?

Thanks for any help.

marlon

http://id.mind.net/~zona/mstm/physic...Diagrams1.html

or google for standing wave animation

It does not contain 3 photons, you need three photons in order to go from the ground state to this excited state.

These phonons do not make any field oscillate, they are the actual result of the oscillations of a massless spin 1 field. Well, actually they arise as the oscillations of a massive spin 1 field (with certain energy value that corresponds to the given energy), and then the mass is taken to be ZERO.

Easy, that is just classicak wave dynamics. Just look at how standing waves are generated and what vibration modi are possible. I refer to the given sites and googling-task

marlon

Sterj

thanks marlon

These are only 2 dimensional oscillations. I know how they look like. I coudn't find 3 dimensional oscillations.

My problem is: Are photons wave packets or are they what you said above. I mean, can we say that photons are wave packets in QFT?

marlon

Well, it's not that difficult to envision: just think of a superposition of the wave in the animation in both x,y and z direction, or perpendicular to the depicted plane.

Well, not really to be exact. The wave packet thing comes from QM. QFT goes one step further in that sense that the actual equation of such a wave has creation and annihilation operators (in the canonical formalism, that is) that actually create and annihilate a photon. The 'wave' just describes the creation of a photon that will propagate through space during a certain time and with certain energy. Just look at the meaning of Feynman diagrams. Photons (as all other particles in QFT) really arise as fluctuations or vibrations of FIELDS that are quantized. hence the name QFT

marlon

Meir Achuz

QFT is usually described in perturbation theory usiing Feynman diagrams.
In this case, the created photons are plane (or spherical) wave trains.
They don't really propagate in space, because having a defiinite momentum, the wave train is infinite and exists thoughout all space. They are best seen in the momentum representation. A physical photon with a finite extent in space would be a linear combination of these photon states forming wave packet, just as a classical EM wave packet is formed. So photon wave packets can be wave packets in QED, formed in the same way wave packets are formed in EM.

marlon

We need to make a clear distinction between real and virtual photons here
Again this can lead to a discussion on semantics

Though it is really important to realize that photons arise as the actual vibrations of the spin 1-field (you can call this the actual wave indeed). But it needs to be said that when the field 'goes' from one vibration-mode to another, the change in energy corresponds to the actual photon. Not the wave itself...

That is the big difference between QM and QFT

marlon

Sterj

thanks you both.

I tried to calculate a harmonic oscillation in free space. I did this by letting one wave come from the right side and one from the left (superposition between these two waves gives me an oscillation):

psi(x,t)=cos(x-t)+sin(x+t)

That's the expression with no energy. For the expression with energy I did this:

p(x,t)=exp[i(kx-wt)]
u(x,t)=exp[i(kx+wt)]

now k=p/h(bar) (1)
and E=hf
p=hf/c

thus: p=E/c and with (1) and the definition m=E/(c*h(bar)) the wavefunction p(x,t) goes into:

p(x,t)=exp[i(mx-wt)]

And if I take the superposition of these two waves I get the oscillation:
psi(x,t)=p(x,t)+u(x,t)

And if the oscillation is in the ground state the factor m transforms because

E=h(bar)w/2

And the w in the wavefunction is the same w as the w in the energy expression for the ground state.

Is this right?