Energy conservation in vertices in QED

1. Jan 16, 2005

An elementary electron-> electron + gamma process can not occur by itself in the vaccum because if we assume the photon invariant mass zero and the electron invariant mass positive, energy and momentum can not be conserved at the same time. The delta E may be computed and from the energy time incertinty principle we compute delta t, the maxium time allowed for the energy violation. After this time interval, whether the electron or the photon must interact further in order to have a process allowed in the vaccum.

If on the other hand we assume the momentum and the energy conservation simultaneously, the electron invariant mass positive and the photon invariant mass unknown, we compute the latter and find it pure imaginary. What would be the physical sense to this.

So, is energy conserved in elementary vertices or not?

2. Jan 16, 2005

marlon

Hi,

energy is always conserved.
The thing with the Heisenberg-uncertainty stating that energy conservation can be violated for a short time is indeed correct. But you need to look at this from the right perspective : the socalled "net energy conservation law" is always respected. By this, i mean that if you were to compare the energy before and just after the proces : it will be the same. thus, energy is conserved. However during the process energy conservation can be violated for a short while but this will be restored just after the violation so the in total it seems like the violation never occured. This is the picture...The imaginary values will only occur during the violation of conservation laws. But taking into account the previous explanation, you can disregard them and keep on calculating as if the conservation laws are respected. Which they ARE, if you look at the ENTIRE process...

marlon

Last edited: Jan 16, 2005
3. Jan 16, 2005

dextercioby

I guess you're having in mind the elementary process of QED,the vertex with 2 fermionic lines and one photonic one.Let's separate this vertex from the scattering process itself (a priori described by a very comlicated F diagram)and discuss it and only it.

Let's assume we have a photon which is emitted by an incoming electron.I state that this process is not a physical one,unless,one of the three (actually 2,but there are 3 propagators in all) particles is not physical,else it is VIRTUAL.
PROOF:1.Because it is an elementary process,we can assume that the total 4momentum is conserved.If we assume that and the fact that ALL THE PARTICLES ARE REAL,MEANING THEY ARE ON THEIR MASS SHEET,we have a contradition.4-momentum is not conserved.
2.Conserving of 4-momentum implies that one of the particles (let's pick the photon,like in a Bhabha or Moeller scattering) is virtual,which means IT IS NOT ON ITS MASS SHEET.
In the case of the photon,u get indeed negative eigenvalue for $\hat{P}^{2}$,which would mean a tachyonic state...
ENDPROOF.

Therefore,u can chose:either consider the three particles as being real,on their mass sheet,and then energy momentum will not be conserved in vertex,or take energy-mementum to be conserved and end-up with one virtual particle.

MY ADVICE IS TO TAKE THE SECOND ALTERNATIVE.

Take the Bhabha diagram of scattering.The incoming particles are real and the photon they change is taken as virtual/nonphysical.The trick is that taking the photon as nonphysical will not alter your results,as in the S matrix element between the 2 states (final and initial) no element from the description of the photon (neither spin,nor helicity,nor momentum,nor energy) interviens.U can check that one... :tongue2:

Take 4momentum to be always conserved in fundamental vertices in QFT.It's only way you can know which particles are real (on their mass sheet) and which are not (are virtual/are not on their mass sheet).

This vertex is the typical example when speaking about virtual particles.

Daniel.

4. Jan 16, 2005

thank you for your replies. i agree with you. further question: would a virtual electron also have a purely invariant imaginary mass?

5. Jan 16, 2005

dextercioby

Not necessary.Think about the Compton scattering.Or you could just read the post below... :tongue2:

Daniel.

PS.Check on the same vertex.This time the photon is real.The incoming electron is real and the outgoing electron is virtual.Like in the Compton scattering.

Last edited: Jan 16, 2005
6. Jan 16, 2005

dextercioby

Lemme prove my assertion according to which in a Compton scattering the "middle" electron is virtual,however with a positive eigenvalue of $$\hat{P}^{2}$$

Consider the vertex as before.Two fermionic lines and one photonic one.An incoming electron and an incoming photon,both of which are real.

$$p_{2}=k+p_{1}$$(1)

The incoming electron is real,so:
$$p_{1}^{2} =m^{2}$$(2)

The incoming photon is real,so:
$$k^{2}=0$$ (3)

Square (1) and get
$$p_{2}^{2}=k^{2}+p_{1}^{2}+2kp_{1}$$ (4)

Use (2) and (3) to find:
$$p_{2}^{2}=m^{2}+2k^{\mu}p_{1,\mu}=m^{2}+2k^{0}p_{1,0}-2\vec{k}\cdot \vec{p}_{1}$$(5)

In the reference frame of the incoming electron
$$\vec{p}_{1}=\vec{0}$$ (6)

So,(5) becomes
$$p_{2}^{2}=m^{2}+2k^{0}p_{1,0} >m^{2}$$ (7)
,therefore,the outgoing electron is not on his mass sheet,then it can be considered virtual.

Computing the Compton scattering matrix element,will lead you the same conclusion as before:virtual particles do not modify the physical observables of the system.

Daniel.

Conservation of