# Quadrature in quantum optics

As you've seen, the word quadrature is overladen with many, none-too-precise meanings.

Here the "quadratures" loosely refer to the position and momentum observables:

$$\hat{x} = \frac{1}{\sqrt{2}}(a + a^\dagger)$$ $$\hat{p} = \frac{i}{\sqrt{2}}\,(a - a^\dagger)$$

where $a,\,a^\dagger$ are the lowering/ raising operators and I've normalized the two observables so that $[\hat{x},\,\hat{p}] = i\,\mathrm{id}$. These two (or measurements coming from these observables) are loosely called "quadratures" because, for a coherent / squeezed state $\psi$ the mean measurements $\langle\psi|\hat{x}|\psi\rangle$ and $\langle\psi|\hat{p}|\psi\rangle$ are sinusoidally-with-time varying quantities which are in *phase quadrature*, *i.e.* a quarter cycle out of phase.

The squeezed states are the most general states that saturate (*i.e.* actually achieve equality in) the Heisenberg product. A coherent state is a special case for which the normalized momentum and position uncertainties are equal, and their preroduct is the uncertainty bound $\hbar/2$. A squeezed state has the same uncertainty product, but the uncertainty on one of position or momentum measurements is smaller than the uncertainty on the other, so one achieves smaller uncertainty than for the corresponding coherent state at the expense of the other. One can generalize the above comments for any "generalized" position and momentum, defined, for any $\phi\in\mathbb{R}$, by:

$$\hat{x} = \frac{1}{\sqrt{2}}(e^{i\,\phi}\,a + e^{-i\,\phi}\,a^\dagger)$$ $$\hat{p} = \frac{i}{\sqrt{2}}\,(e^{i\,\phi}\,a - e^{-i\,\phi}\,a^\dagger)$$

which are conjugate in the QM sense of fulfilling the CCR and the means of whose measurements vary sinusoidally with time, again, in phase quadrature.