Scalar product of free field and conjugate momentum

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.

• Have you proven the plain quantum mechanics statement for $\hat x$ and $\hat p$ ? – Cosmas Zachos Apr 17 at 21:47
• I have a long time ago. – Krishna Apr 17 at 22:08
• Yes. it would help in our question.. .. .. .. ........ . – Cosmas Zachos Apr 18 at 0:38
• Possible duplicate: physics.stackexchange.com/q/41880/2451 – Qmechanic Apr 18 at 5:47

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 $$\langle q | p \rangle \propto \exp\left(i\sum_{i=1}^N q_i p_i \right)\;\longrightarrow\; \exp\left(i\int\mathrm{d}^3x\, \phi(x) \pi(x) \right)\,.$$