# 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?

• Do you know the effect of $(\Delta x)^2$ etc on | z > , etc for the momentum and the oscillator operators? – Cosmas Zachos May 1 '17 at 14:25
• @CosmasZachos, do you speak about $\hat{X}_1 = \frac{\hat{a}+\hat{a}^\ast}{2}$ and $\hat{X}_2 = \frac{\hat{a}-\hat{a}^\ast}{2i}$? It can be easy got that $\left(\Delta X_1\right)^2 = \frac{1}{4}e^{-2 |z|}$. – Ivan May 1 '17 at 15:02
• Right; can you manipulate these to oscillator eigenoperators? – Cosmas Zachos May 1 '17 at 15:05
• @Ivan Another hint: For $z = r e^{i\theta}$ the unitary squeeze operator $$S(z) = \exp\Big[ \frac{z^*a^2 - z{a^\dagger}^2 }{2}\Big]$$ generates transformations of the form $$S^\dagger(z) a\, S(z) = a\,\cosh r - a^\dagger e^{i\theta} \sinh r \\ S^\dagger(z) a^\dagger\, S(z) = - a e^{-i\theta} \sinh r + a^\dagger \,\cosh r$$ – udrv May 1 '17 at 18:01
• Using the provided hints, I've just got the following answer $g^{(2)} = 1 + 2 \frac{\cosh^2{r}}{\sinh^2{r}} - \frac{1}{\sinh^{2}{r}}$. There is a strange answer for me because $g^{(2)} > 1$ but I expected $g^{(2)} < 1$ – Ivan May 2 '17 at 13:26

## 2 Answers

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 $$g^{(2)}\left(z\right) = \frac{2\sinh^4 r + \sinh^2 r\cosh^2 r}{\sinh^4 r} = 3 + \frac{1}{\sinh^2 r} = 3 + \frac{1}{\left<n\right>} > 1$$

• I fixed it, but really the second formula has not been used. BTW there is a really strange answer for me because for a non classical state I expect $g^{(2)} <1$, for instance as for Fock states $g^{(2)} = \frac{n-1}{n} < 1$ – Ivan May 2 '17 at 16:03
• Sure, but you don't take expectation values at the ground state in that latter case. – Cosmas Zachos May 2 '17 at 16:28

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?