• 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.


PO is now at the Physics Department of Bielefeld University!

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

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,345 answers , 22,720 comments
1,470 users with positive rep
818 active unimported users
More ...

  Auxiliary field and loop expansion

+ 1 like - 0 dislike

Something bugs me with the use of auxiliary fields in QFT. On one side I understand that they are nothing more than Lagrange multipliers and should be replaced by their equation of motions in the action, and this works perfectly on-shell (by definition) but what about this replacement at the quantum level, where the equations of motions may not be satisfied? Let's say I have an action with one physical and one auxiliary field, I do two hbar expansion (loop expansion if you prefer): the first after getting rid of my auxiliary field, the second keeping track of it, will I, for sure, find the same calculable quantities in those loops?

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user toot
asked May 21, 2012 in Theoretical Physics by toot (445 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

Calculating observables in a quantum field theory is a subtle process that must be done correctly and "correctly" means to be careful about the right physical interpretation and required physical properties of the theory at each moment (or at least the final moment when we release the results).

You are implicitly talking about theories that are constructed out of their classical limit. The right guide to construct the quantum theory and its observables is to do everything to obey the actual physical criteria, not to do some mindless operations of "quantization". When we do things correctly, we are really solving the puzzle "what is the right full quantum theory that has a certain classical limit" and we must realize that there is no fully universal and mechanical "recipe" to find the answer. One has to preserve the unitarity and gauge symmetries (decoupling of unphysical polarizations of particles), Lorentz symmetry, and other things we expect to hold in the quantum theory. This dictates how everything is treated; it also implies that we must do regularization, renormalization, and we may discover some new parameters that enter into counterterms etc.

When it's done correctly, every classical starting theory produces the same quantum theory or the same set of quantum theories (or it can be proved to be incompatible with quantization, e.g. in the presence of gauge anomalies). A particular formalism is just a tool, not the master.

In particular, when you talk about auxiliary fields, they are unphysical degrees of freedom in the quantum theory for pretty much the same reasons as they are unphysical in the classical theory. First, it is not really true that quantum mechanically, the fields don't obey the equations of motion. They do. In the Heisenberg picture, the operators obey the Heisenberg equations of motion. They're formally the same equations as those in classical physics, with hats added (operator equations). In particular, auxiliary fields obey non-differential i.e. non-dynamical, algebraic equations and they're unphysical. They don't produce new degrees of freedom. (One must first construct the Hilbert space and auxiliary degrees of freedom, having no canonical momenta because of the absence of derivatives, never make the Hilbert space any bigger.)

In the Feynman path integral approach, the path integral over auxiliary fields is an infinite-dimensional ordinary integral that treats each spacetime point separately from others. So it may be simply integrated over them whenever you calculate the amplitudes, whether they're scattering amplitudes or Green's functions or something else. What is special about the auxiliary fields is that they don't have derivative terms in the action; they don't propagate. For that reason, they don't produce any propagators in the Feynman diagrams, either. Equivalently, you may say that the propagators for the auxiliary fields are "infinitely short" in all the Feynman diagrams so they get attached to the vertices. The attachment of these short propagators to vertices is a visual way to express the fact that you're solving them; you are just expressing the auxiliary fields that enter vertices as functions of the other fields. (These "shrunk propagators" appear e.g. if you describe a supersymmetric theory by the Feynman rules in the superspace and reinterpret the Feynman diagrams in the language of Feynman diagrams for component fields.)

You may imagine that the auxiliary fields are very massive fields. Why? Because the coefficients of their mass terms (terms without derivatives but I mean not only the bilinear ones here) are much larger than the coefficients of their kinetic terms (with derivatives) and the ratio is nothing else than the masses (or squared masses for bosons). This analogy tells you that the auxiliary fields, much like the very massive fields, are first "integrated out" of the Feynman path integral.

