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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?

Qmechanic
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user34801
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4 Answers4

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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.

Community
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MarkWayne
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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.

xaxa
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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.

Trimok
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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.

Philip Gibbs - inactive
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