Spin and quantum statistics

[Sriman Dutta]

Hi everyone, 

While reading Quantum Mechanics from Griffiths, I came upon a point where the author writes that the relationship between spin of a particle and its characteristic statistical behavior is explained by relativity. 

I'm in complete darkness regarding this. Can someone please explain how this is explained? And also it will be great if someone can cite some source or mathematically explain the phenomenon. 

 

Thanks in advance!

[joigus]

4 hours ago, Sriman Dutta said:

Hi everyone, 

While reading Quantum Mechanics from Griffiths, I came upon a point where the author writes that the relationship between spin of a particle and its characteristic statistical behavior is explained by relativity. 

I'm in complete darkness regarding this. Can someone please explain how this is explained? And also it will be great if someone can cite some source or mathematically explain the phenomenon. 

 

Thanks in advance!

The rigorous theorem requires heavy-duty maths: field operators, states and vacuum, and representations of the Lorentz group.

A relatively easy-to-follow discussion is provided by John Baez:

https://math.ucr.edu/home/baez/spin_stat.html

In a nutshell, what the spin-statistics tells you can be stated as this: The splitting of particle classes into bosons and fermions, as characterized by their exchange properties (statistics):

\[\varphi_{1}\left(x\right)\varphi_{2}\left(x'\right)=\varphi_{2}\left(x\right)\varphi_{1}\left(x'\right)\] (bosons)

or,

\[ \varphi_{1}\left(x\right)\varphi_{2}\left(x'\right)=-\varphi_{2}\left(x\right)\varphi_{1}\left(x'\right) \] (fermions)

Can be obtained also from their properties under the Lorentz group (the rotation factor of it) by rotating the whole universe around the midpoint between them. If the vacuum is Lorentz invariant, we won't be doing anything to it, and particles will just be exchanged if the rotation is of angle \( \pi \).

The name spin-statistics is because "spin" has to do with properties under rotations, and "statistics" has to do with properties under exchange. "Bosons under rotations are also bosons under exchange; conversely for fermions" is the content of the theorem.

I hope that helped.

[Sriman Dutta]

On 11/29/2020 at 5:58 PM, joigus said:

The rigorous theorem requires heavy-duty maths: field operators, states and vacuum, and representations of the Lorentz group.

A relatively easy-to-follow discussion is provided by John Baez:

https://math.ucr.edu/home/baez/spin_stat.html

In a nutshell, what the spin-statistics tells you can be stated as this: The splitting of particle classes into bosons and fermions, as characterized by their exchange properties (statistics):

\[\varphi_{1}\left(x\right)\varphi_{2}\left(x'\right)=\varphi_{2}\left(x\right)\varphi_{1}\left(x'\right)\] (bosons)

or,

\[ \varphi_{1}\left(x\right)\varphi_{2}\left(x'\right)=-\varphi_{2}\left(x\right)\varphi_{1}\left(x'\right) \] (fermions)

Can be obtained also from their properties under the Lorentz group (the rotation factor of it) by rotating the whole universe around the midpoint between them. If the vacuum is Lorentz invariant, we won't be doing anything to it, and particles will just be exchanged if the rotation is of angle \( \pi \).

The name spin-statistics is because "spin" has to do with properties under rotations, and "statistics" has to do with properties under exchange. "Bosons under rotations are also bosons under exchange; conversely for fermions" is the content of the theorem.

I hope that helped.

I see.

Is the approach to the mathematical proof requires group theory?

Also imagine if there are two particles are in positions 1 and 2, then their exchange in positions is equivalent to rotation of the world around them keeping them constant. But how can that change the wavefunction ? Like I can view the two particles are seen one side, and from other side, their wavefunction values are different (a minus sign pops up!) ?

 

[joigus]

25 minutes ago, Sriman Dutta said:

Is the approach to the mathematical proof requires group theory?

It does, because elementary particles are characterised in QFT as irreducible representations of the Poincaré group (Lorentz groups and translations).

If you study the Lorentz group for its own sake, you will be led purely mathematically to objects that represent faithfully the properties of spinors, because they rotate by SU(2), and locally SU(2) looks like SO(3).

That as to rotations. The version that includes boosts (the whole Lorentz group) are objects that boost and rotate with SL(2,C), which looks locally like SO(1,3) (Lorentz group).

[Sriman Dutta]

I see.

Thanks a lot for the information!

[joigus]

59 minutes ago, Sriman Dutta said:

I see.

Thanks a lot for the information!

Oh, you're most welcome. It's such a pleasure to have a sensible conversation for a change...