# QED: virtual photons -spin conservation?

1. May 30, 2005

### Kruger

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???

2. May 30, 2005

### dextercioby

What spin conservation are u talking about...?

Daniel.

3. May 30, 2005

### Kruger

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.

4. May 30, 2005

### Meir Achuz

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.

5. May 30, 2005

### dextercioby

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.

6. May 30, 2005

### dextercioby

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.

7. May 30, 2005

### marlon

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

8. May 31, 2005

### Kruger

Ok, thanks. I know this angular momentum operator L.

9. May 31, 2005

### Kruger

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. May 31, 2005

### Meir Achuz

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. May 31, 2005

### Meir Achuz

"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. May 31, 2005

### Kruger

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. Jun 1, 2005

### Kruger

Is the textbook wrong? I think no.

14. Jun 1, 2005

### marlon

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. Jun 1, 2005

### Kruger

what's with quantum electrodynamics? The also say that only one virtual photon is created.

16. Jun 1, 2005

### Meir Achuz

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. Jun 2, 2005

### Kruger

"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. Jun 2, 2005

### masudr

He didn't say, "one photon can be created from the vacuum," he said, "one of nothing."

19. Jun 2, 2005

### Meir Achuz

English is a tricky language.

20. Jun 7, 2005

### reilly

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