# What is the point of complex fields in classical field theory?

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?

• Could you please justify/elaborate point 3? I don't think it is correct, both a complex scalar field and two real scalar fields can be Lorentz invariant. It is really the action that is required to be invariant, the fields themselves are automatically, since they are by definition "scalar" fields (as opposed to vector fields, tensor fields etc.). – Wolpertinger Jan 29 '16 at 21:41
• Complex fields, just like the use of a complex numbers to represent voltages or currents in electronic ac circuits is a mathematical convenience device. There is no pretense that the physical system actually has a complex valued voltage, current or field value. We are also not saying that the complex valued wave function has a physical meaning. It's simply an intermediate calculation device that gives the correct result. – CuriousOne Jan 29 '16 at 21:43
• @Numrok I'm not sure about that point, too. The idea is that a complex number is by definition a en.wikipedia.org/wiki/Scalar_(physics) quantity, so invariant under coordinate transformation. But a vector of two real quantities is not a scalar. So it is not invariant under coordinate transformations. – asmaier Jan 29 '16 at 22:03
• @asmaier you are not dealing with a vector of two real quantities. You are dealing with a set of 2 real scalars. – Wolpertinger Jan 29 '16 at 22:08
• @Numrok But isn't a set of two real scalars a vector? The two scalars also have the same units and it is often said, that complex numbers can be represented as vectors. – asmaier Jan 29 '16 at 22:15

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.

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.

• Actually, that's the same mechanism for the Weinberg rotation in Standard Model. – Wein Eld Jan 29 '16 at 22:10

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.