The low-energy effective description of a given QFT is an expansion of the form $$ \mathcal L_\mathrm{IR}\sim\sum_{n,m} \lambda_{n,m}\phi^n\partial^ m\phi $$ where we include all terms that are compatible with the symmetries of the original UV theory.

The usual narrative is that $\mathcal L_\mathrm{IR}$ captures all the physics below some scale, in other words, for sufficiently low energy the effective description should be indistinguishable from the original theory.

Now, the effective description typically has higher derivative terms, thus by Ostrogradsky's theorem it has no ground state: the Hamiltonian is unbounded from below. Given this, what does it even mean for $\mathcal L_\mathrm{IR}$ to describe low-energy physics? There is no notion of low-energy in $\mathcal L_\mathrm{IR}$, as this theory contains states at all energies, both positive and negative. There is no distinguished choice of origin, so it appears there is no way of defining "low-energy"!

The original UV theory has a ground state, and there is a well-defined notion of low-energy physics: it is all physics whose energy scale is between the ground state and some UV cutoff. The effective description does not admit a ground state, so how do we even specify an energy range for which it is supposed to be valid (and agree with the UV theory)?


1 Answer 1


Here's a slick way to derive the Ostragradsky instability in a higher derivative theory, which we can use to illustrate the main point.

Let's consider the following higher derivative theory for a scalar field$^\star$:

\begin{equation} S = \int {\rm d}^4 x \left(-\frac{1}{2} (\partial \phi)^2 + \frac{1}{M^8} (\square \phi)^4 \right) \end{equation} where $M$ is a mass scale.

First, to see the ghost, let's perturb around a background, $\phi=\bar{\phi}+\varphi$, with $\square{\bar{\phi}}=M^3$.

To first order in $\varphi$ around this background, the equation of motion is \begin{equation} \left(\square + \frac{\square^2}{M^2}\right)\varphi = 0 \end{equation} and the propagator for $\varphi$ in momentum space is \begin{eqnarray} G &=& \frac{1}{k^2 + \frac{k^4}{M^2} - i \epsilon} \\ &=& \frac{M^2}{ k^2 ( k^2 + M^2) - i\epsilon} \\ &=& \frac{1}{k^2 - i\epsilon} - \frac{1}{k^2 + M^2- i\epsilon} \end{eqnarray} So we can see that there are two degrees of freedom, and the propagator for one has the wrong sign, indicating a ghost.

However, we can view this Lagrangian as an effective field theory for one scalar degree of freedom, $\phi$, below the scale $M$. In this case, we should proceed perturbatively, with $k\ll M$. We will never actually produce the ghost degree of freedom, on shell, since the mass of the ghost is $M$. When quantizing the theory, we should not include a creation operator for the ghost.

Returning to the original Lagrangian, and forgetting about the background, the way to proceed consistently is to treat it as a theory with one healthy scalar degree of freedom. We write a propagator for a massless scalar field. In terms of Feynman diagrams, we treat the $(\square \phi)^4$ interaction as a 4-$\phi$ vertex, with powers of momentum associated with the legs.

Of course a true effective field theory should include an infinite tower of higher derivative interactions. You can view this as a toy model only including the first few terms. Where this picture will break down, is if $k$ becomes of order the cutoff. Then we cannot consistently ignore the infinite tower of interactions. The infinite tower of interactions is non-local, and signals the onset of a new degree of freedom.

I used Section 2 of the following reference in writing this up: https://arxiv.org/abs/1401.4173

Another useful example to have in mind, is a top down approach, where a integrating out massive particle can generate an infinite tower of interactions below the mass scale of the particle. Truncating this infinite tower at finite order will apparently lead to an Ostragradsky ghost, which is however not present in the UV theory.

Let's start with

\begin{equation} S = \int {\rm d}^4 x \left(-\frac{1}{2} (\partial \phi)^2 - \frac{1}{2} (\partial \psi)^2 - \frac{1}{2} M^2 \psi^2 + \lambda \phi^3 \psi \right) \end{equation}

Then we can integrate out $\psi$ by replacing $\psi$ in the action with the formal solution to the equations of motion \begin{equation} \psi = \frac{\lambda \phi^3}{\square+M^2} \end{equation} This leads to a non-local term in the action. However, at energy scales below $M$, we can Taylor expand this expression in powers of $(\square/M^2)$ to generate a tower of interactions \begin{equation} S = \int {\rm d}^4 x \left(-\frac{1}{2} (\partial \phi)^2 + \frac{\lambda\phi^3}{M^2} \left[1 - \frac{\square}{M^2} + \left(\frac{\square}{M^2}\right)^2 + \cdots \right] \phi^3 \right) \end{equation} If you truncate this infinite tower at a finite order in $k/M$ (which you would do for an EFT approach), you would find interactions like $(\partial^2 \phi)^6$, which naively would lead to an Ostragradsky instability. However, this is really just an artifact of not including the full set of interactions of the UV complete theory above $M$. Turning this around, given the EFT for $\phi$, we won't know the correct UV completion in terms of $\psi$, but we can develop a consistent EFT that works at scales below $M$ by proceeding perturbatively.

$^\star$ There's a technical footnote on my first example. Technically, since the interaction $(\square\phi)^4$ is proportional to the lowest order equations of motion, you can push this interaction to higher order using a field redefinition. However, you can't completely remove the interaction, and the algebra is simpler this way, so I am going to ignore this subtlety in the post.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.