In the book "Quantum Field theory and the Standard Model" by Matthew Schwartz, the author states:

In quantum field theory, we generally work in the Heisenberg picture, where all time dependence is in operators such as $\phi$ and $a_p$. For free fields, the creation and annihilation operators for each momentum $\bf{p}$ in the quantum field are just those of a simple harmonic oscillator. These operators should satisfy $a_p(t) = e^{-i\omega_p t}a_p$ and $a_p^\dagger(t) = e^{i\omega_p t}a_p^\dagger$, where $a_p$ and $a_p^\dagger$ are time independent. Then we can define a quantum scalar field as

$$\phi_0({\bf{x}}, t)=\int \dfrac{d^3 p}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_p}}(a_p e^{-ipx}+a_p^\dagger e^{ipx})$$

with $p^\mu = (\omega_p, {\bf{p}})$ and $\omega_p = |{\bf{p}}|$.

Now I first thought the author was presenting the KG field as just one example. I also thought that this form of writing $\phi$ was only for the KG field, after all it was developed in order to make the field satisfy the massless KG equation.

By this statement, it seems that the author implies this is valid for all quantum scalar fields. Is this true? I mean, this exact same expansion, with creation and annihilation operators defined in a Fock Space, is valid for all quantum scalar fields?

If so, what distinguishes one field from the other?

And why it would be reasonable for this to be so general as that, if it was derived from a very simple case?

  • $\begingroup$ Which book? Which author? Also, the passage you quoted does not seem to imply to me that the mode expansion is valid for all fields. It's saying that if you are given some creation/annihilation operator $a_p,a_p^\dagger$, then you may define the field like that. The point is that you are usually not given these operators for non-free fields. $\endgroup$
    – ACuriousMind
    Apr 5, 2017 at 14:39
  • $\begingroup$ I forgot to mention the book and author, I did it now. What confuses me is: this mode expansion was derived to make the field satisfy one specific PDE, namely, $(\Box + m^2)\phi = 0$. But as far as I know from Classical Field Theory, each field has its own equation of motion coming from its own lagrangian. My question might be better phrased as: are all free scalar fields the same as the Klein-Gordon field? Because that is the impression I get from this excerpt. $\endgroup$
    – Gold
    Apr 5, 2017 at 17:43

2 Answers 2


Note that the excerpt you cite says:

For free fields, the creation and annihilation operators for each momentum $\mathbf p$ in the quantum field are just those of a simple harmonic oscillator.

So the author is implying that all free, i.e. non-interacting, scalar fields are the same (apart from their mass), which is true because the differences between fields are down to their differing interactions.

The problem is that the states of an interacting field are not Fock states. Indeed we don't know what the states of an interacting field are since we cannot solve the equations for them and instead we have to fall back to using perturbation theory. That means we don't know what the creation and annihilation operators are either.

  • $\begingroup$ Thanks John Rennie. So when it comes to QFT, the only scalar free field is the Klein-Gordon field? This confuses me a little, because I imagined that there could be other equations of motion for free fields in QFT other than the KG equation. $\endgroup$
    – Gold
    Apr 5, 2017 at 17:45
  • $\begingroup$ @user1620696 Well, in addition to their differing interactions, you still have differences between the spin transformations and commutation relations between the free fields, even in the non-interacting case. But all of that is baked into the definitions of the creation and annihilation operators $\endgroup$ Apr 5, 2017 at 18:00
  • $\begingroup$ So, the KG field is different from the Dirac field, beause it is spin 0, and therefore a boson, which has consequences. $\endgroup$ Apr 5, 2017 at 18:01
  • $\begingroup$ And the photon field is different from both, because it transforms under rotations, while the KG field does not. $\endgroup$ Apr 5, 2017 at 18:03
  • $\begingroup$ @JerrySchirmer: the question did refer to free scalar fields. Neither the Dirac field nor the photon is a scalar field. $\endgroup$ Apr 5, 2017 at 18:38

As John Rennie's answer noted, interacting cases such as $\phi^4$ theory don't have this form. You can show the integral representation you quoted implies $\hat{\phi}$ has zero vacuum expectation value, which should tell you the Highs field is something different; indeed, this is due to such a quartic potential term, which leads to a nonlinear differential equation whose solutions don't comprise a vector space. That somewhat scuppers the attempt to provide such an integral representation.

It's also worth noting that, in a general curved spacetime, scalar fields satisfying a homogeneous linear equation such as KG have an analogous integral representation in which the ladder operators' coefficients are the classical solutions, which cannot be expected to be $\exp\pm ip\cdot x$ in most spacetimes.


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