# The concept of particle in a quantum field

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

• related (duplicate?): One particle states in an interacting theory, What's a particle anyway?. Apr 27, 2017 at 9:26
• 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 $p^2 =m^2$ Apr 27, 2017 at 9:57
• @AccidentalFourierTransform Thank you for your references! They help a lot. Apr 27, 2017 at 10:08
• @innisfree: For example, how to define a particle for a field with a strange Hamiltonian $H=\int d^3x[\pi^4+(\nabla \pi\cdot\nabla\phi)^2+m^2\phi^4]$? Apr 27, 2017 at 10:12
• @innisfree I think OP means you Fourier transform the free field Hamiltonian to find the harmonic oscillator form, which is true. Apr 27, 2017 at 10:42

• 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? Apr 27, 2017 at 12:45
• 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. Apr 27, 2017 at 12:48
• 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 $v\in V$. Apr 27, 2017 at 12:55