Okay, So i'm doing some research that involves the Schrodinger representation in quantum field theory. The ground state wave functional for the Klein Gordon field is a generalized gaussian in position space (basically a lattice of harmonic oscillators). I would like to calculate the equal time correlation functions $\langle \phi (\vec x)\phi (\vec y)\rangle_0$. I know that the answer is $$D(\vec x-\vec y)=\int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec k\cdot (\vec x-\vec y)}}{2\omega_k},$$ borrowing mostly from the notation of P&S. In the formalism that I am using I have something like $$\Psi[\phi]=N\times \exp\bigg[-\frac{1}{2}\int d^3x\int d^3y\; \phi(\vec x)G(\vec x-\vec y)\phi(\vec y)\bigg]\\=N\;exp\bigg[-\frac{1}{2}\int\frac{d^3k}{(2\pi)^3} \omega_k|\tilde\phi(\vec k)|^2\bigg].$$

I know that the factor $\phi(x)\phi(y)=\int\frac{d^3p}{(2\pi)^3}e^{i\vec p\cdot\vec x}\int\frac{d^3p'}{(2\pi)^3}e^{i\vec p'\cdot\vec y}\tilde\phi(\vec p)\tilde\phi(\vec p')$, can be expanded in a Fourier transform. I am trying to evaluate the correlation function,

$$\langle \phi (\vec x)\phi (\vec y)\rangle_0=\int D\phi \phi (\vec x)\phi (\vec y)|\Psi[\phi]|^2\\ \to N\int\frac{d^3p}{(2\pi)^3}e^{i\vec p\cdot\vec x}\int\frac{d^3p'}{(2\pi)^3}e^{i\vec p'\cdot\vec y}\tilde\phi(\vec p)\tilde\phi(\vec p')\exp\bigg[-\int\frac{d^3k}{(2\pi)^3} \omega_k|\tilde\phi(\vec k)|^2\bigg]$$

I can split up the Fourier transforms into real and imaginary parts and find that the functional integration vanishes unless (check ch. 9 of P&S for a similar calculation) $\vec p=-\vec p'$. THe problem I'm having is that I am consistently missing a factor of two in the final result (after proper normalizations and so forth of course) which I think comes from the fact that I'm adding the integrals

$$\bigg(\tilde\phi^R(\vec p)\bigg)^2+\bigg(\tilde\phi^I(\vec p)\bigg)^2.$$

Even if you normalize the splitting up into real and imag parts, this gets compensated by the argument of the exponential. I was trying to determine if there is an overcounting factor associated with the double integral, but I don't see that being the case either. Anyone have some similar problems with calculations like this?

EDIT: Some remarks on this Formulation of QFT: This is an alternative way to approach QFT where you take the Klein Gordon Hamiltonian,$$H=\frac{1}{2}\int d^3x\;\pi^2-\bigg(\vec\nabla\phi\bigg)^2+m^2\phi^2$$ The canonical momentum $\pi$ of the fields is taken to be the generator of translations in the configuration space (so a functional derivative operator) $$\pi\to-i\frac{\delta}{\delta\phi}$$

And solve a functional Schrodinger equation. This formalism is not manifestly covariant, but all observable quantities (to my knowledge) are in agreement with the Fock space representation.

The normalization is found using the functional generalization of techniques used in QM. THe normalization is not well defined by it cancels for calculation of correlation functions.

This method is especially well suited for an informational perspective of QFT since the wave functional is related to the probability distribution of the field configuration in the normal way. My research is related to rewriting aspects of QFT in terms of information theoretic aspects. Renormalization will be seen (hopefully) as a mixture of concepts like coarse graining and sufficiency.

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    $\begingroup$ Why are you using this strange $\Psi$ functional here? The correct way is to calculate $\left< \phi(x) \phi(y) \right> = N \int D\phi \cdot \phi(x) \phi(y) e^{i S[\phi]}$, which should give you the correct propagator. $\endgroup$ Commented Oct 20, 2014 at 8:14
  • $\begingroup$ It seems like you have a error in the normalization. How do you compute $N$? $\endgroup$ Commented Oct 20, 2014 at 8:18
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    $\begingroup$ @SolenodonParadoxus This is an alternative formulation of QFT. There's an interesting review of it by Jackiw. $\endgroup$
    – thedoctar
    Commented May 10, 2019 at 10:22

1 Answer 1


I can't be certain, but I think the problem is that you're treating the real and imaginary parts of $\tilde{\phi}$ as full and independent fields. If $\phi$ is a real field, as suggested by the form of your vacuum functional, then the real and imaginary parts of $\tilde{\phi}$ are constrained to be even parity and odd parity, respectively, and thus each carry half of the information about the form of the original $\phi$. For this reason, you might be better off summing the real and imaginary parts of $\tilde{\phi}$, effectively constructing a high dimensional version of the Hartley transform. It's either that, or finding a way to properly constrain the $\tilde{\phi}^R$ and $\tilde{\phi}^I$ functional integrals to only be over even and odd functions, respectively.


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