I am looking for a generic treatment or a concrete example where canonical quantization is performed without using free fields. For a scalar field

$$\phi(x,t) \sim \sum_k \phi_k{(t)} \, u_k(x) + \text{h.c.}$$

$$\pi(x,t) \sim \sum_k \pi_k(t) \, u_k(x) + \text{h.c.}$$

$$[\phi_k, \pi_l] = -i\delta_{k,l}.$$

This means that the $u_k(x)$ are not plane waves but some different orthonormal functions.

The idea behind is this

  • the free theory is partially not useful, e.g. for the non-perturbative IR sector of QCD
  • for solitons, e.g. the Sine-Gordon-Model, the $u_k(x)$ describe „distorted waves“ in a non-trivial kink-background
  • creation and annihilation operators would create / annihilate these non-trivial excitations
  • the basic rules for quantisation should be identical
  • details re normal ordering, asymptotic states etc. would look different or could even be just formal, e.g. if the functions $u_k(x)$ are not explicitly known

Does anybody know a specific example?

  • $\begingroup$ are you looking for a specific example? because in general you find that a field, just as a scalar field, is given by plane waves if the creation and destruction operators are harmonic oscillators. In this case the time dependece of these opertors is absorbed by the Fourier basis terms, hence you got plane waves. But if the dynamic is not trivial the creation and destruction operator could be inharmonic oscillators and you don't end up with a free field. If this satisfy you I coud try explaining it with more detail $\endgroup$
    – Ratman
    Jun 2, 2020 at 8:56
  • $\begingroup$ Thanks, that‘s clear, I am just looking for examples. But anyway, a more detailed discussion could be interesting. $\endgroup$
    – TomS
    Jun 2, 2020 at 19:28

1 Answer 1


You can of course change 'basis' when expressing canonical fields. The answer is perhaps simpler for creation/annihilation operators (which are related to canonical fields via $\phi \sim a + a^\dagger$ and so on).

To be precise $a^\dagger_f$ adds, in the Fock space, a particle with wavefunction $f$ with the correct symmetry (antisymmetric for Fermions, symmetric for Bosons). And analogously $a_g$ "destroys" or removes a particle with wavefunction $g$.

Often we consider $a^\dagger_x$ (or $a^\dagger_k$), i.e. the operator that adds a particle at position $x$ (or with momentum $k$). This actually introduces extra mathematical complications as these functions are not normalizable (i.e. do not belong to the Hilbert space). In any case physicists can get along without worrying too much about this.

In any case when raising and lowering operators are properly defined they satisfy:

$$ [a_f , a_g^\dagger] = \langle f| g\rangle $$

for bosonic particles, where $\langle f| g\rangle$ is the scalar product in the Hilbert space. Likewise we have

$$ \{a_f , a_g^\dagger \} = \langle f| g\rangle $$

in case of fermions. Now you can easily see the effect of creating arbitrary wavefunctions by exploiting the fact that $a_f$ is conjugate linear in $f$and $a^\dagger_g$ is linear in $g$.

Pick your favorite basis $\{e_i\}$ in the Hilbert space. You can expand

\begin{align} f &= \sum_i f_i e_i \\ g &= \sum_i g_i e_i \end{align}

and obtain for example

\begin{align} a_f &= \sum_i f_i^* a_{e_i} \\ a_g^\dagger &= \sum_i g_i a_{e_i}^\dagger \end{align}

and of course we have (say for bosons)

$$ [a_{e_i} , a_{e_j}^\dagger] = \delta_{i,j} $$

  • $\begingroup$ Thanks a lot, but this is clear to me; I am just looking for examples. $\endgroup$
    – TomS
    Jun 2, 2020 at 19:26
  • $\begingroup$ Well, in condensed matter $a_k^\dagger$ creates Wannier states or Bloch states depending on context.. $\endgroup$
    – lcv
    Jun 2, 2020 at 19:35
  • $\begingroup$ thanks for the hint $\endgroup$
    – TomS
    Jun 2, 2020 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.