# I Spin vs Helicity conservation

1. Mar 6, 2018

### Silviu

I am a bit confused about spin conservation at relativistic energies. I am reading a QFT book by Peskin and at a point he specifies that "In the nonrelativistic limit the total spin of the system is conserved". Later when we go to the relativistic limit (there is the interaction between a photon and an electron this time), the spin is not conserved anymore (I attached a pic of what I mean).
So the spin is not always conserved? It is just the helicity? And what is the general statement (a conservation law shouldn't care about the energy at which the experiment takes place)? What is always conserved here? Moreover in the pic I attached he specifies that "the total spin angular momentum of the final state is one unit less than that of the initial state". I am not sure I understand. The spin doesn't point along the same direction before and after. I agree they have different values, but I am not sure I understand why it is a difference of 1? What is the common axis they use for both before and after? Thank you!

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2. Mar 9, 2018

### ChrisVer

spin alone has no reason to be conserved- it's the total angular momentum (orbital + spin, as in $J= S + L$) that is conserved (as is stated in the caption).
The orbital angular momentum is changed as well.

3. Mar 9, 2018

### Silviu

But why is the change in the spin 1 in this case? What axis is considered?

4. Mar 9, 2018

### ChrisVer

Just by looking at the pic, I would say that in the first you have a +1 (photon) as it shows rotation clockwise - 1/2 (fermion) as it shows rotation anti-clockwise = 1/2 initially.
In the final state I see a counter-clockwise rotation for the photon -1 and a clockwise rotation for the fermion 1/2 , so -1/2...
difference is = -1
the absolute difference will be 1 even if you assigned the signs differently.

5. Mar 10, 2018

### Silviu

Thank you for your reply. I understand this, but how can you compare, as they are not along the same axis? In the first case you have 1/2 along the initial axis and in the final case you have -1/2 along a different axis? They are obviously not the same, but why is the difference exactly 1?

6. Mar 10, 2018

### ChrisVer

I don't know what axis you are talking about... I just looked at where the arrows point (if you want an axis you can pick one that is the same for the final and initial state). It won't change the fact that the arrow initially goes clockwise and finally counterclockwise and vice versa.

7. Mar 10, 2018

### Silviu

What I mean is: let's assume the angle is (close to) 90 degrees. Call the initial direction x and the final direction y. So initially we have spin -1/2 along the x direction and in the end spin 1/2 along the y direction. Don't we need to move them to the same axis (either both x or both y) in order to compare the change in spin?

8. Mar 10, 2018

### ChrisVer

Initially you have a spin +1 for the photon along the z-axis (rotation clockwise)... finally you have a rotation counterclockwise, which means that the spin is -1 on the z-axis.
Similarly for the fermion.

9. Mar 10, 2018

### Silviu

What do you call z-axis? The direction of photon changes. Does this mean you can rotate your coordinate system before and after? Isn't that usually fixed?

10. Mar 10, 2018

### ChrisVer

nop

11. Mar 10, 2018

### Silviu

So how can you have +1 before and -1 after on the z axis, if the axis is fixed but the direction of the photon changes?

12. Mar 11, 2018

### ChrisVer

Why would it? how is the momentum related to the spin in your case (not helicity)?

13. Mar 11, 2018

### Silviu

I am so lost now... isn't the spin along the direction of momentum, both before and after? This means that if the momentum direction changes, spin direction changes, too, isn't this right?

14. Mar 12, 2018

### ChrisVer

The spin direction changes (it flips), why would the z-axis change?

15. Mar 12, 2018

### Silviu

Ok, so let's say initially the photon travels along the positive $\hat{z}$, which means that the spin of the photon is +1 along the $\hat{z}$. After the interaction, let's say that the photon will move along $-\hat{z}+\hat{x}$. Based on the picture, as the spin still points along the direction of the photon, the spin will be +1 along $-\hat{z}+\hat{x}$. By the same argument the spin of the electron will be 1/2 along $\hat{z}-\hat{x}$. So the total spin in the end will be +1/2 along $-\hat{z}+\hat{x}$ while initially it was +1/2 along $\hat{z}$. Is this right?