• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

149 submissions , 124 unreviewed
4,020 questions , 1,442 unanswered
4,923 answers , 20,951 comments
1,470 users with positive rep
533 active unimported users
More ...

  Polar Decomposition of a Complex Scalar Field

+ 9 like - 0 dislike

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, \begin{equation} {\cal L} = - m ^2 \left| \phi \right| ^2 - \frac{ \lambda }{ 4} \left| \phi \right| ^4 + \left| \partial _\mu \phi \right| ^2 \end{equation}

We can then make a field transformation, \begin{equation} \phi (x) = \rho (x) e ^{ i \theta (x) } \end{equation} 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, \begin{equation} {\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 \end{equation} 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?

This post imported from StackExchange Physics at 2014-04-18 17:37 (UCT), posted by SE-user JeffDror
asked Apr 15, 2014 in Theoretical Physics by JeffDror (650 points) [ no revision ]
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)?

This post imported from StackExchange Physics at 2014-04-18 17:37 (UCT), posted by SE-user Hunter
@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.

This post imported from StackExchange Physics at 2014-04-18 17:37 (UCT), posted by SE-user JeffDror
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.

This post imported from StackExchange Physics at 2014-04-18 17:37 (UCT), posted by SE-user JSchwinger
@JSchwinger: I agree it isn't clear how that would work or whether it can be done at all...

This post imported from StackExchange Physics at 2014-04-18 17:37 (UCT), posted by SE-user JeffDror

This is a field parametrization which is only used for the low-energy excitations near a vacuum, it isn't meant to properly cover the whole space non-perturbatively. This is also essentially a duplicate of this question: http://www.physicsoverflow.org/13796/nonlinear-field-redefinitions , except asking about U(1) instead of SU(2).

I would not call it a duplicate ...

I think there are a couple of important subtleties outlined in the question. This field redefinition seems to have to some very strange properties...

1 Answer

+ 4 like - 0 dislike

According to established tradition, a field may take values in arbitrary manifolds. Thus the fields defined by the polar decomposition of a scalar field are perfectly fine fields: $\rho(x)$ is a field with values in the manifold of positive reals, and $\phi(x)$  is a field with values in the circle.

$\rho^2 \partial^2 \phi$ is a good kinetic term, as it is positive definite and quadratic in the fields. More is not required.

The fields represented as a linear combination of creation and annihilation operators are _very_ special, namely free fields with Heisenberg commutation relations. Most fields (including all fields in the standard model) cannot be written in this simple form.

It is just for pedagogic reasons (and since it suffices for pertirbation theory of $\phi^4$ theory) that textbooks start with Heisenberg fields, which may leave the wrong impression that there are no other fields, and that kinetic terms must have the special form of a free field theory.

answered Apr 19, 2014 by Arnold Neumaier (12,640 points) [ no revision ]

Thanks for your response. I find this very surprising, to say the least. I have a couple of follow up questions. 

1)Can you define a propagator for a field with a $\rho^2 \partial_\mu \theta \partial^\mu \theta $? This seems to me like a four-point function. How could you then calculate scattering amplitudes if a particle "can't propagate"?

2)If the field is restricted to be only positive values. Then how can we quantize the field? Normally we say the mass of the field is the second derivative of the field. But in this case the $\rho$ field doesn't even seem differentiable at $0$ due to the requirement at $\rho>0$.

1. To do perturbation theory, one expands around a local minimizer (not around zero!!) of the classical action plus (possibly space-time dependent) counterterms, then one gets a quadratic term with a good operator as propagator.

2. Mass is a property of the spectrum of a theory; the bare mass appearing in perturbative quantization has no meaning (it is renormalized away). Quantization can always be done via a path integral, though on general manifolds one must be careful to get the measure right. There is no single uniformly valid recipe, but cf. Weinberg, Vol. 1, Chapter 7.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights