# Terms allowed in Lagrangian density

+ 5 like - 0 dislike
856 views

I never fully understood why the Lagrangian density (let's say for a scalar field) was restricted to have only first order derivatives of the field in time and space in QFT. One reason I can think of is to ensure locality. But that doesn't restrict one from having a finite number of higher-order derivatives of the field, does it? Thanks!

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user user34801

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user Neuneck
Possible duplicates: physics.stackexchange.com/q/4102/2451 , physics.stackexchange.com/q/18588/2451 and links therein.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user Qmechanic

+ 4 like - 0 dislike

First, it's not true that the Lagrangian density is restricted to have first-order space and time derivatives. The example of the scalar field, \begin{align} \mathcal{L}=\tfrac{1}{2}\partial_\mu\phi\partial^\mu\phi - \tfrac{1}{2}m^2\phi^2 + \mathcal{L}_{int}(\phi) \end{align} is a clear counterexample.

Now, following Weinberg, QFT I, section 10.7 on the Kallen-Lehmann representation, we can show (I won't reproduce the full derivation here) that inclusion of higher-order derivatives in the $\mathcal{L}-\mathcal{L}_{int}$ (the free Lagrangian) is inconsistent with the positivity postulate of quantum mechanics. The mathematical statement is that the exact two-point function -- the $\phi$-propagator -- must behave as \begin{align} \Delta'(p) \underset{p^2\to\infty}\longrightarrow \frac{1}{p^2}. \end{align}

From the perspective of effective field theory, however, it is possible to have derivative couplings of any finite order in a local theory (infinite orders are inherently non-local), as long as we introduce a dimensional (usually mass) scale of the appropriate dimension at each order. (See Motl's answer to Why are differential equations for fields in physics of order two? for a full explanation.)

The answers given earlier are all incorrect for one or more reasons. See the comments.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user MarkWayne
answered Jun 13, 2013 by (270 points)
I suppose, that, by "inconsistent with the positivity postulate of quantum mechanics", you mean that the Hamiltonian may have a negative spectrum. If so, it is not a coherent quantization... Now, it seems maybe that a distinction must be done between fundamental theories and effective theories. But fundamental theories certainly suffer from this problem (it is also known as Ostrogradski instability , which has a classical part and a quantum part )

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user Trimok
Thanks for the response! I understand the problem better now. I read Motl's answer and while I understand everything he says from a mathematical perspective, I'm still confused about the physical grounds for his argument. Is it always true that the distance-scale "L" in Motl's answer that appears in the coefficients of the derivative terms is a microscopic scale that can be neglected when you raise it to an integer power? Thanks!

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user user34801
No, Trimok, that is not what "positivity postulate" means. Consider having a look at the reference to Weinberg that I provided. user34801: The higher-order terms are always associated with a heavy, unmeasured particle (think, roughly speaking, of the W boson in Fermi's 4-fermion theory of the weak interaction). The more derivatives, the more inverse powers of this heavy mass must appear at the vertex. In this way, the higher-order terms are strongly suppressed with order.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user MarkWayne
@user34801: it's self-serving to point this out now but if you are helped by one answer in way that is clearly more useful than others, you should accept that answer.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user MarkWayne
thanks for pointing that out!

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user user34801
+ 1 like - 0 dislike

This is a quantum problem.

People don't know how to quantize fields in Lagrangian densities wich appears with a total derivative order greater than 2, while keeping a coherent theory.

For instance, the term $\partial_i \Phi \partial^i \Phi$ has a total derivative order of 2, and we know how to quantize this field, and get a coherent theory.

But it is no more true with higher total derivative orders.

In fact, the Euler-Lagrange equations are more complex, so the definition of the canonical momenta associated to $\Phi$ (in our example) is more complex too.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user Trimok
answered Jun 12, 2013 by (955 points)
-1: Effective field theory is a completely consistent way of quantizing higher order fields.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user MarkWayne
Are you able to explicitely exhibit a Lagrangian density with higher-order derivatives, and a coherent quantization ?

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user Trimok
The chiral effective theory of mesons is an obvious example (see work by Meissner and collaborators). General relativity may be quantized as an effective theory (Holstein and others have worked this out.)

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user MarkWayne
Well, Ok, effective field theory is a special case, see this ref especially page 28,29. But my caveat is correct about fundamental theories. Of corse, if you begin with the initial Lagrangian (without higher-order derivatives), and integrate out some heavy fields then use a pseudo-lagrangian (with higher-order derivatives) which reproduces, at low energy, the same scattering amplitudes, I concede it is an other context. It is a very particular process.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user Trimok
I'll have to disagree again. There is no distinction between "fundamental" and "effective" theories. All realistic modern theories are "effective," the standard model in particular. So EFT is not a special case, at all. The only special cases are the renormalizable theories. Perhaps this is what you mean by "fundamental."

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user MarkWayne
You can see the point I'm making regarding the generality of EFT's if you look at the reference to Meissner's talk you supplied above. In particular, pages 16 & 17.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user MarkWayne
The standard model is renormalizable. And there is certainly always a gap between a fundamental theory, and a realistic theory. More technically, in the effective Lagrangian with high-order derivatives, there is no "standard quantization", because it would involve a conjugate moment which would be too complex because or the high-order derivatives. So, I see always a difference between a "pure" Quantum Field Theory, and an effective field theory. Nethertheless, I have learned something, and it is the most important...

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user Trimok
+ 1 like - 0 dislike

In classical physics there is no particular reason why higher derivatives of the fields cannot be used. One example is general relativity where the Einstein-Hilbert action has first and second derivatives of the metric field.

In quantum field theory the range of Lagrangians that can be quantised consistently is quite limited and does not include anything with higher derivatives, but it is not sure that this is an unsurmountable limitation.

Text books probably only treat the first derivatives because it keeps the analysis simple and covers most basic example.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user Philip Gibbs
answered Jun 12, 2013 by (650 points)
This is true as long as the higher-order terms are included in the interaction part of the Lagrangian.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user MarkWayne
+ 0 like - 1 dislike

This is an assumption.
For instance, in classical mechanics Lagrangian depends only on velocities (first derivatives of position) of particles, thus you have second order diff. equation of motion (Newton's law). Likely, you get second order partial diff. equations from Lagrangian [pedantic mode] if high-order derivatives in Lagrangian are not an exact divergence [/pedantic mode].

You can get equations based on this assumption and check it experimentally, so far they do agree.

This post imported from StackExchange Physics at 2014-07-13 04:40 (UCT), posted by SE-user xaxa
answered Jun 12, 2013 by (-10 points)
 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): Email me at this address if my answer is selected or commented on: 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$\hbar$ysicsOve$\varnothing$flowThen 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.