Degree of second-order coherence and squeezed state Degree second order coherence is determined by the following expression
$$g^{(2)}(z) =\frac{\left<z\right|{a}^{+}{a}^{+}{a}{a}\left|z\right>}
{\left<z\right|{a}^{+}{a}\left|z\right>^2}. $$
I want to calculate it for the case when $\left|z\right>$ is a squeezed state, i.e. $$\left|z\right> = {S}\left(z\right)\left|0\right>,$$ where $${S}\left(z\right) =e^{\frac{1}{2}z^{*}{a}^2 -
\frac{1}{2}z{a}^{+2}}$$ is the squeeze operator, $\left|0\right>$ - the vacuum state, $z \in \mathbb{C}$. 
I expect that $g^{(2)}(z) < 1$ but want to have the exact answer. Could anybody help me?
 A: The answer can be got using the following expressions for $z = r e^{i \theta}$ (thanks @udrv for the hint):
$$S^{+} a S = a \cosh r - a^{+} e^{i \theta} \sinh r$$
and
$$S^{+} a^{+} S = - a e^{-i \theta} \sinh r + a^{+} \cosh r$$
Thus 
$$\left<n\right> = \left<z\right|a^{+}a\left|z\right> = \left<0\right|S^{+}a^{+}SS^{+}aS\left|0\right> = \sinh^2 r$$
For $\left<a^{+}a^{+}aa\right>$ I could get $$\left<z\right|a^{+}a^{+}aa\left|z\right> = \left<0\right|S^{+}a^{+}SS^{+}a^{+}SS^{+}aSS^{+}aS\left|0\right> = \left<\phi\right.\left|\phi\right>,$$
where 
$$\left|\phi\right> = S^{+}aSS^{+}aS\left|0\right> = -e^{i \theta}\sinh r \cosh r \left|0\right> + \sqrt{2} e^{2 i \theta}\sinh^2 r \left|2\right>.$$ 
Therefore $$\left<\phi\right.\left|\phi\right> = 2\sinh^4 r + \sinh^2 r\cosh^2 r.$$
And finally $$g^{(2)}\left(z\right) = \frac{2\sinh^4 r + \sinh^2 r\cosh^2 r}{\sinh^4 r} = 3 + \frac{1}{\sinh^2 r} = 3 + \frac{1}{\left<n\right>} > 1$$
A: Does anyone have an intuition as to why in the limit $z,r \to 0 $ the $g^2(z) $ does not approach 1 (as for vacuum) but instead blows up to infinity?
