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As far as I'm aware you can solve for the wave functional $\Psi[\phi]$ of a field using the Schrodinger equation $$i\hbar\frac{\partial \Psi}{\partial t}=H\Psi.$$

  1. Should $H$ here be the Hamiltonian, or the Hamiltonian density?
  2. If it's the Hamiltonian, is there a version of this equation for the Hamiltonian density?
  3. Would assuming a certain form for the functional like $\Psi\sim e^{-\omega \phi^2}$ be of any use in simplifying the equation?

I'm pretty new to QFT so excuse me if the question is poorly formulated.

Cosmas Zachos
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Gianni Boschetti
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    Linked. 1. Hamiltonian 2.yes, but why??? it would be quite misguided. 3. for quadratic hamiltonians, only, yes, but, again, work with functionals, not functions. Have you understood your Jackiw or Hatfield textbook review? – Cosmas Zachos Jan 18 '23 at 21:20
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    2- I know the Hamiltonian density for this field, so I wanted to use canonical quantization to get an equation for $\Psi$ that contains $\phi$ and $d\Psi / d\phi$. If i use the total Hamiltonian I have to deal with the integral in 3d space. Then again I'm new to QFT so probably should check those text books. Thanks. – Gianni Boschetti Jan 18 '23 at 21:36
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    "2- I know the Hamiltonian density for this field..." So... integrate over all space to get the Hamiltonian... what is the problem? – hft Jan 18 '23 at 23:42
  • Related https://physics.stackexchange.com/q/525851/226902 and https://physics.stackexchange.com/q/142180/226902 – Quillo Mar 11 '23 at 04:24

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Recall, from Hatfield's textbook (QFT of point particles and strings) & Jackiw's review that the functional equation you are emulating is just that, a functional equation the extension of an infinite sum of canonical pairs $[q_i,p_j]=i\hbar \delta_{ij}$ to $$ [\phi(x),\pi(y)]\propto \delta(x-y). $$ So, just as the Hamiltonian in QM deals with all degrees of freedom, just so in QFT, $$ H\psi[\phi]= \int\!\! d^3 x \left (-\frac{\delta^2}{\delta\phi(x)^2} +\phi(x)O\phi(x)+... \right )\psi[\phi], $$ where the ellipses (...) suggests cubic and higher terms in the potential, rarely used. $O$ is a normally nonlocal operator, i.e., $$ O \phi(x) = \int \! d^3y ~O(x-y) \phi(y), $$ such as $O=m^2-\nabla^2$, etc. I've left the time dependence implicit throughout.

It is then evident that the ground state of the quadratic Hamiltonian is $$ \propto e^{-\tfrac{1}{2} \int d^3z ~\phi(z) \sqrt{O} \phi(z) }, $$ but you must attend to the δ-functions.

  • When confused, try to consider uncoupled oscillators, i.e. $O$ a constant.

  • PS if you insist on using functions instead of functionals, you can always convert the latter to the former by sticking in gonzo gratuitous delta functions, $$ H'[\phi]=\int d^3 x ~~{\cal H}(x) \delta (x-y)= {\cal H}(y), $$ but why??

A more explicit definition for $\sqrt O$ is clearest in 1d space. For $$ \sqrt{O}\phi(x)\equiv \int\! dy~ K(x-y)\phi(y). ~~~\leadsto \\ O\phi[x]= \sqrt{O}\sqrt{O}\phi(x) =\int\! dydz~ K(x-y) K(y-z)\phi(z) ~~~\implies \\ O(x-z)= \int\! dy ~~K(x-y) K(y-z), $$ the equation defining the kernel of the square root.

Cosmas Zachos
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    To clarify one point that confused me at first passing, the reason $(m^2-\nabla^2)$ is non-local is that its integration kernel is $(m^2-\nabla^2)\delta(x-y)$; namely, $\nabla^2$ has to hit both the delta function and $\phi(x)$, in $O\phi(x)$. – ɪdɪət strəʊlə Jan 18 '23 at 22:30
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    Yes; gradients are nonlocal. You may simulate them with a discrete number of point canonical variables, poorly. – Cosmas Zachos Jan 18 '23 at 22:33