The action of free scalar field theory is as follows: $$S=\int d^4 x \frac{\dot{\phi}^2}{2}-\frac{\phi(m^2-\nabla^2)\phi}{2}.$$

I have been thinking to redefine field as $\phi'(x)=\sqrt{m^2-\nabla^2}\phi(x)$, where I assume that I can separate the operator $(m^2-\nabla^2)$ as $\sqrt{m^2-\nabla^2}\sqrt{m^2-\nabla^2}$ and integrate one of them by part (by infinite expansion of the square root of Helmholtz operator). The hamiltonian becomes

$$H=\int d^3x\frac{\pi'(m^2-\nabla^2)\pi'}{2}+\frac{\phi'^2}{2}$$

One can check the equation of motion and solutions are consistent to the original free field theory (as it should be), the things I am worried are:

1. Is it fair to do the infinite expansion+integration by part?
2. Since there are infinite series in the field redefinition, is it nonlocal? According to Wavefunction in quantum mechanics and locality and Why are higher order Lagrangians called 'non-local'?, it seems to be nonlocal. However, since there is no time derivatives in the square root, the initial data at every space point should be two, which corresponds to one dof at every point. If the infinite expansion is valid, one can know the field configuration $\phi'(x)$ if we know $\phi(x)$, it's not clear to me that whether we need all those initial data for the infinite series of derivative.

The above simple example might be too trivial, the question I am investigating is whether the following action stable or not, $$H=\int d^3x \frac{1}{4}\pi \frac{1}{(1+2\beta \nabla^2)}\pi-\phi(\beta\nabla^2\nabla^2+\nabla^2)\phi,$$ where $\beta$ is a constant. I was thinking use the non-local field redefinition $$\pi':=\frac{1}{\sqrt{1+2\beta\nabla^2}}\pi, \phi':=\sqrt{1+2\beta\nabla^2}\phi$$ to rewrite the hamiltonian as $$H=\int d^3x \frac{1}{4}\pi'^2 -\phi\frac{(\beta\nabla^2\nabla^2+\nabla^2)}{1+2\beta\nabla^2}\phi,$$ which is guaranteed to be no-ghost. I agree the Hamiltonian I start with is rather unusual but it is a reduced Hamiltonian of a constrained higher derivative theory.

It is not obvious to see how to analyze this by usual canonical transformation, is there any good way to analyze this or could anyone tell me what will go wrong with this kind of field redefinition?

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+1 In my opinion, you should also write the interaction terms. –  drake Mar 12 '13 at 23:04

1) Lagrangian Formulation. The Lagrangian density for a massive free scalar in the $(+,-,-,-)$ convention is

$$\tag{1} {\cal L}~=~\frac{1}{2}d_{\mu}\phi ~d^{\mu}\phi-\frac{1}{2}m^2\phi^2.$$

The corresponding Euler-Lagrange equation is the massive Klein-Gordon equation $$\tag{2} (d_{\mu}d^{\mu}+m^2)\phi~=~0.$$

The momentum is $$\pi ~:=~ \frac{\partial \cal L}{\partial\dot{\phi}}~=~\dot{\phi}.$$

2) Hamiltonian Formulation. The Hamiltonian density is

$$\tag{3} {\cal H}~=~\frac{1}{2}\pi^2 +\frac{1}{2}({\bf \nabla}\phi)^2+\frac{1}{2}m^2\phi^2.$$

The Hamiltonian equations of motion are

$$\tag{4} \dot{\pi}~=~({\bf \nabla}^2-m^2)\phi, \qquad \dot{\phi}~=~\pi.$$

Note that both $\phi$ and $\pi$ satisfy the massive Klein-Gordon equation (2).

3) Canonical transformation.

$$\tag{5} \widetilde{\phi}~:=~ \pi, \qquad \widetilde{\pi}~:=~ -\phi.$$

The new Hamiltonian density is precisely OP's Hamiltonian density

$$\tag{6} {\cal H}~=~\frac{1}{2}\widetilde{\phi}^2 +\frac{1}{2}({\bf \nabla}\widetilde{\pi})^2+\frac{1}{2}m^2\widetilde{\pi}^2.$$

Note that both $\widetilde{\phi}$ and $\widetilde{\pi}$ satisfy the massive Klein-Gordon equation (2). So the new Hamiltonian density (6) can be reproduced without any non-local field redefinition.

$$\tag{7} \widetilde{\phi}~:=~\sqrt{m^2-{\bf \nabla}^2}\phi, \qquad \sqrt{m^2-{\bf \nabla}^2}~:=~m\sum_{n=0}^{\infty} \begin{pmatrix}1/2 \\ n \end{pmatrix} \left(-\frac{{\bf \nabla}^2}{m^2} \right)^n,$$

is a spatially non-local (but temporally local) field redefinition.

Since the field redefinition (7) contains no time-derivatives, Cauchy data still amounts to specify $\widetilde{\phi}$ and its time-derivative on a Cauchy surface. No need for higher time-derivatives.

And Yes, the field redefinition (7) formally leads to the Hamiltonian density (6). However, as mention in section 3, already a local canonical transformation (5) could achieve the same formulation. OP's non-local field redefinition (7) corresponds to a non-local canonical transformation

$$\tag{8} \widetilde{\phi}~:=~(m^2-{\bf \nabla}^2)^{\frac{1}{2}}\phi,\qquad \widetilde{\pi}~:=~(m^2-{\bf \nabla}^2)^{-\frac{1}{2}}\pi.$$

5) OP asks in a comment if such kind of non-local field redefinition (7) is allowed in the quantum field theory in general? It depend on who's the arbitrator. A mathematician would probably focus on whether the detailed derivation/proof makes sense, while a physicist would likely be content if the final result/goal makes sense.

6) OP considers in an update (v5) the manifestly positive (but non-renormalizable) Hamiltonian density

$${\cal H}_{\beta} ~=~\frac{1}{2}\left((1+2\beta{\bf \nabla}^2)^{-\frac{1}{2}}\pi\right)^2 +\frac{1}{2}\left((1+\beta{\bf \nabla}^2)^{\frac{1}{2}}{\bf \nabla}\phi\right)^2$$ $$\tag{9} ~=~\frac{1}{2}\widetilde{\pi}^2 +\frac{1}{2}\left(\sqrt{\frac{1+\beta{\bf \nabla}^2}{1+2\beta{\bf \nabla}^2}}{\bf \nabla}\widetilde{\phi}\right)^2~\geq~0,$$ with non-local canonical transformation

$$\tag{10} \widetilde{\phi}~:=~(1+2\beta{\bf \nabla}^2)^{\frac{1}{2}}\phi,\qquad \widetilde{\pi}~:=~(1+2\beta{\bf \nabla}^2)^{-\frac{1}{2}}\pi.$$

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Thanks, I know how to get the Hamiltonian by your eq.(5), I guess my question is whether this kind of non-local field redefinition is allowed in the quantum field theory, and if not, how is it going wrong? –  Howard Mar 12 '13 at 18:17
I think you could use sigma matrices to write $m^2 -\nabla^2$ as a product of two $local$ differential operators, which are going to look a lot like Dirac operators (with gamma matrices replaced by sigma matrices). Of course you would need two scalar fields obeying the Klein-Gordon equation to make sense of that, but perhaps you could disentangle the dynamics afterwards. Or you could just use $\phi$ and $\pi$ and do this instead. Qmechanic's answer is certainly the most reasonable way to go about this though. –  alexarvanitakis Mar 12 '13 at 19:33
Why do you adress the OP always in third person? –  NikolajK Mar 12 '13 at 22:14