# Eigenstate of field operator in QFT

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Why don't people discuss the eigenstate of the field operator? For example, the real scalar field the field operator is Hermitian, so its eigenstate is an observable quantity.

This post imported from StackExchange Physics at 2014-04-21 16:24 (UCT), posted by SE-user user34669
asked Apr 21, 2014
These states are called coherent states, and you will find a discussion in any decent introductory text book on QFT.

This post imported from StackExchange Physics at 2014-04-21 16:24 (UCT), posted by SE-user Thomas
@Thomas coherent state is the eigenstate of the annihilation operator. This is not the eigenstate of the field operator.

This post imported from StackExchange Physics at 2014-04-21 16:24 (UCT), posted by SE-user user34669

## 1 Answer

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As $\phi(f)$ and $\pi(f)$, which are self adjoint, satisfy the same commutation relations of $X$ and $P$, the closure of the space generated by polynomials of the former pair of operators applied to $\lvert 0\rangle$ is isomorphic to $L^2(R)$, Therefore the spectrum of $\phi(f)$ and $\pi(f)$, is purely continuous and coincides to $R$ and there are no proper eigenvectors, but they are just formal ones and isomorphic to $\lvert x\rangle$ and $\lvert p\rangle$.

This post imported from StackExchange Physics at 2014-04-21 16:24 (UCT), posted by SE-user V. Moretti
answered Apr 21, 2014 by (2,075 points)

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