- #1

- 624

- 11

- I
- Thread starter Silviu
- Start date

- #1

- 624

- 11

- #2

- 268

- 72

For particles with mass, helicity isn't a good quantum number. So, it doesn't have to be conserved.

From previous postings I believe you have QFT book by Schwartz. Regarding your second question, I recommend to read section 5.3 of that book (##e^+e^-\rightarrow \mu^+\mu^-##), where he in a non-technical manner explains that the spin polarization isn't conserved. It is also useful to chapter 13 (QED) where it is discussed in terms of QFT. In principle, the final spins are usually not measured. Instead, you measure the differential cross section with respect to an angle ##\theta##, where determines the directions of the outgoing particles.

- #3

- 624

- 11

So, if helicity is not a good quantum number for massive particles and in the massless case helicity is the same as chirality, why do we need helicity in the first place. Why don't we just use chirality?For particles with mass, helicity isn't a good quantum number. So, it doesn't have to be conserved.

Yes, that is the book I am using. So in 5.3 as far as I can tell is mainly using helicity (as he assumes the particles are massless). Even so, as I said in the first post, he reaches the conclusion that the initial and final state spins must add up to 1 or -1 so that the spin is conserved at each vertex (or in other words, the photon can be produce having spin 1). Is this understanding correct? Then if we have electron electron scattering, by using the same arguments, how do you conserve the spin at a given point? In that case the photon doesn't come from both particles at the same time, but it gets transferred from one to another.From previous postings I believe you have QFT book by Schwartz. Regarding your second question, I recommend to read section 5.3 of that book (##e^+e^-\rightarrow \mu^+\mu^-##), where he in a non-technical manner explains that the spin polarization isn't conserved. It is also useful to chapter 13 (QED) where it is discussed in terms of QFT. In principle, the final spins are usually not measured. Instead, you measure the differential cross section with respect to an angle ##\theta##, where determines the directions of the outgoing particles.

- #4

- 268

- 72

One the other hand, you can expand your fields (representing the particles) in terms of components having positive and negative helicity.

You then consider all possible combinations and sum over the helicity, when the cross section is computed.

Depending on the kinematics, some combinations of helicities of the initial and final states will vanish or be neglible, e.g. in the ultra-relativistic limit.

Obviously, for m=0 you will only have one combination. So, using a helicity basis is typically convenient.

On the slides at https://www.hep.phy.cam.ac.uk/~thomson/lectures/partIIIparticles/Handout4_2009.pdf , this is explained quite well.

I'm not sure if I understand your 2nd question correctly. At each vertex you have 1 photon and two electrons. The spin of the photon is 1 and each of the electrons has spin 1/2. So, denoting the spin of the incoming electron by ##s_1##, the one of the outgoing one by ##s_2## and the spin of the photon by ##s_\gamma##, you need to satisfy ##|s_1-s_\gamma|\leq s_2 \leq |s_1+s_\gamma|## , since spin is a vector quantity. Obviously, the projections (polarizations) must obey the additative law.

- Last Post

- Replies
- 14

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 455

- Last Post

- Replies
- 0

- Views
- 805

- Last Post

- Replies
- 5

- Views
- 897

- Last Post

- Replies
- 3

- Views
- 778

- Replies
- 0

- Views
- 656

- Last Post

- Replies
- 4

- Views
- 705

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 807

- Last Post

- Replies
- 15

- Views
- 5K