What Is the Significance of 720 Degree Symmetry in Fermion Rotations?

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It is interesting that our elementary fermions have 1/2 spin, meaning it takes a full 720 degree rotation to bring them back to their original state and these fermions constitute ordinary matter, eg. quarks, and electrons.

Classical nature, however, does not have a 720 degree symmetry, but only a 360 degree symmetry. How did we lose that 720 degree symmetry? Is it the presence of bosons as force-mediators? (eg. photons, gluons etc)
 
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Sometimes it does. Search for Dirac's scissors.
 
I think it's due to the tininess of hbar/2. The deviation of hbar/2 due to spin in macroscopic matter distribution is suppressed by large (integral) orbital angular momentum which makes it impossible to see the tiny quantum effect.
 
It is interesting that our elementary fermions have 1/2 spin, meaning it takes a full 720 degree rotation to bring them back to their original state
This is incorrect. Indeed when you rotate a spinor by 360 degrees the sign of the wavefunction changes. But a fermion wavefunction, being a spinor quantity, is a double-valued function: it has an implicit ± sign in front. Meaning that ψ and -ψ both represent the same state. Not just "different by a phase", they are really the same in all respects. So rotating a fermion by 360 degrees, despite appearances, does bring it back to the same state.
 
Thanks, Bill_K

Since whenever we evaluate the wavefunction, we are always multiplying ψ by ψ* so any minus signs would be eliminated, of course. That is much more clear in my mind now.

However, it is still curious that such an important symmetry for massive particles is absent in ordinary spacetime that they exist in.
 
zen loki said:
it is still curious that such an important symmetry for massive particles is absent in ordinary spacetime that they exist in.
It is not absent; it is a consequence of spacetime symmetry SU(3,1) ~ SU(2) * SU(2). As I said, fermionic effects are dominated by macroscopic effects. In addition - as said correctly - in bilinear combinations the sign just cancels
 
Right, Tom, it is a consequence of SU(3,1) symmetry.

I wonder if there is any possible conceivable macroscopic effect where a 360 degree turn yeilds a different result than a 720 degree turn. I say this, but I mean excepting things like Dirac scissors.
 
zen loki, As I've already pointed out, your original statement that fermions are invariant under 720 degree rotations rather than 360 degrees is false, so I don't know why you keep asking the same question. :confused:
 
Bill_K said:
This is incorrect. Indeed when you rotate a spinor by 360 degrees the sign of the wavefunction changes. But a fermion wavefunction, being a spinor quantity, is a double-valued function: it has an implicit ± sign in front. Meaning that ψ and -ψ both represent the same state. Not just "different by a phase", they are really the same in all respects. So rotating a fermion by 360 degrees, despite appearances, does bring it back to the same state.
I don't think so; in the same way you could argue that a vector object is invariant w.r.t. a 180° rotation
 
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Bill, I can see that they are the same and I appreciate the clarity you have brought. And I can not think of any exceptions to what we have covered here.

And while I understand the concept here, I am trying to push my understanding and see if there are any exceptions, or even if any exceptions are possible. I doubt such a thing should be possible as that would be a very strange wavefunction operator.
 

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