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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 $H=\int d^3x[\pi^4+(\nabla \pi\cdot\nabla\phi)^2+m^2\phi^4]$?

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    $\begingroup$ related (duplicate?): One particle states in an interacting theory, What's a particle anyway?. $\endgroup$ Apr 27, 2017 at 9:26
  • $\begingroup$ Can you give an example of the kind of theory you're talking about? The motivation for your first equation isn't just that's it's a Fourier transform - it's the general solution to the K.G. equation if $p^2 =m^2$ $\endgroup$
    – innisfree
    Apr 27, 2017 at 9:57
  • $\begingroup$ @AccidentalFourierTransform Thank you for your references! They help a lot. $\endgroup$
    – StupidBird
    Apr 27, 2017 at 10:08
  • $\begingroup$ @innisfree: For example, how to define a particle for a field with a strange Hamiltonian $H=\int d^3x[\pi^4+(\nabla \pi\cdot\nabla\phi)^2+m^2\phi^4]$? $\endgroup$
    – StupidBird
    Apr 27, 2017 at 10:12
  • $\begingroup$ @innisfree I think OP means you Fourier transform the free field Hamiltonian to find the harmonic oscillator form, which is true. $\endgroup$
    – gautampk
    Apr 27, 2017 at 10:42

1 Answer 1

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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.

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  • $\begingroup$ Thank you for answering! For example, given the commutation relation $[\phi(\boldsymbol r),\pi(\boldsymbol r')]=i\delta(\boldsymbol r-\boldsymbol r')$, how to construct the creation and annihilation operators? $\endgroup$
    – StupidBird
    Apr 27, 2017 at 12:45
  • $\begingroup$ The commutation relations are not properly defined (mathematically) like that. They are defined starting from a symplectic space $(V,\sigma)$, by means of the Weyl relations (unitary representation of the Heisenberg group): $W(v)W(w)=e^{-i\sigma(v,w)}W(v+w)$. This is roughly speaking the exponentiated version of the relations you write (the "right" version for a series of mathematical reasons). The set of operators $\{W(v),v\in V\}$ generate a C*-algebra that can be represented on Hilbert spaces by means of the GNS construction. $\endgroup$
    – yuggib
    Apr 27, 2017 at 12:48
  • $\begingroup$ Given now a regular state (a state for which the $\mathbb{R}$-action $\lambda\mapsto \omega(W(\lambda v))$ is continuous for any $v\in V$), it is possible to define in its GNS space $H_\omega$ the generator of the unitary operator $\pi_{\omega}(W(v))$. Such generator is the field operator calculated in $v\in V$. Splitting into real and imaginary part (roughly speaking) one gets the creation and annihilation operators of the classical state $v\in V$. $\endgroup$
    – yuggib
    Apr 27, 2017 at 12:55
  • $\begingroup$ The mathematics is so deep for a physics student... But at leat I may get the message -- there exists some way to systematically define the creation/annihilation operators. Thanks! $\endgroup$
    – StupidBird
    Apr 27, 2017 at 13:07

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