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Let the Lagrangian be a functional of $\hat{\phi}$ and $\partial_{\mu}\hat{\phi}$, i.e. $\hat{L} = L(\hat{\phi},\partial_{\mu}\hat{\phi})$, where $\hat{\phi}$ is an operator.

The conjugate momenta is defined as $$\hat{\pi} = \frac{\delta \hat{L}}{\delta \partial_0{\hat{\phi}}}. \tag{1}$$

Let the eigenfunctions and eigenvectors of $\hat{\phi}$ and $\hat{\pi}$ at $t=0$, be defined as:

$$\hat{\phi}(0,{\bf x})|\phi\rangle = \phi({\bf x})|\phi\rangle\qquad\text{and} \qquad\hat{\pi}(0,{\bf x})|\pi\rangle = \pi({\bf x})|\pi\rangle.\tag{2}$$

Question: Based on (1) and (2), is it possible to derive that:

$$\langle \phi|\pi\rangle = \exp\left ( \frac{i}{\hbar}\int \phi({\bf x})\pi({\bf x}) d^3x \right )~?\tag{3}$$

Is the relation (3) valid for any arbitrary Lagrangian $L$?

Qmechanic
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Angela
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1 Answers1

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  1. Quantization of a theory starting from a classical Lagrangian density ${\cal L}(\phi,\partial\phi)$ is a non-trivial process. Let us assume that it is possible for OP's theory.

  2. There might be operator ordering issues when performing a Legendre transformation from the Lagrangian to the Hamiltonian formulation at the quantum level. To avoid this, in what follows, it will be more relevant to use the momentum operator $\hat{\pi}$ from the Hamiltonian formulation.

  3. Granted the equal-time CCR $$ [\hat{\phi}({\bf x}),\hat{\pi}({\bf y})]~=~i\hbar\delta^3({\bf x}\!-\!{\bf y})\hat{\bf 1},$$ and the definition (2), then one may derive the overlap $$\langle \phi |\pi\rangle ~=~\exp\left\{ \frac{i}{\hbar}\int_{\mathbb{R}^3} \!d^3{\bf x}~\phi({\bf x})\pi({\bf x})\right\} \tag{3}$$ (up to a phase factor, which is conventionally put to 1), cf. e.g. this and this related Phys.SE posts.

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Qmechanic
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