Understanding Spinor's Helicity

[kelly0303]

Hello! The free Dirac hamiltonian doesn't commute with the z component of the spin operator $S_z$, but it commutes with the helicity operator $h=S\cdot\hat{p}$. This means one can know at the same time the energy and helicity of a particle, but not its spin along the z-axis. I am a bit confused about this. One can measure the momentum of a free particle and hence get its energy. So energy and momentum are simultaneous observables for a free particle. But so is helicity, too. So if I know the momentum of the particle, and choose my z axis to be in that direction, and I know it's helicity, too, don't I know (by the definition of helicity) the spin of the particle along the z, axis i.e. $S_z$? What is wrong with my logic? Thank you!

 

[Dr.AbeNikIanEdL]

As far as I can tell there is nothing wrong, this is just not very useful for analysing any interesting situation, where you would usually have some other significant direction that dictates how you want to choose your coordinate system. For example a magnetic field in some direction, that you might want to choose to be the z direction rather than the direction of motion.

 

[kelly0303]

As far as I can tell there is nothing wrong, this is just not very useful for analysing any interesting situation, where you would usually have some other significant direction that dictates how you want to choose your coordinate system. For example a magnetic field in some direction, that you might want to choose to be the z direction rather than the direction of motion.

I am not sure I understand. In this situation I gave (which indeed is not very useful in practice), I am able (if my logic is correct) to measure the spin and energy of the particle at the same time. But their operators don't commute. Shouldn't it be impossible to measure both, no matter how I set up the axis? For example, in the momentum position case, no matter what system I choose, or how I place my axis (or what I do whatsoever) I will never be able to precisely measure both. Why could I do that here?

 

[Dr.AbeNikIanEdL]

No, you are able to measure the projection of the spin along the direction of motion of the particle. Calling that direction z does not really change anything. If you go through the math you will see that Hamiltonian and $S_z$ operator do commute if $p_x=p_y=0$ (which is just a special case of saying that helicity commutes with the Hamiltonian).