# Energies of this field Hamiltonian?

Consider the simplest eigenfunctionals of the (free) field Hamiltonian

$$H= \frac{1}{2} \int d^3 x ~( \Pi (x)^{2}+(\nabla\Phi)^{2} +m^2\Phi ^{2}).$$

Assuming this Hamiltonian and the functional quantization

$$\Pi \sim \frac{-i\delta}{\delta \Phi}$$

what would be the eigen-energies?

Well, they are all infinite, since free scalar field theory is simply a compactly repackaged array of quantum harmonic oscillators. You are simply supposed to follow the systematic translation rules for the Schrödinger representation from QM to QFT carefully, e.g. following Jackiw 1988 or Lüscher 1985.

As for one oscillator, the energy eigen-functionals are Gaussian in form. I'll give you a jump-start: the ground state eigen-functional $\langle\Phi|0\rangle$ is just $$\Psi[ \Phi] =\exp\left( -\frac{1}{2}\int d^3 x\,\Phi\left( x\right) \sqrt{m^{2}-\nabla_{x}^{2}}~ \Phi\left( x\right) \right).$$ Boundary conditions are assumed such that the $\sqrt{m^{2}-\nabla_{x}^{2}}$ kernel in the exponent is naively self-adjoint. "Integrating by parts" one of the $\sqrt{m^{2}-\nabla_{z}^{2}}$ kernels, functional derivation $\delta\Phi\left( x\right) /\delta\Phi\left( z\right)=\delta^3\left( z-x\right)$ leads to
$$\frac{\delta}{\delta\Phi\left( z\right) }\Psi [\Phi] =-\left( \sqrt{m^{2}-\nabla_{z}^{2}}\,\Phi\left( z\right) \right) \Psi[ \Phi] ,$$ so that $$\frac{\delta^{2}\qquad }{\delta\Phi\left( w \right) \delta\Phi \left( z\right) }\Psi[ \Phi] =\left( \sqrt{m^{2}-\nabla_{w}^{2}}\, \Phi\left( w\right) \right) \left( \sqrt{m^{2}-\nabla_{z}^{2}}\, \Phi\left( z\right) \right) \,\Psi[ \Phi]\\ - \sqrt {m^{2}-\nabla_{z}^{2}}\,\delta^3\left( w-z\right) \,\Psi[ \Phi]~.$$

Note that the divergent zero-point energy density, $$E_0= { 1\over 2 } \lim_{w\rightarrow z} \sqrt {m^{2}-\nabla_{z}^{2}}~ \delta^3\left( w-z\right),$$ may be handled rigorously using $\zeta$-function regularization. It is but the sum of all zero-point energies of the infinity of oscillators. The price of operator-valued distributions.

Leaving this zero-point energy present leads to the standard energy eigenvalue equation, again through integration by parts, $$\frac{1}{2} \int d^3 z\,\left( - \frac{\delta^{2}} {\delta\Phi\left( z\right) ^{2} }+\Phi\left( z\right) \left( m^{2}-\nabla_{z}^{2}\right) \Phi\left( z\right)-2E_0 \right) ~\Psi[ \Phi]=0 ~,$$ that is the lowest eigenvalue of H is $\int d^3 x ~E_0$.

You might proceed to form functional ladder operators, etc... and pursue the finer aspects of Schrödinger functional theory to the bitter end...

• I'm not a fan of $\int \mathrm{d}^3x\, \Phi(x) \sqrt{m^2-\nabla_x^2} \Phi(x)$ because it makes the integration kernel look local. A version without this drawback is $$\propto\int \mathrm{d}^3x\,\mathrm{d}^3y\, \Phi(x)\, (m^2-\nabla_x^2)\, K_1(|x-y|)\, \Phi(y),$$ where $K_1$ is the modified Bessel function of the second kind. n.b. $$\delta^3(x-y) \propto \int\mathrm{d}^3z\, K_1(|x-z|)\, (m^2-\nabla_z^2)\, K_1(|z-y|).$$ – Sean E. Lake Mar 23 '18 at 19:27
• Absolutely! I guess the nasty square root of the gradient might warn one of nonlocality. I left it like this to gloss over ferocious problems of distributions, simply to evoke its conceptual origin! Jackiw in the Swieca lectures linked through the comments gives it a kernel name as well... My task here is not to address the UV and IR subtleties, but to evoke the broad formal analogy. – Cosmas Zachos Mar 23 '18 at 19:36
• In another life, I could make a pedagogical connection to this one, but it's a real stretch.... – Cosmas Zachos Mar 23 '18 at 23:13
• Oh, bother, I screwed it up. It's supposed to be $K_1(|x-y|)/|x-y|$, not just $K_1$. – Sean E. Lake Mar 24 '18 at 0:24