# Klein-Gordon Field Quantization and Bose-Einstein Statistics in Peskin & Schroeder

I am trying to understand how Klein-Gordon particles obey Bose-Einstein statistics from Peskin & Schroeder's QFT textbook (page no. 22). The excerpt is given below:

From this passage it is clear to me that the two-particle states: $$a_\textbf{p}^\dagger a_\textbf{q}^\dagger|0\rangle, a_\textbf{q}^\dagger a_\textbf{p}^\dagger|0\rangle$$ are equal due to the commutation relation: $$[a_\textbf{q}^\dagger, a_\textbf{p}^\dagger] = 0$$ derived earlier in the chapter. They represent to a two-particle system of total energy $$\omega_\textbf{p} + \omega_\textbf{q}$$ and total momenta $$\textbf{p} + \textbf{q}$$. But the following sentence is not clear to me:

Moreover, a single mode $$\textbf{p}$$ can contain arbitrarily many particles...

Also, I don't see how the statements of this paragraph fit with the fact that for a two-particle Bosonic (or, Fermionic) system, if we swap the particles the wave function does not (or, does) change. Therefore, I couldn't follow the argument of why Klein-Gordon particles obey Bose-Einstein statistics. Could you please help me to understand this?

• With fermions, the modes' creation operators anticommute.
– J.G.
Commented Nov 1, 2021 at 7:43

So if $$[a_\textbf{q}^\dagger, a_\textbf{p}^\dagger] = 0$$ and therefore $$a_\textbf{p}^\dagger a_\textbf{q}^\dagger|0\rangle=a_\textbf{q}^\dagger a_\textbf{p}^\dagger|0\rangle$$

means that under particle exchange the two-particle state stays unchanged or $$|\bf p,q\rangle=|\bf q,p\rangle$$ if the two particle state is represented by $$|\bf p,q\rangle$$. This is not the case if the particles are fermions, since there is a sign change under particle exchange $$|\bf p,q\rangle=-|\bf q,p\rangle$$ and the creation operators anti-commute.

Note that the statement "Moreover, a single mode p can contain arbitrarily many particles" is consistent for bosons and not fermions, since there is no limit to particle numbers with bosons but there obviously is with fermions due to the Pauli exclusion principle.

I don't see how the statements of this paragraph fit with the fact that for a two-particle Bosonic (or, Fermionic) system, if we swap the particles the wave function does not (or, does) change.

This statement is true for bosons and not fermions for which the state changes sign under the exchange of particles.

He has basically stated that particles obeying the Klein-Gordon equation (the KG equation describes spin zero bosons), or "Klein-Gordon particles" will obey Bose-Einstein statistics using a mathematical argument explaining that the state describing the system does not change under particle exchange.

• Thanks for explaining. As the book doesn't compare with the case of Fermions, I didn't have any rigorous clue how particle exchange (i.e., swapping the momentum indices) corresponds to the Bosons. Commented Nov 1, 2021 at 6:49
• That’s Ok. Good luck with your studies. Commented Nov 1, 2021 at 6:50

$$\newcommand{\Ket}[1]{\left|#1\right>}$$

In the Klein-Gordon field, a creation operator $$a_p^\dagger$$ represents the creation of a single particle in mode $$p$$. The statement that a single mode can contain arbitrarily particles is simply a statement that, just like the quantum harmonic oscillator, $$(a_p^\dagger)^n \Ket{0}$$ does not vanish and gives a unique state for every $$n$$ (representing a state with $$n$$ particles in mode $$p$$).