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I am not able to understand that why it is very special or convenient to work out path integral quantization of bosons and fermions in coherent states. Is it possible to work things in a very general state that could be present in the Hilbert space or any other state (but not coherent state)

Qmechanic
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user135580
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2 Answers2

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You can do QFT in any basis, but the coherent state basis is a particular convenient choice. That's because if you have a normal-ordered second-quantized Hamiltonian of the form $$H = \int d^3x\, \mathcal{H}(a^\dagger(x), a(x)),$$ then it's very easy to convert it into an equivalent path-integral Lagrangian, which is much easier to deal with, which is simply $$L = \int d^3x\, \mathcal{L}(\bar{\psi}, \psi), \\ \mathcal{L}(\bar{\psi}, \psi) := \bar{\psi} \partial_\tau \psi - \mathcal{H}(\bar{\psi}, \psi).$$

I.e. just take the function form of $\mathcal{H}$ and replace every $a^\dagger$ with a $\bar{\psi}$ and every $a$ with a $\psi$, where $\bar{\psi}$ and $\psi$ are c-numbers so you no longer need to keep track of their ordering. Without this universal recipe, for each Hamiltonian you wanted to consider, you'd need to either work in canonical QFT and keep careful track of operator ordering, or to rederive the more convenient Lagrangian equivalent from scratch every time. You never need to work in the coherent state basis, but it's very convenient to do so because the Hamiltonian and Lagrangian densities match up very closely.

tparker
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Actually, the coherent states are used for quantization using the Feynman Integral mostly in undergraduate books on QM. In real applications one usually uses fields in the spacetime coordinate basis, because in this basis Lagrangian is manifestly local and Lorentz invariant. For evaluating amplitudes in the perturbation theory using the Feynman diagrams, it is convenient to work with fields in the momentum space, because in this basis propagator has a mach simpler form, and it is trivial to implement the condition of conservation of momentum.

Andrey Feldman
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