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Fermion Field of Standard Model

+ 2 like - 0 dislike
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Why fermion field is treated as anti-commuting and boson field as truly classical in standard model?

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user Curious
asked Oct 4, 2012 in Theoretical Physics by UnknownToSE (505 points) [ no revision ]

2 Answers

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If you think anti-commuting field is too arbitrary and do not like anti-commuting field, you may ask do we really need to use anti-commuting field to describe fermions? Can a theory with only bosons have fermionic excitations emerging at low energies? The answer is yes! So we do not need to use anti-commuting fields to describe fermions, and this is true in any dimensions.

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user Xiao-Gang Wen
answered Oct 11, 2012 by Xiao-Gang Wen (3,319 points) [ no revision ]
I thought this was only true in 2D?

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user josh314
A theory with only bosons can have fermionic excitations emerging at low energies. This is true in any dimensions.

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user Xiao-Gang Wen
@Xiao-GangWen I would be interested to read more about only boson-theory. Do you have any reading recommendations or the technical term for this theory?

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user JakobH
@JacobH: See arxiv.org/abs/1210.1281 and arxiv.org/abs/cond-mat/0302460

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user Xiao-Gang Wen
+ 2 like - 0 dislike

First of all, the Standard Model doesn't treat bosonic fields as classical. They're quantum mechanical i.e. non-classical, they're just not anticommuting or Grassmann-odd. Second, a consistent theory just requires the relationship between spin and statistics, see e.g. the

http://en.wikipedia.org/wiki/Spin-statistics_theorem

Combining integer spin with Fermi statistics leads to ghost or energy or the norm that isn't positively definite, and vice versa (half-integer spin with Bose statistics). It was proved by Pauli.

However, your very question didn't actually talk about the integer vs half-integer spin at all. It was talking about the relationship between fermions and anticommuting fields. This is almost a tautology. A fermion is a particle whose wave function for many particles is antisymmetric, $\psi(x_1,x_2)=-\psi(x_2,x_1)$ etc., so the fields that create these particles must be anticommuting, $a^\dagger(x_1) a^\dagger(x_2)=-a^\dagger(x_2)a^\dagger(x_1)$. The multiparticle state in QFT is written as $$ |\text{2 fermions}\rangle = \int d^3 x_1 d^3 x_2\, \psi(x_1,x_2) a^\dagger(x_1)a^\dagger(x_2)|0\rangle $$ Because the wave function $\psi$ is antisymmetric, only the antisymmetric combination $a^\dagger(x_1)a^\dagger(x_2) - a^\dagger(x_2)a^\dagger(x_1)$ contributes to the state, and in fact, only this combination is nonzero. The sum – the anticommutator – vanishes. That's why the antisymmetry of $\psi$ is "automatic": if there were a non-antisymmetric part of $\psi$, it would vanish in the integral above because the product of the creation operators is antisymmetric.

The same for bosons and "commuting", without the minus sign.

The answer to your "why" question is that your statement is really a tautology, pretty much a definition of bosons and fermions, up to the possibly confusing comments about "one antisymmetry" implying the "other antisymmetry" above. Of course, you could also ask why one uses commuting or anti-commuting fields to describe particles at all. Well, Nature just works in this way. Quantum fields naturally reduce to multi-body quantum mechanics with the automatic symmetry or antisymmetry – and they may give rise to automatically Lorentz-invariant theories, too (something that would be hard in the "non-relativistically styled" multiparticle quantum mechanics).

This post imported from StackExchange Physics at 2014-04-11 15:50 (UCT), posted by SE-user Luboš Motl
answered Oct 4, 2012 by Luboš Motl (10,248 points) [ no revision ]

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