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I see a lot of books/lectures about classical field theory making use of complex scalar fields. However why complex fields are used in the first place is often not really motivated. Sometimes one can read as a footnote that a complex field is in principle mathematically equivalent to two real fields (see also What is the difference between complex field and two scalar fields?), but then the author often goes on using a complex field anyway.

This is confusing, because from quantum mechanics one learns that a complex quantity is not measurable. This is of course not the case in classical field theory, where both the real and the imaginary part must be simultaneously measurable quantities.

I heard physically motivated reasons for using complex fields like:

  1. A complex scalar field represents different particles than a vector of two real fields. But this argument doesn't make sense in classical field theory, it is (if at all correct) only relevant in quantum field theory.

  2. Only a complex field can represent charged particles, real fields are necessarily neutral.

  3. A complex scalar field is a scalar and so it is by definition Lorentz invariant. A vector of two real fields is not Lorentz invariant and so one must use a complex field.

But I'm unsure which of these reasons (if any) is really valid. What is the point of using complex fields in classical field theory?

asmaier
  • 10,250

3 Answers3

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Two real scalar fields $\phi_1$ and $\phi_2$ satisfying an $SO(2)$ symmetry and one complex scalar field $\psi$ are equivalent. However, the latter is more convenient because the particles made by $\psi$ and $\psi^\dagger$ are each others' antiparticles. In the real case, the fields that have this property are $\phi_1 \pm i \phi_2$, so once you change basis from $\phi_1$ and $\phi_2$ to $\phi_1 \pm i \phi_2$ you've reinvented the complex scalar field.

This is explained nicely starting from p.53 in Sidney Coleman's QFT notes.

As you said, a complex quantity is not measurable in QM. And indeed $\psi$ is not an observable, which feels strange because quantum fields are often motivated at the start of a QFT course as local observables. Unfortunately this motivation isn't quite right, as we rarely measure quantum fields directly. For example, the number density, charge density, and current density for a charged complex scalar field are all field bilinears like $\psi^\dagger \psi$, and hence real.

knzhou
  • 107,105
5

In my viewpoint, the free complex field theory $$\tag 1 \mathcal{L}=\partial_\mu\phi\partial^\mu\phi^*-\frac{1}{2}m^2\phi\phi^*$$ is actually equivalent to the free double real field theory $$\tag 2 \mathcal{L}=\partial_\mu\phi_i\partial^\mu\phi_i-\frac{1}{2}m^2\phi_i\phi_i,$$ where $i=1,2$. To see this, simply write $\phi=\phi_1+i\phi_2$ and you can obtain Eq.(2) from Eq.(1). Eq.(1) is simply a compact form of Eq.(2). If you like you can define $\psi=\left(\phi_1,\phi_2\right)^T$ and have the form $$\tag 1 \mathcal{L}=\partial_\mu\psi^T\partial^\mu\psi-\frac{1}{2}m^2\psi^T\psi$$.

The double real field Lagrangian is, of course, Lorentz invariant. It also can represent charged particles since it also has $O(2)$ invariance and we can identify the corresponding Noether charge as electric charge.

BUT, there indeed a very crucial difference between the complex field and double real field. $\phi$ and $\phi^*$ ''have'' the opposite charge while $\phi_1$ and $\phi_2$ do not (please convince yourself by looking into $\phi=\rho e^{i\theta}$ and $\phi^*=\rho e^{-i\theta}$ and the transformation of O(2) is actually a rotation for the $\theta$). Since the observables are the charges, so $\phi$ and $\phi^*$ represent the particles with opposite charge while $\phi_1$ and $\phi_2$ can not.

Wein Eld
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1

Complex fields naturally lend themselves to an associated charge and current density, and this is the main reason for their introduction in physical theories.

Consider the Lagrangian density $L = c^2\phi\phi^*-\mu_0^2c^2\phi\phi^*$. A transformation of the type $\phi' = \phi e^{i\epsilon}$, $\phi^{*'} = \phi^* e^{-i\epsilon}$ corresponds in infinitesimal form to a gauge transformation of the form$\delta x = 0$, $\delta \eta_\rho = \epsilon c_\rho \eta_\rho $ (no summation on $\rho$) where the $c_\rho$ are constants and $c = i$, $c^* =-i$. If the

Lagrangian density is invariant under this transformation, then there is a conservation equation of the form $\frac {d\Theta^\nu} {dx^{\nu}} = 0$, where $\Theta^\nu = c_\rho \frac {\partial{L}} {\partial{\eta_{\rho,\nu}}}\eta_\rho$.

$ \frac{d\Theta^\nu}{dx^\nu} = 0$ is in the form of an equation of continuity

with $\Theta^\nu$ in the role of a current density $j^\nu$. (Noether's theorem).

The given Lagrangian density is invariant under this transformation. Hence, there is an associated current density for the Klein-Gordon field that can be given as $ j_\mu = iq\left( \frac {d\phi} {dx^\mu} \phi^* - \phi \frac {d\phi*} {dx^\mu}\right)$, in agreement with the conventional QM current density.

The entire derivation of the conserved charge current density depends upon the fact that the field is complex.

Ref: Goldstein Classical Mechanics Chapter 13.

Mephistopheles
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