Scalar product of free field and conjugate momentum

Question

Given $[\Phi (x), \Pi(y)] = \delta^{3}(x-y)$,$ $ $\Phi|\phi\rangle = \phi(x)|\phi\rangle$ and $\Pi|\pi\rangle = \pi(x)|\pi\rangle$, I am trying to prove $\langle\phi|\pi\rangle \sim e^{i\int d^{3x}\pi(x)\phi(x)}$. Any hint is appreciated.

Answer 1

Consider the wave functional $\boldsymbol{\Psi}[\phi] \equiv \langle \phi | \Psi\rangle$. In this representation $\Phi(x)$ is the multiplication operator and $\Pi(x)$ the derivative. As one can infer from their commutation relation (you forgot an $i$ by the way) $$ [\Phi(x),\Pi(y)] = i \delta^3(x-y)\,,\quad\Longrightarrow\quad \Phi(x) \boldsymbol{\Psi}[\phi] = \phi(x) \boldsymbol{\Psi}[\phi] \,,\qquad \Pi(x) \boldsymbol{\Psi}[\phi] = \frac1i\frac{\delta}{\delta\phi(x)}\boldsymbol{\Psi}[\phi] \,. $$ The analogy with the quantum mechanical case is ($\Phi(x),\Pi(y),\phi,\boldsymbol{\Psi}[\phi] \longrightarrow \hat{q}_i,\hat{p}_j,q,\psi(q)$): $$ [\hat{q}_i,\hat{p}_j] = i \delta_{ij}\,,\quad\Longrightarrow\quad \hat{q}_i \psi(q) = q_i \psi(q) \,,\qquad \hat{p}_i \psi(q) = \frac1i\frac{\partial}{\partial q_i}\psi(q) \,. $$ Then, from the fact that $\boldsymbol{\pi}[\phi] \equiv \langle \phi|\pi\rangle$ is an eigenfunctional of $\Pi(x)$, you get a functional differential equation $$ i\pi(x)\boldsymbol{\pi}[\phi] = \frac{\delta}{\delta\phi(x)} \boldsymbol{\pi}[\phi] \,. $$ The solution is easily seen to be the one that you want to prove, up to a multiplicative constant.

This is not a proof, but also the naive continuum limit of the quantum mechanics result works $$ \langle q | p \rangle \propto \exp\left(i\sum_{i=1}^N q_i p_i \right)\;\longrightarrow\; \exp\left(i\int\mathrm{d}^3x\, \phi(x) \pi(x) \right)\,. $$