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Is there any way in quantum field theory to derive the Pauli exclusion principle for fermions?

I looked at the wiki page but it just says,

In relativistic quantum field theory, the Pauli principle follows from applying a rotation operator in imaginary time to particles of half-integer spin.

but there it gives no references for mathematical or even physical basis of this sentence.

Is there any introductory reference for that statement? or can anybody explain it, please?

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    $\begingroup$ Go through Zee's qft book for detailed and clear explanation. Basic reasoning is that if you don't use proper statistics then energy of the system will not be bounded from below. This will actually happen due to lack of Lorentz invariance of the different time-ordered product used in the construction of S-matrix. $\endgroup$ – ved Nov 23 '16 at 12:21
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Pauli exclusion principle is a consequence of the Fermi statistics for free fermionic fields. I am going to provide a sketch of the derivation here.

First, consider the bosonic case. The space of states free bosonic quantum field (Fock space) is constructed by applying the bosonic creation/annihilation operators

$$ a(p), \; a^{\dagger}(p). $$

The commutation relations read

$$ \left[ a(p), \; a^{\dagger}(q) \right] = \delta_{pq}, $$

where $\delta_{pq}$ is the appropriate delta function distribution. In what follows we will consider a single degree of freedom labeled by the momentum $p$ and build a subspace of the Fock space spanned by various excitations of this degree of freedom. We have thus

$$ \left[ a, \; a^{\dagger} \right] = 1, \quad [a, a] = [a^{\dagger}, a^{\dagger}] = 0.$$

One starts with the Poincare vacuum state $\left| 0 \right>$ and applies $a^{\dagger}$ multiple times to obtain a generic state. A typucal element of the basis would look like

$$ \left| n \right> \sim \left( a^{\dagger} \right)^n \left| 0 \right>. $$

Its physical interpretations is the following: we have exactly $n$ particles with momentum $p$ when the system is in the state $\left| n \right>$.

Now consider the Fermionic case. In this case we have the anticommutation relation:

$$ \left\{ a, \; a^{\dagger} \right\} = a a^{\dagger} + a^{\dagger} a = 1, $$ $$ \frac{1}{2} \left\{ a^{\dagger}, \; a^{\dagger} \right\} = \left( a^{\dagger} \right)^2 = 0. $$

We start with the fermionic Poincare vacuum $\left| 0 \right>$. But the same construction no longer holds, because we have $\left| 2 \right> = 0$ because of the second anticommutation relation. We have thus a 2-dimensional Hilbert space for any Fermionic degree of freedom with basis elements

$$ \left| 0 \right> = a \left| 1 \right>, \quad \left| 1 \right> = a^{\dagger} \left| 0 \right>. $$

We interpret these states as "there is no particle with momentum $p$" and "there is a particle with momentum $p$". This is exactly the Pauli exclusion principle: there couldn't be more than one particles filling the first-quantized state with momentum $p$.

Now to how this is connected to the rotations. The result I'm referring to is called the spin-statistics theorem, it states that

  1. quantum fields with integer spins can only be consistently quantized using Bose-Einstein statistics (commutation relations), consequently they don't obey the exclusion principle
  2. quantum fields with half-integer spins can only be consistently quantized using Fermi-Dirac statistics (anticommutation relations), consequently they obey the exclusion principle

By "consistently quantized" I mean that the resulting free quantum field theory would satisfy the Wightman axioms, in particular, the energy spectrum would be bounded from below.

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  • $\begingroup$ Thanks... it really helped... but i think there is some kind of cycling deduction here... between being fermion and obeying the fermi dirac statistics(pauli-exclusion principle)... right? $\endgroup$ – P.A.M Nov 27 '16 at 15:29
  • $\begingroup$ @P.A.M not at all, this is just a definition. The point is that integer-spin fields are bosons (obey Bose-Einstein, commute) and half-integer spins are fermions (obey Fermi-Dirac, anticommute). $\endgroup$ – Prof. Legolasov Nov 27 '16 at 23:57
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    $\begingroup$ now, we can ask:" why are they fermions!!!? $\endgroup$ – P.A.M Nov 28 '16 at 14:19
  • $\begingroup$ You mean why is the assertion of the spin-statistics theorem valid? It is a well-known fact, you can find proofs in many QFT textbooks. For an introduction, see en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem $\endgroup$ – Prof. Legolasov Nov 28 '16 at 14:54

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