I am learning QFT and fermion statistics.

I am confused about whether the Pauli Exclusion Principle is a fundamental rule or it can be deduced from QFT?

I saw a sentence from wiki but I don't understand.

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

  • $\begingroup$ I may be wrong, but I believe it is a consequence of the commutation relations fundamental to a fermion field. The Commutation relations being kind of the "keystone" to quantum field theory. $\endgroup$ – R. Rankin Dec 14 '16 at 6:15

A fermion is a state $|\vec p,\sigma\rangle$ with half-integer spin, i.e., such that $$ J|\vec p,\sigma\rangle=\sigma|\vec p,\sigma\rangle\qquad\text{with}\qquad \sigma\in\mathbb N+\frac12 $$ where $J$ is the angular momentum operator (generator of rotations).

Therefore, upon a rotation by an angle $2\pi$ around an arbitrary axis, $$ U(2\pi)|\vec p,\sigma\rangle=\mathrm e^{i\pi}|\vec p,\sigma\rangle=-|\vec p,\sigma\rangle $$

Finally, if you have a two-fermion system, interchanging them is equivalent to rotating the system by an angle $2\pi$, and therefore $$ |\vec p_1,\vec p_2\rangle=U(2\pi)|\vec p_2,\vec p_1\rangle=-|\vec p_2,\vec p_1\rangle $$

To make this suggestive argument precise, one needs the CPT theorem, which can be proven in an axiomatic framework using just a couple of properties of quantum fields. For more details, see PCT, Spin and Statistics, and All That, by Streater and Wightman.

Spin, Statistics, CPT and All That Jazz by Baez might be of interest too (see the last entry for an explanation of what a rotation in imaginary time has to do with the spin and statistics).

  • $\begingroup$ Yeah! It makes sense. Thank you very much. I am going to see that book. By this argument the Pauli exclusion can then be a result of spin and the rotations. $\endgroup$ – ZHANG Juenjie Dec 15 '16 at 0:19
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    $\begingroup$ A follow-up question. $\endgroup$ – rob Jul 27 at 14:08

The spin-statistics theorem is a consequence of causality in a relativistic QFT. In order the theory to be causal, the commutator of the fields

$\left[ \Phi (x), \Phi(x^{\prime}) \right ]$

must vanish outside the light-cone, namely, for $(x-x^{\prime})^2>0$.

See the detailed discussion of the subject in Ch. 5 of Weinberg's first volume.

  • $\begingroup$ Well, for this I also know and it is also in Peskin & Schroeder chapter 2. But this only says that for fermions it should be the anti-commutation relations but it doesn't say anything about the pauli exclusion principle. $\endgroup$ – ZHANG Juenjie Dec 14 '16 at 8:19
  • $\begingroup$ I just did a very rough scratch of that chapter. It seems that it mainly explains the reason why some types of particles cannot be constructed while however it doesn't say the reason of Pauli exclusion principle. $\endgroup$ – ZHANG Juenjie Dec 14 '16 at 8:31
  • $\begingroup$ @ZHANGJuenjie Pauli principle is just a consequence of ant-symmetry of fermionic wave functions. Let's consider the wave function of two fermions $\Psi(x_1, x_2)$. It must be anti-symmetric under the exchange of the fermions, thus $\Psi(x_1, x_2)=\Psi_1(x_1) \Psi_2(x_2)-\Psi_1(x_2) \Psi_2(x_1)$. This implies that $\Psi(x_1,x_2)=0$ if the fermions are in the same state. $\endgroup$ – Andrey Feldman Dec 14 '16 at 8:46
  • $\begingroup$ Hello. but why it must be anti-symmetric under the exchange of the fermions. This is on Feynman Physics lectures, and I guess that we know the result and then constructed the wave function this way, $\endgroup$ – ZHANG Juenjie Dec 15 '16 at 0:16

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