Quantcast
  • 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.

News

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

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Why is a Lagrangian density enough to specify a quantum field theory?

+ 7 like - 0 dislike
2822 views

A Lagrangian density, a polynomial in a classical field and its derivatives, can be used to specify a dynamics of a conservative classical field, without too much restriction on differentiability of the initial classical field configuration.

A quantum field, however, is a higher-order mathematical structure —an initial quantum field state specifies an infinite system of incompatible probability measures over classical field configurations— so why do we not need a higher-order object than a Lagrangian density to specify the dynamics?

If we work with a crude regularization, a cutoff \(\Lambda\), for example, effectively using the quantum field in convolution with a smooth cutoff function \(f_\Lambda(x)\), \(\hat\phi_\Lambda(x)=[\hat\phi\star f_\Lambda](x)\), the regularized quartic Lagrangian density would be \(:\!\frac{\lambda}{4!}(\hat\phi^\dagger_\Lambda(x)\hat\phi_\Lambda(x))^2\!:\), but we could also use a quartic Lagrangian density \(:\!\sum_i\frac{\lambda_i}{4!}(\hat\phi^\dagger_{\Lambda_i}(x)\hat\phi_{\Lambda_i}(x))^2\!:\), or a continuum version, \(:\!\int\frac{\lambda(z)}{4!}(\hat\phi^\dagger_{\Lambda(z)}(x)\hat\phi_{\Lambda(z)}(x))^2\mathrm{d}z\!:\). Is there a proof somewhere that these and all comparable constructions will be no different from the single \(\Lambda\) case, no matter how we take the limits \(\Lambda_i,\Lambda(z)\rightarrow \infty\) and no matter how we take the coupling, mass and field renormalizations? ADDED: From an effective field theory perspective the finite cutoff case is of interest in itself, for which case the multiple cutoff field examples are distinct from the single cutoff field example. [Preferably take this last paragraph to be illustrative, not as part of the question.]

asked Jul 4, 2014 in Theoretical Physics by Peter Morgan (1,230 points) [ revision history ]
edited Jul 4, 2014 by Peter Morgan

2 Answers

+ 5 like - 0 dislike

It is not always enough to specify a Lagrangian density.

For example, in this paper http://arxiv.org/abs/1305.0318 it is shown that one must make choices about what sorts of non-local operators one wishes to consider in the theory, it being not consistent with quantization constraints to include all possible ones.

One could maybe argue that this is not quite an example, since in (my paper with Anton Kapustin) http://arxiv.org/abs/1308.2926 we find that this can be interpreted as a choice of topological action for a (discrete) 2-form field built using $\pi_1$ of the gauge group. EDIT: So maybe some Lagrangian densities are better than others?

answered Jul 7, 2014 by Ryan Thorngren (1,925 points) [ revision history ]
edited Jul 8, 2014 by Ryan Thorngren
+ 3 like - 0 dislike

One needs a Lagrangian density plus a working renormalization scheme. The latter must settle all ambiguities hidden in going from the classical Lagrangian formulation to the quantum field theory.

Ordinary quantum mechanics is just 0+1-dimensional quantum field theory. Here giving a Lagrangian is essentially equivalent to giving a classical Hamiltonian, and the corresponding quantum Hamiltonian is determined by it only up to operator ordering, i.e., up to adding an arbitrary commutator (which vanishes in the classical limit). 

Higher-dimensional QFTs have in principle the same sort of ambiguities, but the requirement of locality places stronger restriction on which ambiguities lead to local quantum field theories.

answered Jul 12, 2014 by Arnold Neumaier (15,787 points) [ no revision ]

I find "a Lagrangian density plus a working renormalization scheme" a very helpful way to state what is needed. Many thanks.

But any theory can be renormalizable, if one wants it strongly. If you obtain a function $f(x)$ from your theory and you need $g(y)$ instead to agree with experiment, you just discard $f(x)-g(y)$ from your $f(x)$ and you get the desirable $g(y)$. You just need to know the right answer $g(y)$. If you do not know it, your theory will remain non renormalizable.

Renormalization appeared in CED and it did not settle "ambiguities".

Your recipe for renormalization has little to do with what the term means in QFT.

One can discard terms only if they vanish in an appropriate limit.

@ArnoldNeumaier: Then comment my post, please. (If something vanishes itself, no need to discard it.)

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:
p$\varnothing$ysicsOverflow
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
...