# Squeeze operator

In quantum physics, the squeeze operator for a single mode of the electromagnetic field is[1]

${\displaystyle {\hat {S}}(z)=\exp \left({1 \over 2}(z^{*}{\hat {a}}^{2}-z{\hat {a}}^{\dagger 2})\right),\qquad z=re^{i\theta }}$

where the operators inside the exponential are the ladder operators. It is a unitary operator and therefore obeys ${\displaystyle S(\zeta )S^{\dagger }(\zeta )=S^{\dagger }(\zeta )S(\zeta )={\hat {1}}}$, where ${\displaystyle {\hat {1}}}$ is the identity operator.

Its action on the annihilation and creation operators produces

${\displaystyle {\hat {S}}^{\dagger }(z){\hat {a}}{\hat {S}}(z)={\hat {a}}\cosh r-e^{i\theta }{\hat {a}}^{\dagger }\sinh r\qquad {\text{and}}\qquad {\hat {S}}^{\dagger }(z){\hat {a}}^{\dagger }{\hat {S}}(z)={\hat {a}}^{\dagger }\cosh r-e^{-i\theta }{\hat {a}}\sinh r}$

The squeeze operator is ubiquitous in quantum optics and can operate on any state. For example, when acting upon the vacuum, the squeezing operator produces the squeezed vacuum state.

The squeezing operator can also act on coherent states and produce squeezed coherent states. The squeezing operator does not commute with the displacement operator:

${\displaystyle {\hat {S}}(z){\hat {D}}(\alpha )\neq {\hat {D}}(\alpha ){\hat {S}}(z),}$

nor does it commute with the ladder operators, so one must pay close attention to how the operators are used. There is, however, a simple braiding relation, ${\displaystyle {\hat {D}}(\alpha ){\hat {S}}(z)={\hat {S}}(z){\hat {S}}^{\dagger }(z){\hat {D}}(\alpha ){\hat {S}}(z)={\hat {S}}(z){\hat {D}}(\gamma )\qquad {\text{where}}\qquad \gamma =\alpha \cosh r-\alpha ^{*}e^{i\theta }\sinh r}$ [2]

Application of both operators above on the vacuum produces squeezed states:

${\displaystyle {\hat {D}}(\alpha ){\hat {S}}(r)|0\rangle =|\alpha ,r\rangle }$.