1
$\begingroup$

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?

Improve this question
$\endgroup$
6
  • $\begingroup$ Do you know the effect of $(\Delta x)^2$ etc on | z > , etc for the momentum and the oscillator operators? $\endgroup$ – Cosmas Zachos May 1 '17 at 14:25
  • $\begingroup$ @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 $\left(\Delta X_1\right)^2 = \frac{1}{4}e^{-2 |z|}$. $\endgroup$ – Ivan May 1 '17 at 15:02
  • $\begingroup$ Right; can you manipulate these to oscillator eigenoperators? $\endgroup$ – Cosmas Zachos May 1 '17 at 15:05
  • 1
    $\begingroup$ @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 $$ S^\dagger(z) a\, S(z) = a\,\cosh r - a^\dagger e^{i\theta} \sinh r \\ S^\dagger(z) a^\dagger\, S(z) = - a e^{-i\theta} \sinh r + a^\dagger \,\cosh r $$ $\endgroup$ – udrv May 1 '17 at 18:01
  • $\begingroup$ 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 $g^{(2)} < 1$ $\endgroup$ – Ivan May 2 '17 at 13:26
1
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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 $g^{(2)} = \frac{n-1}{n} < 1$ $\endgroup$ – Ivan May 2 '17 at 16:03
  • $\begingroup$ Sure, but you don't take expectation values at the ground state in that latter case. $\endgroup$ – Cosmas Zachos May 2 '17 at 16:28
-1
$\begingroup$

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?

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.