# Terms allowed in Lagrangian density [duplicate]

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!

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## marked as duplicate by Qmechanic♦Oct 22 '14 at 0:08

Check out physics.stackexchange.com/q/4102 maybe you'll find your answer there. – Neuneck Jun 12 '13 at 6:29
Possible duplicates: physics.stackexchange.com/q/4102/2451 , physics.stackexchange.com/q/18588/2451 and links therein. – Qmechanic Jun 12 '13 at 19:07

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.

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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 ) – Trimok Jun 13 '13 at 9:14
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! – user34801 Jun 13 '13 at 13:32
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. – MarkWayne Jun 13 '13 at 20:21
@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. – MarkWayne Jun 14 '13 at 4:02
thanks for pointing that out! – user34801 Jun 14 '13 at 6:35

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.

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-1: Effective field theory is a completely consistent way of quantizing higher order fields. – MarkWayne Jun 13 '13 at 5:43
Are you able to explicitely exhibit a Lagrangian density with higher-order derivatives, and a coherent quantization ? – Trimok Jun 13 '13 at 8:38
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.) – MarkWayne Jun 13 '13 at 20:17
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. – Trimok Jun 14 '13 at 2:23
It's good that you've learned something. But your last comments here are statements that don't appear to be well defined at best and are probably incorrect, at worst. I don't say this to beat you down -- but you might want to refine these statements for your own benefit. – MarkWayne Jun 14 '13 at 17:22

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.

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This is true as long as the higher-order terms are included in the interaction part of the Lagrangian. – MarkWayne Jun 13 '13 at 5:44

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.

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-1: It isn't an assumption. Please see my answer. – MarkWayne Jun 13 '13 at 5:44