Why are complex fields in the Lagrangian?

I know that a complex field has twice the number of degrees of freedom of a real field, and that fields (in QFT) aren't observables so we don't really care if they are real.

But why the need for complex fields? Is there stuff that doesn't work unless there's a complex field?

• Strictly speaking they are not compulsory. You can still do with real fields by taking multiplets, but sometimes it is more convenient to use complex numbers instead. – Phoenix87 May 4 '15 at 18:23
• Essentially a duplicate of physics.stackexchange.com/q/11396/2451 and links therein. – Qmechanic May 4 '15 at 18:32

There is no non-trivial one-dimensional representation of $\mathrm{U}(1)$ on a scalar field $\mathbb{R}^4\to\mathbb{R}$, but on complex fields $\mathbb{R}^4\to\mathbb{C}$, we have the one-dimensional "phase" representations by $$\phi\mapsto\mathrm{e}^{e\mathrm{i}\chi}\phi$$ for $e\in\mathbb{Z},\chi\in\mathfrak{u}(1)\cong\mathbb{R}$ for $\mathrm{U}(1)$ parametrized as $\chi\mapsto \mathrm{e}^{\mathrm{i}\chi}$ (the unit circle in the complex plane).

Since $\mathrm{U}(1)$ is the archetypical example of a continuous (gauge) symmetry (think of electromagnetism), complex scalar fields are an important (toy) model in QFT.

Of course, every complex scalar field may equivalently be replaced by two real scalar fields being its real and imaginary part, so they are not actually needed, but using only real fields may complicate the actual calculations and notations immensely.

When switching from a complex scalar $\phi$ to two real ones $\mathrm{Re}(\phi),\mathrm{Im}(\phi)$, we observe that $$\mathrm{e}^{e\mathrm{i}\chi}\phi = (\cos(e\chi) + \mathrm{i}\sin(e\chi))(\mathrm{Re}(\phi) + \mathrm{i}\ \mathrm{Im}(\phi))$$ and so, writing the real vector $\widetilde{\phi} = \left( \begin{matrix} \phi_1 := \mathrm{Re}(\phi) \\ \phi_2 := \mathrm{Im}(\phi)\end{matrix}\right)$, we see that the complex one-dimensional representation of $\mathrm{U}(1)$ turns into a two-dimensional real one with $$\widetilde{\phi}\mapsto R_e(\chi)\widetilde{\phi}$$ with the rotation matrix $$R_e(\chi) := \left(\begin{matrix}\cos(e\chi) & -\sin(e\chi) \\ \sin(e\chi) & \cos(e\chi)\end{matrix}\right)$$ which is now looking more like a representation of the real 2D rotations $\mathrm{SO}(2)$ (the usual one for $e = 1$). As a real representation, this is irreducible (you cannot diagonalize all rotation matrices at once), so you cannot reduce the degrees of freedom and still have a non-trivial representation of $\mathrm{U}(1)\cong\mathrm{SO}(2)$. Two real d.o.f. are the minimum to have some kind of non-trivial continuous symmetry going on, since $\mathrm{U}(1)$ is the simplest Lie group apart from the un-exciting $\mathbb{R},+$.

• and why do we need two real fields? As in, why 2 degrees of freedom? Why not 3? – SuperCiocia May 4 '15 at 20:20
• @SuperCiocia: Because a complex number $z$ is equivalently described by two real numbers $\mathrm{Re}(z)$, $\mathrm{Im}(z)$. – ACuriousMind May 4 '15 at 20:25
• Yes I know that, I meant why do we need two degrees of freedom for our field theories? Why not 3 or 4? – SuperCiocia May 4 '15 at 20:27
• @SuoerCiocia: Mainly because, as I say, there is no non-trivial representation of $\mathrm{U}(1)$ (or any other relevant Lie group, for that matter) on one real degree of freedom. You can do a theory of a real scalar, but it will be boring (in particular, it won't have electromagnetism). – ACuriousMind May 4 '15 at 20:32

What type of fields are you using?

If you are working with spinor fields, the representation of Lorentz transformations is complex. So even if the field is real in some reference frame, if you switch to another reference frame it will become complex. There's no way to avoid complex spinor fields.

Actually, you can do without complex fields, at least in some general and important cases, and I don't mean replacing a complex field with two real fields. Schroedinger noted that, in the case of a scalar field interacting with electromagnetic field (the klein-Gordon-Maxwell electrodynamics, or scalar electrodynamics), you can use the so-called unitary gauge, where the scalar field is real. You can also write an equivalent Lagrangian with a real field (please see, e.g., Eq.14 in my article http://akhmeteli.org/akh-prepr-ws-ijqi2.pdf (published in Int'l J. Quantum Information) - the Lagrangian was derived by Takabayashi). What about spinor fields? @Bosoneando, e.g., believes that "There's no way to avoid complex spinor fields". Surprisingly, there is. I showed in http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in J. Math. Phys.) (see also http://arxiv.org/abs/1502.02351) that three out of for complex components of the Dirac spinor in the Dirac equation can be algebraically eliminated in a general case. The remaining component can be made real by a gauge transform.

• Hi, @akhmeteli, I'm sorry for the late reply. I hadn't seen your answer before. I have to tell you that your arXiv paper is wrong: You can't take derivatives when you're solving a differential equation. Not all solutions of (5) are solutions of (1). – Bosoneando Jul 11 '16 at 23:41
• Imagine you want to solve $i\partial_x y=y$, whose solution, $y=C e^{-ix}$, is complex. What you're trying to do is $y = i \partial_x y = i \partial_x(i\partial_x y) = -\partial_x^2 y$, so the solution is $y=A\cos x + B\sin x$, which is real if $A$ and $B$ are. BUT it is not, for general $A$ and $B$, a solution of the original equation. – Bosoneando Jul 11 '16 at 23:41
• Also, you don't address the main point of my answer: the spinor representation of the Lorentz group are complex. If you require the spinor to be real, you're singling out a reference frame and breaking Lorentz invariance. – Bosoneando Jul 11 '16 at 23:42
• @Bosoneando: "You can't take derivatives when you're solving a differential equation. Not all solutions of (5) are solutions of (1)." While I agree that "Not all solutions of (5) are solutions of (1)", it does not mean my paper is wrong. Moreover, I explicitly wrote in my preprint: "the set of solutions of equation (5) used to derive equation (27) is broader than the set of solutions of the Dirac equation (cf. [4])." You should specifically show what is wrong in my article, otherwise I'll have to consider your critique unfounded. – akhmeteli Jul 12 '16 at 3:36
• @Bosoneando: "which is real if A and B are. BUT it is not, for general A and B, a solution of the original equation." Again, you should show what specifically is wrong in my preprint. So far I don't see how this is relevant. – akhmeteli Jul 12 '16 at 3:40