How do I show that squeeze states are non-classical? I am trying to show that squeezed states are non classical by showing that the Glauber-Sudarshan $P$ function takes negative values. A squeezed state is one for which one of the quadratures $\Delta X_1 < \frac{1}{2}$. In Introductory quantum optics by Gerry and Knight on page 152 they show that for the quadrature $X_1$ we have
$$ \langle (\Delta X_1)^2 \rangle = \frac{1}{4} \bigg\{1+ \int P(\alpha)[(\alpha +\alpha^*)-(\langle a \rangle- \langle a^\dagger \rangle)]^2d^2\alpha\bigg\}.$$
I am trying to derive this from the fact that for an observable $A$ and a density operator $\rho$,$\langle A \rangle = \mathrm{tr}(A \rho)$. We can express the density operator in terms of the Glauber-Sudarshan $P$ function as
$$ \rho = \int d^2 \alpha P(\alpha)|\alpha \rangle \langle \alpha |$$
where $ \{|\alpha \rangle \}$ are coherent states. So working through the algebra I find
\begin{equation}
\begin{split}
\langle (\Delta X_1)^2\rangle & = \mathrm{tr}\bigg[ \rho (\Delta X_1)^2 \bigg] \\
& = \mathrm{tr} \bigg[ \int \mathrm{d}^2 \alpha P(\alpha) |\alpha \rangle \langle \alpha | (\Delta X_1)^2 \bigg] \\
& = \int \mathrm{d}^2 \alpha P(\alpha)  \langle \alpha | (\Delta X_1)^2 |\alpha \rangle  \\
&= \frac{1}{4} \int \mathrm{d}^2 \alpha P(\alpha)
\end{split}
\end{equation}
where I used the fact that $\langle \alpha | (\Delta X_1)^2 |\alpha \rangle = \frac{1}{4}$ for a coherent state $|\alpha \rangle$. The density matrix has unit trace so $\int \mathrm{d}^2 \alpha P(\alpha)=1$ and I get $(\Delta X_1)^2 =\frac{1}{4}$ which does not agree with Gerry and Knight. Where have I gone wrong? 
 A: The notation used by Gerry and Knight is different from what you are thinking. Their definition is given as follows:
\begin{equation}
\label{definition}
\langle (\Delta X_1)^2\rangle = \langle X^2_1 \rangle - \langle X_1 \rangle^2
\end{equation}
Evaluating each term of the last equation, one have
$$\langle X^2_1 \rangle = \frac{1}{4}\left(1+\int d^2 \alpha \, P(\alpha) [(\alpha+\alpha^*)^2]\right) \, ,$$
and,
$$\langle X_1 \rangle^2 = \frac{1}{4}\int d^2 \alpha \, P(\alpha)[(\alpha+\alpha^*)(\langle a\rangle + \langle a^{\dagger} \rangle)] \, ,$$
where $\langle a\rangle \equiv \int d^2 \alpha \, P(\alpha) \, \alpha$ and $\langle a^\dagger \rangle \equiv \int d^2 \alpha \, P(\alpha) \, \alpha^*$. Now, summing and subtracting the term $(\langle a\rangle + \langle a^{\dagger} \rangle)^2$ in the last equation and computing $\langle (\Delta X_1)^2\rangle$ reads
\begin{equation}
\langle (\Delta X_1)^2 \rangle = \frac{1}{4} \bigg\{1+ \int P(\alpha)[(\alpha +\alpha^*)-(\langle a \rangle- \langle a^\dagger \rangle)]^2d^2\alpha\bigg\}.
\end{equation}
Because of the term inside the square bracket is always positive and from the condition that \langle $(\Delta X_1)^2 < 1/4$, follows that $P(\alpha)$ must be negative at least in some regions of the phase space.
