QED: virtual photons -spin conservation?

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Discussion Overview

The discussion revolves around the conservation of spin in quantum electrodynamics (QED) when virtual photons are involved. Participants explore the implications of virtual photon emission by electrons and the associated conservation laws, including angular momentum and spin conservation, within the context of electromagnetic interactions.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants argue that when an electron emits a virtual photon, the conservation of spin must be considered, suggesting that two photons may need to be created to satisfy this conservation.
  • Others question the specific type of spin conservation being referenced, with some asserting that total angular momentum is conserved rather than spin alone.
  • One participant notes that the spin states of particles are entangled, leading to the use of averages in calculations, while another emphasizes the importance of orbital angular momentum in these discussions.
  • There is a contention regarding the creation of virtual photons from the vacuum, with some stating that at least two particles must be created to conserve momentum, while others suggest that one photon can be created if there is an existing electron.
  • Participants discuss the implications of the Heisenberg Uncertainty Principle (HUP) on the existence of virtual photons and the associated energy-time relationship, with differing interpretations of how this applies to single versus pairs of photons.
  • One participant raises concerns about the validity of textbook explanations regarding virtual photon creation and the HUP, leading to further debate about the accuracy of these sources.
  • There is a mention of charge conjugation invariance and its role in preventing the creation of a single virtual photon from the vacuum without external influence.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the conservation of spin and momentum in the context of virtual photon interactions. The discussion remains unresolved, with no consensus reached on several key points.

Contextual Notes

Participants highlight limitations in understanding related to the definitions of spin and angular momentum, as well as the assumptions underlying the creation of virtual particles from the vacuum.

Kruger
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I'm talking of quantum electro dynamics where we try to explain the electromagnetic "force" with virtual photons. Now an electron can "send" a photon that has more energy then it self (this photon is virtual because it can only exist because of the HUP). But spin must always be conserved. But it isn't. If spin want's to be conserved there must be created two photons?
 
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What spin conservation are u talking about...?

Daniel.
 
spin conservation of the whole system.
Imagine:
no photon is created cause of "field interaction" (electromagnetic force) --> system has spin of 0

a photon is created cause of "field interaction" (two electrons for example act via coloumb force --> system has spin of 1. Now, the two system's have not the same total spin.
 
An electron has so spin 1/2. So when it emits a photon, the spin addition is
1/2 --> 1/2=1+1/2. The photon and the electron that emitted it are in a spin eigenstate with total spin 1/2.
 
Incidentally,the total angular momentum is conserved in a Feynman diagram and not the spin.Note that the in and out particles are eigenstates of the free field hamiltonian which commutes with the total angular momentum operator,therefore they are not in a spin eigenstate,but a J eigenstate.Therefore,the spin states are entangled,and that's why we use spin averages and spin sums when computing physical quantites,like diff.cross sections.

Daniel.
 
Meir Achuz said:
An electron has so spin 1/2. So when it emits a photon, the spin addition is
1/2 --> 1/2=1+1/2. The photon and the electron that emitted it are in a spin eigenstate with total spin 1/2.


Not necessarily.Spin is not conserved.You should also watch out for L (orbital ang.momentum) states and depict them for every particle (if possible).


Daniel.
 
dextercioby said:
Not necessarily.Spin is not conserved.You should also watch out for L (orbital ang.momentum) states and depict them for every particle (if possible).


Daniel.

Very true, also the strong force does not respect 'spin conservation'

Just look at the spin of a proton and realize that is built out of three quarks that each must have s_z = 1/2 or -1/2

Clue is to realize that spin does NOT add linearly, my friends...Just recall Clebsch Gordan...As dexter pointed out, it is L (or should i say J) that needs to be conserved because or dear QM and QFT needs to be invariant under rotations...

marlon
 
Ok, thanks. I know this angular momentum operator L.
Thanks for your help.
 
Consider the following:
We have no fermionic particle in the system. A virtual photon is created via the HUP (from vacuum). What's with momentum conservation?
 
  • #10
I try to answer a simple question in simple terms.
Bringing in quarks does not help, especially since, for the predominantly S wave bound state of 3 quarks in a proton, spin is conserved.
Cbelbsch-Gordan coefficients in the addition 1/2+1/2+1/2=1/2 are calclulated on the basis of conservtion of spin angular momentum.
There is no difference in the strong-EM-or weak forces in the conservtion of angular momentum. In the absence of orbital, that means spin.
I think Marlon might be using one component of spin to mean "spin".
Quarks having different s_z is required by conservation of spin.
 
  • #11
"Consider the following:
We have no fermionic particle in the system. A virtual photon is created via the HUP (from vacuum). What's with momentum conservation?"

You have just proven that no single particle can be "created from the vacuum".
At least (and usually) two must be created to conserve momentum.
 
  • #12
You have just proven that no single particle can be "created from the vacuum".

Yes. My textbook says something different. It says that a photon with energy d(E) via the HUP can exist for time d(t) as following:
d(t)d(E)=h/4pi
and with the photon pair it should yield the following:
d(t)d(E)=h/8pi

Is my textbook wrong?
 
  • #13
Is the textbook wrong? I think no.
 
  • #14
Kruger said:
Yes. My textbook says something different. It says that a photon with energy d(E) via the HUP can exist for time d(t) as following:
d(t)d(E)=h/4pi
and with the photon pair it should yield the following:
d(t)d(E)=h/8pi

Is my textbook wrong?

what probably they mean is this dE represents the uncertainty of one photon's energy. If you have two photons you will have twice the uncertainty on the energy. If dE is doubled, the dt must be half the original value in order to keep on respecting the HUP. But in this case you measure the energy of each photon apart...Normally you should just get one dE for the total pair.But the picture stays the same though

marlon
 
  • #15
what's with quantum electrodynamics? The also say that only one virtual photon is created.
 
  • #16
One virtual photon can be created if there is already an electron to conserve momentum. One of nothing can be created from the vacuum.
 
  • #17
"One of nothing can be created from the vacuum"

That can't be true. Momentum conservation would not be sadisfied. There have to be created a pair of photons.
 
  • #18
He didn't say, "one photon can be created from the vacuum," he said, "one of nothing."
 
  • #19
English is a tricky language.
 
  • #20
There are lot's of reasons why one virtual photon cannot be created from the vacuum. Charge conjugation invariance precludes such a transformation, as a photon has - charge conjugation while the vacuum has + CC. Further, with all the known interactions, a virtual photon is most likely to be created along with a virtual electron-positron pair. These transitions conserve both linear and angular momentum. The standard Feynman diagram for the process has the three particles annihilate each other at the end of the process.

Noway can one photon be created from the vacuum (without some external influence.)
Regards,
Reilly Atkinson
 
  • #21
There are lot's of reasons why one virtual photon cannot be created from the vacuum.

I would say:

There are lot's of reasons why ONLY one virtual photon cannot be created from the vacuum.
 
  • #22
In English, "one" generally means one.
 

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