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In QFT, take Klein-Gordon field as axample, the concept of particle is introduced only after making the Fourier transformation
$\phi (\boldsymbol x)=\int \frac{d^3\boldsymbol p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_\boldsymbol p}}(a_\boldsymbol p e^{i\boldsymbol p \cdot \boldsymbol x}+a_\boldsymbol p^\dagger e^{-i\boldsymbol p \cdot \boldsymbol x})$
$\pi (\boldsymbol x)=\int \frac{d^3\boldsymbol p}{(2\pi)^3} (-i)\sqrt{\frac{\omega_\boldsymbol p}{2}} (a_\boldsymbol p e^{i\boldsymbol p \cdot \boldsymbol x}-a_\boldsymbol p^\dagger e^{-i\boldsymbol p \cdot \boldsymbol x})$
and find that the energy and momentum operator are diagonal in $(a_\boldsymbol p^\dagger,a_\boldsymbol p)$:
$H=\int \frac{d^3\boldsymbol p}{(2\pi)^3}\omega_\boldsymbol p(a_\boldsymbol p^\dagger a_\boldsymbol p+\frac{1}{2})$
$\boldsymbol P=\int \frac{d^3\boldsymbol p}{(2\pi)^3}\boldsymbol p a_\boldsymbol p^\dagger a_\boldsymbol p$
Then (also considering the commutation relations) it is said that $a_\boldsymbol p^\dagger$/$a_\boldsymbol p$ creates/annihilates a particle with energy $\omega_\boldsymbol p$ and momentum $\boldsymbol p$.

However, in a general quantum field, it is not guaranteed that there exists a pair of creation/annihilation operator that can diagonalize the energy or momentum operator. In this case, how to define a "particle" in this quantum field? Is it true that the definition of a particle relies on the particular form of the Hamiltonian, and not all quantum field have the concept of "particle"?

Update:
For example, how to define a particle for a field with a strange Hamiltonian $H=\int d^3x[\pi^4+(\nabla \pi\cdot\nabla\phi)^2+m^2\phi^4]$?

StupidBird
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1 Answers1

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In any regular representation of the Weyl algebra of canonical commutation relations it is possible to define creation and annihilation operators corresponding to every "one-particlec state" (point in the infinite dimensional classical phase space). These operators may not all share common domain of definitions (i.e. it may not be possible to "excite" a given state with different arbitrary particles), but nonetheless they can all be defined.

So as long as the theory contains some algebra of canonical commutation relations (as it should be desirable from a physical standpoint), and the state which carries the GNS representation is regular, the concept of creation and annihilation of a field excitation (particle) can be defined. The Hamiltonian in this picture does not come directly into play, however it is expected that the state on which to construct the GNS representation should be the vacuum of the theory. Henceforth for any theory with a regular vacuum the "particle" point of view should make sense.

yuggib
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