Sorry for the lengthy question, pretty much the whole text is the standard derivation of the solution of the KG equation which I included to illustrate my doubts, and some questions are at the end. The Klein-Gordon equation is

$$ (\partial^2+m^2)\phi(x)=0\tag{1}$$ where $\partial^2=\partial_\mu\partial^\mu$. Taking a Fourier transform of $\phi$

$$\phi(x)=\int \frac{d^4 k}{(2\pi)^4}\phi(k)e^{ikx} \tag{2}$$

and inserting it into the equation we have

$$ \int \frac{d^4 k}{(2\pi)^4}(k^2-m^2)\phi(k)e^{ikx}=0\tag{3}$$

which implies

$$(k^2-m^2)\phi(k)=0 \implies \phi(k)=2\pi f(k)\delta(k^2-m^2) \tag{4}$$

for some function $f$. Define $\omega=\sqrt{\mathbf{k}^2+m^2}$ where $k=(k_0, \mathbf{k}$), then $$\delta(k^2-m^2)=\frac{1}{2\omega}\left[\delta(k_0-\omega)+\delta(k_0+\omega)\right]\tag{5}$$

And our solution becomes

$$\phi(x)=\int \frac{d^4 k}{(2\pi)^3}\frac{1}{2\omega}\left[\delta(k_0-\omega)+\delta(k_0+\omega)\right]f(k)e^{ikx}\tag{6}$$

performing the integration on $k_0$

$$\phi(x)=\int \frac{d^3 k}{(2\pi)^3}\frac{1}{2\omega}(f(\omega,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+i\omega t}+f(-\omega,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}-i\omega t})\tag{7}$$

In the second term, we can change variable $\mathbf{k}\rightarrow-\mathbf{k}$ and get

$$\phi(x)=\int \frac{d^3 k}{(2\pi)^3}\frac{1}{2\omega}(f(\omega,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+i\omega t}+f(-\omega,-\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}-i\omega t})\tag{8}$$

Now: all the sources I can find go on and impose that the field must be real, and write something like

$$\phi(x)=\int \frac{d^3 k}{(2\pi)^3}\frac{1}{2\omega}(a(\omega,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+i\omega t}+a^\dagger(\omega,\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}-i\omega t})\tag{9}$$

and this is presented as the "general solution to the Klein-Gordon equation"


  1. Why should we impose that the field is real, i.e. $f(-\omega, -\mathbf{k})=f(\omega, \mathbf{k})^\dagger$? I see no mathematical reason to do so, and I don't see why a complex field would pose physical problems. I know about the harmonic oscillator interpretation, but I'd say that this interpretation is a consequence of the fact that the field is real and not vice versa. Why do we eliminate complex field solutions?

  2. Are the $2\pi$ factors important? In equation $(4)$ I introduced a factor $2\pi$ just so that the final solution would coincide with the one given in standard sources, but again, I don't see another reason to add it (actually the whole "the solution must be a function times delta" is a bit sketchy, how to see this?)

  3. Some give the solution with $\sqrt{\omega}$ instead of $\omega$ in the denominator (Peskin & Schroeder for instance) in analogy with the harmonic oscillator. To try and get this I thought of defining $a(\omega,k)=\sqrt{\omega}f(\omega,k)$ instead of $a(\omega,k)=f(\omega,k)$. Does this make sense? Do the two solutions have any physical difference?

  • $\begingroup$ 1. It depends on whether it is a real or a complex Klein-Gordon field. $\endgroup$
    – Qmechanic
    Commented Nov 24, 2018 at 12:48
  • $\begingroup$ @Qmechanic I see, so we divide the two cases and in this case we chose the real one, which I guess should be some special case of the complex one? What's the physical meaning of requiring a real field? $\endgroup$ Commented Nov 24, 2018 at 12:55

1 Answer 1


Answering the three questions in order:

1) Both the real scalar field and the complex scalar field are important and physically useful, and refer to different classes of physical objects. Imposing that $\phi$ is real means that you are interested in the real scalar field solutions of the Klein-Gordon equation. For example, a real scalar field represents a spin-0 particle which is its own antiparticle, and a complex scalar field represents a spin-0 particle that is not its own antiparticle.

2) The factor of $2\pi$ comes from the fact that $x$-space and $k$-space are related by a Fourier transform. Where to put the factors of $2\pi$ is a matter of convention. You can either require:


$$\tilde{f}(k)=\int dx\;e^{ikx}f(x)$$

or you can require:

$$f(x)=\int dk\;e^{-ikx}\tilde{f}(k)$$

$$\tilde{f}(k)=\int \frac{dx}{2\pi} e^{ikx}f(x)$$

or you can even require:



There are actually an infinite number of choices for conventions for defining this Fourier transform. Mathematicians tend to like the third "symmetric" convention, while in QFT we tend to use the first one. For a more in-depth treatment of this, see Fourier transform standard practice for physics.

3) The choice of scaling is arbitrary, as the calculation of any observable will not be affected by this (again, as long as the choice of convention is consistent throughout the calculation).

  • $\begingroup$ Thanks for the answer! A few follow ups: 1) I started to review KG when I was asked to find solutions to the Proca equation, which is equivalent to 4 KG fields. Is it correct to say that, in light of what you said, a massive real Proca field describes something like W and Z vector bosons, while a massive complex Proca field describes some vector boson that is not its own antiparticle? I cannot think of an example. Should there also be a complex photon field, as a massless Proca field, that describes a massless spin 1 boson that is not its own antiparticle? $\endgroup$ Commented Nov 24, 2018 at 13:24
  • $\begingroup$ @user2723984 First of all, the $W$ is not its own antiparticle, and whether the $Z$ is depends on your definition of "its own antiparticle." Second, the massive vector bosons in the Standard Model are not describable by massive Proca fields. First of all, the Proca Lagrangian with $m\neq 0$ breaks gauge invariance. Second, the $W$ and $Z$ are products of electroweak symmetry breaking, and the fields that generate the physical $W$ and $Z$ bosons are a mix of the electromagnetic field and a set of weak fields. $\endgroup$ Commented Nov 24, 2018 at 13:34
  • $\begingroup$ Ok sorry I messed that up, I'm new to QFT. Then forget about what I said about W and Z, I'll the standard model for when I understand something about QFT ;) What about the rest? Is this dichotomy always true, i.e. that a real field describes something that is its own antiparticle and a complex field doesn't? $\endgroup$ Commented Nov 24, 2018 at 13:37
  • $\begingroup$ @user2723984 The simplest place where you'll find a physical situation involving a complex scalar field is in scalar QED: en.wikipedia.org/wiki/Scalar_electrodynamics. This approximates the interaction between charged pions and photons at energies where the internal structure of the pion and its pseudoscalar nature are not relevant. As for real scalar fields, the only physical fundamental real scalar field known to exist is the Higgs field (though you might find other examples in condensed matter physics). $\endgroup$ Commented Nov 24, 2018 at 13:38
  • 1
    $\begingroup$ @user2723984 It's more than just "something that is its own antiparticle;" specifically, it's a spin-0 particle which is its own antiparticle. Likewise, a complex scalar field describes a spin-0 particle which is not its own antiparticle. $\endgroup$ Commented Nov 24, 2018 at 13:44

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