# Amplitude of single-mode field in a cavity

I have two contradictory result about the amplitude of single mode field when it is quantized

• In C. Gerry and P. Knight's Introductory quantum optics, the amplitude is given by $$\sqrt{\frac{2\omega^2}{V\epsilon_0}},$$ which puts the hamiltonian of radiation in the cavity at $$H=\frac{1}{2}(p^2 + \omega^2q^2)$$ and the electric field as $$E=\sqrt{\frac{2\omega^2}{V\epsilon_0}} q(t)\sin{kz} \ \ \hat{x}$$

• There is also another suggestion in Alain Aspect's lecture Quantum Optics - Canonical quantization of a single mode on youtube, where Aspect suggests that amplitude is $$\sqrt{\frac{\hbar\omega}{V\epsilon_0}},$$ which puts the hamiltonian at $$H=\hbar\omega|\alpha|^2$$ and the electric field as $$E = i \sqrt{\frac{\hbar\omega}{V\epsilon_0}} \alpha(t)\sin{kz}\ \ \textbf{e}$$

Which one is correct? or is there what make me confused?

There isn't a "correct" or an "incorrect" choice. The two are just two different, equivalent ways to handle the book-keeping of the different constants involved in the formalism, and different conventions in terms of working with real-valued quadratures (like Gerry and Knight's $$q$$ and $$p$$) or complex-valued ones (like Aspect's $$\alpha$$).