With $\hat{a}$ being the annihilation operator for bosons, we can define the displacement operator and squeezing operator as:
$$\hat{D}(\alpha)=\exp(\alpha\hat{a}^{\dagger}-\alpha^{*}\hat{a})\qquad \hat{S}(\xi)=\exp(\frac{1}{2}\xi^{*}\hat{a}^{2}-\frac{1}{2}\xi{\hat{a}^{\dagger}}^{2})$$
In the book Quantum Optics written by Scully and Zubairy, the sqeezed coherent state is defined as:
$$|\alpha,\xi\rangle=\hat{S}(\xi)\hat{D}(\alpha)|0\rangle \tag{2.7.10}$$
Exercise 2.5 of this book, says:
An alternatte definition of a squeezed coherent state is
$$|\alpha,\xi\rangle=\hat{D}(\alpha)\hat{S}(\xi)|0\rangle $$
I don't figure out how these two definitions are equivalent. Could anyone help to give a rigid proof?
Best Answer
Both of them give squeezed states, but not the same squeezed states (for example, you can calculate and see that their average photon numbers are different)