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

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1445 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 (1,125 points) [ revision history ]
edited Jul 4, 2014 by Peter Morgan

2 Answers

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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 (1,605 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 (12,385 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.)

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