# Why there is the requirement for derivatives no higher than second order in free quantum field equations? [duplicate]

Why there is the requirement for derivatives no higher than second order in free quantum fields equations?

We can get the equations for the free fields of an arbitrary spin by using the requirements of linearity of the equations, derivatives no higher than second order. But why there is the second requirement? Is it connect with initial conditions $A_{\mu}(0), \partial^{\mu}A_{\nu}(0)$ for the field? For example, Schrodinger equation is first-order time-derivative equation, because we have the postulate that if we know the $\psi (\mathbf r , t)$ in the present, we know it in the future.

• Possible duplicates: physics.stackexchange.com/q/18588/2451 , physics.stackexchange.com/q/4102/2451 and links therein. – Qmechanic Sep 26 '13 at 16:52
• @Qmechanic . But I asked the special question: "Is it connect with initial conditions $A_{μ}(0),\partial_{μ}A_{ν}(0)$ for the field?". – user8817 Sep 26 '13 at 16:57
• Any equation can be put into first order form by introducing new variables, so the restriction has to do with another requirement. The true requirement is that the space of solutions must have a norm and be a unitary representation of the space-time symmetries, the Poincare group. In higher derivative theories one generally finds states with negative norm. – John Sep 26 '13 at 18:21
• @John . How to show that higher derivative theories give states with negative norm? – user8817 Sep 26 '13 at 20:14

Higher order derivatives (usually?) correspond to non-renormalizable terms in the Lagrangian. Every term in the Lagrangian density needs to have units of $[M^4]$, so that the action is unitless (in units where $\hbar = 1$). For example, take a scalar field $\phi$. This has units $[M]$, or equivalently $[1/D]$. Each derivative adds a factor of $[M] = [1/D]$ (think of it as $\delta / \delta x$).

With no derivatives, you can form the standard mass term

$$\frac{1}{2} m^2 \phi^2$$

The coefficient $m$ must thus have units $[M]$, making it a mass.

You can't form a Lorentz scalar with only one derivative, since $\partial_\mu \phi$ is a one-form and without breaking rotational invariance there is no vector to take an inner product with. The next possibility is two derivatives. The only allowable term is

$$\frac{1}{2} \left( \partial_\mu \phi \right) \left( \partial^\mu \phi \right) \underset{\text{parts}}{\sim} - \frac{1}{2} \phi \partial^2 \phi$$

This term doesn't have a unitful coefficient, because the two $\phi$'s and two $\partial$'s make it have units $[M^4]$ already. A term $m \partial^2 \phi$ is equivalent to $0$ by integration by parts, so with no boundary and fields that die off at infinity they don't matter. A term like $m(x) \partial^2 \phi$ would contribute, but breaks translational invariance.

What about three derivatives? Same as with one - there is no four-vector to contract the third derivative with, so this is impossible without breaking rotational (and possible translational) invariance with some preferred four-vector field $v^\mu(x)$.

The next possibility is four derivatives. A generic possibility is

$$\frac{1}{\Lambda^2} \left( \partial^2 \phi \right)^2$$

Because of the four derivatives and two appearances of $\phi$, the term needs a coefficient with units $[M^{-2}]$, which is why I've added $1/\Lambda^2$ where $\Lambda \sim [M]$. An inclusion of this term requires a mass scale $\Lambda$. Like the Fermi four-fermion theory, this would violate unitarity at energies $\sqrt{s} \gg \Lambda$.

You can go through the same exercise for spin-1/2 and spin-1 fields and reach much the same conclusion, although the details will be different.

• Thank you! But how to prove, that, generally, adding the derivative higher than third order-derivative to the lagrangian of the theory leads to it's non-unitarity? – user8817 Sep 28 '13 at 20:18
• That I do not know. I only know a little bit of field theory, i.e. to the level of "We'll show this to tree level or first loop level; person X proved this non-perturbatively / to all orders in perturbation theory / everyone believes it but it's not proven." A full proof that such third-order derivatives aren't possible might exist (barring exotic solutions), but it's likely very advanced. I'm sorry that I don't know where to point you to find this. – jwimberley Sep 30 '13 at 12:38