# Interpreting Hamiltonian of single-mode squeezing

Hamiltonian represents energy. I can understand this when considering about harmonic oscillator, whose Hamiltonian is expressed as: $$\hat{H} = \frac{1}{2m}\hat{p}^2 + \frac{m\omega^2}{2}\hat{q}^2$$ This equation can be interpreted as energy very clearly, because $$p$$ is momentum and $$q$$ is position, so these terms represent kinetic energy and quadratic potential.

However, when considering about the single-mode squeezing of light using spontaneous parametric down conversion in Optical Parametric Amplification(OPA), Hamiltonian is given as follows: $$\hat{H} = i(\hat{a}^2 - \hat{a}^{\dagger 2})$$ Here, $$\hat{a}$$ is an annihilation operator, which means an annihilation of a photon. Solving the Heisenberg motion of equation, I can confirm that the state is indeed squeezed under this Hamiltonian. However, I cannot find the physical interpretation of this squeezing Hamiltonian. I can guess that $$\hat{a}^{\dagger 2}$$ means the creation of two photons due to the interaction of signal light and pump light in the OPA. Similarly, $$\hat{a}^2$$ is annihilation of two photons. But I cannot understand why $$\hat{a}^{\dagger 2}$$ has negative sign and imaginary unit $$i$$. In my understanding the Hamiltonian become $$\hat{H}=\hat{a}^{\dagger 2} - \hat{a}^2$$, because the energy (variation?) is sum of increase on the creation of two photon and decrease on the annihilation of two photon.

I'd like to know the physical interpretation of $$\hat{H} = i(\hat{a}^2 - \hat{a}^{\dagger 2})$$ .

• I'm very curious about the origin of imaginary unit $i$ here. – ytaguchi Nov 2 '20 at 11:28