Quite generally, there is some freedom and "lack of complete rigor" in the semi-mechanical procedures used in quantum field theory. But the general idea, as I said, is to guarantee that we produce a consistent quantum theory with the given classical limit and the required symmetries. For this reason, classically equivalent starting points are meant to produce the same final quantum product, too. Auxiliary fields and various fields redefinitions are examples of the "classical reformulations" that can make the quantization harder or easier. Both examples exist.

You may surely think of artificial examples in which it's simply possible to get rid of the auxiliary fields and obtain a theory whose quantum calculations are easier and spoiled by fewer subtleties. In that case, you should do it. You may also avoid it and get the same quantum results at the end but you will have to deal with unnecessary extra subtleties along the way. However, there are also the opposite examples. Auxiliary fields may simplify one's life.

The Nambu-Goto action, a starting point of string theory, is a great example. This action is proportional to the area of the two-dimensional world sheet spanned by the propagating one-dimensional string in the spacetime (it's multiplied by a constant known as the string tension). The proper area is an integral of a square root of the determinant of the induced metric (from the spacetime to the world sheet) and it is a horribly non-polynomial action that would be hard to quantize. However, one may describe the system in an equivalent way, using an extra degree of freedom, an auxiliary metric tensor on the world sheet (whose equations of motion simply say that it's equal to the induced metric from the $X$ fields, up to an overall scaling of the tensor), and the action actually acquires a transparent quadratic (bilinear) form.

In the right variables, it is a free theory! This bilinear action with the usual Klein-Gordon-like kinetic terms for the scalar fields $X$ describing the embedding of the world sheet in the spacetime (Polyakov action if we also include the Faddeev-Popov ghosts) is easy to be quantized. The final product, perturbative string theory, may be viewed as the quantization of the non-polynomial, square-root-spoiled Nambu-Goto action, too, although we didn't apply a mechanical quantization procedure to the Nambu-Goto action directly and mindlessly.

Auxiliary fields often simplify the calculations and reduce the technical subtleties. After all, one component (and, in some counting, two components) of any gauge field potential may be viewed as an auxiliary field; the number of physical polarizations of the photon is just two, fewer than the 4 components of the potential $A_\mu$. However, using the whole $A_\mu$ including some unphysical components – which become ordinary auxiliary fields if we add some gauge-fixing terms – is a way to preserve the manifest Lorentz symmetry of the Maxwell system and its excitations, photons.

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user Luboš Motl
answered May 21, 2012 by Luboš Motl (10,278 points) [ no revision ]
Thanks for this clear explanation ;)

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user toot
+ 2 like - 0 dislike

Your confusion is legitimate: why can we use the classical equation of motion to reduce the auxiliary field, when in the path integral you integrate over all values of the auxiliary field, including those that don't satisfy the classical equations of motion?

The simplest way to justify this is by the path-integral version of the Fourier transform delta function representation:

$$ \int e^{ikx} dk \propto \delta(x)$$

Similarly, in a path integral:

$$ \int e^{i\int \phi(x) \eta(x)} D\eta \propto \delta(\phi)$$

When you integrate over all $\eta$, you end up setting $\phi$ to zero at each point. This is true for any expression in place of $\phi$, it is obvious when you put the theory on a lattice, since the integral over $\eta$ reproduces the delta function at each site.

This is the general situation when you have a nonpropagating field. The field fluctuates wildly from point to point, because there are no derivative terms to control its value between neighboring lattice sites. The integral over the auxiliary field just sets the thing it multiplies to zero.

It is important to emphasize that that in Euclidean space the auxiliary field terms stay oscillatory if you do the continuation correctly. Some authors make it unclear by ommiting these necessary i's.

In any other circumstance, you can see the way the path integral reproduces the classical equations for an auxiliary field, and it's actually clearer than in classical mechanics, where Lagrange multipliers have an air of black magic.

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user Ron Maimon
answered May 22, 2012 by Ron Maimon (7,720 points) [ no revision ]
Thank you for your intuitive argument =)

This post imported from StackExchange Physics at 2014-05-01 12:21 (UCT), posted by SE-user toot

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).
Please complete the anti-spam verification

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

Your rights