# Polar Decomposition of a Complex Scalar Field

People often write a complex scalar field via polar decomposition. What does this parametrization precisely mean?

To be more explicit consider the following Lagrangian of a complex scalar field with a $U(1)$ symmetry, $${\cal L} = - m ^2 \left| \phi \right| ^2 - \frac{ \lambda }{ 4} \left| \phi \right| ^4 + \left| \partial _\mu \phi \right| ^2$$

We can then make a field transformation, $$\phi (x) = \rho (x) e ^{ i \theta (x) }$$ We typically refer to $\rho (x)$ and $\theta (x)$ as real scalar fields, but this is a strange for a couple reasons. Firstly, the $\rho (x)$ field is positive definite. This is a boundary condition that we don't normally see in QFT. Its especially weird if $\rho$ is quantized around $0$ since it can't go in the negative direction''.

The second reason I think calling $\rho (x)$ and $\theta (x)$ quantum fields is strange is because after we write out the Lagrangian we get, $${\cal L} = - m ^2 \rho ^2 - \frac{ \lambda }{ 4} \rho ^4 + \partial _\mu \rho \partial ^\mu \rho + \rho ^2 \partial _\mu \theta \partial ^\mu \theta$$ The $\theta (x)$ field doesn't have a proper kinetic term and so can't be a propagating field!

If these objects aren't quantum fields then how should I think about them?

• Maybe I'm not understanding it, but isn't $\rho$ just the radius (hence it is positive definite) and $\theta$ the angle of the Higgs potential (the Mexican hat)? – Hunter Apr 15 '14 at 21:19
• @Hunter: Basically, yes. But in this reparameterization are we justified in calling $\rho$ and $\theta$ fields and use them for example in scattering computations? I think the answer is no, but I have seen people refer to them as fields. – JeffDror Apr 15 '14 at 21:56
• I think this is only true for a classical field theory. Otherwise, if you consider $\phi$ as a field composed of creation and annihilation operators, it is not clear that you can make such a decomposition. – Melquíades Apr 15 '14 at 23:16
• @JSchwinger: I agree it isn't clear how that would work or whether it can be done at all... – JeffDror Apr 16 '14 at 11:12
• I think this kind of separation is used to illustrate, for instance, Goldstone theorem (with a Mexican hat potential). With a decomposition $\rho = \rho_0 + \rho'$, you will find (with some redefinition) a kinetic term for $\theta$ – Trimok Jul 16 '14 at 10:22

I think the confusion arises from indistinct usage of the term scalar fields when referring to a physical field, and when referring to scalar functions. I will try to answer your question clarifying separately these elements:

• Scalar field vs Scalar function
• Complex numbers' polar representation
• What this parametrization means...

Scalar field vs Scalar function: In Mathematics a scalar function is any function mapping some vector space to its body (taking several arguments and returning a number, see Vector Space in Wikipedia). In Quantum Field Theory (QFT), the field function describing a boson's field is usually called a scalar field, because it has the mathematical properties of a scalar function, and also because of its transformation properties (see V. Parameswaran Nair-Chapter2 for a simple view, more serious definitions R.Blumenhagen-Chapter2). But most of all, while a scalar field is a scalar function, it describes a quantized field, meaning it will have further mathematical constraints. See also Scalar Field in Wikipedia.

Complex number polar representation: This is one of the several ways we can represent complex numbers in terms of two separate real numbers (very well covered in Complex Number in Wikipedia). Any complex number can be expressed in this representation: if $c=a+ib$ is a complex number with $a$ and $b$ the real and imaginary components, you can always find $\rho=\sqrt{a^2+b^2}$ and $\theta=\arccos{\frac{a}{\rho}}=\arcsin{\frac{b}{\rho}}$, which is easy to infer from the geometrical representation of complex numbers in a 2D Euclidean space. From here you get the restriction of $\rho > 0$, but this does limit this representation, i.e. you can represent all complex numbers.

What this parametrization means: $\phi$ here is both a scalar field in the QFT sense, and also a mathematical scalar field or function since it maps from $\mathbb{R}^4$ to $\mathbb{C}$. The transformation $$\phi_{(x)}=\rho_{(x)} e^{i\theta_{(x)}}$$ is using the polar representation to express $\phi$ in terms of two mathematical scalar functions $\rho$ and $\theta$, each of them mapping from $\mathbb{R}^4$ to $\mathbb{R}$.

But one must bear in mind that $\rho$ and $\theta$ are not physical scalar fields, thus after substitution into the Lagrangian you have only varied the mathematical expression (thus the strange kinetic energy term for $\theta$''). In other terms, although both $\rho$ and $\theta$ make up the physical scalar function necessary to characterize your scalar field $\phi$, they are not separate scalar fields by themselves. So, your mathematical parametrization of $\phi$ does not translate into a physical decomposition of a boson field into 2 other boson fields.