Studying Output States Generated by Optical Beam Splitter and 2-cascaded BS

Abstract

Based on the idea of transition from classical optics to quantum optics we deduce the natural expressions of optical beam splitter (BS) and 2-cascaded BS operators in coherent state representation and also obtain their normally ordered forms. Moreover, we theoretically prepare some desired quantum entangled states and investigate several carried experiments. Finally, our work shall be focused on the following two aspects: the quantum-optical catalysis of the BS and quantum state truncation of the 2-cascaded BS.

Appendix: A The proofs of (12d)

Proof

From (12c) we have

$$ \begin{array}{@{}rcl@{}} \frac{dB(\theta )}{d\theta } &=&-:[(a_{1}a_{1}^{\dagger }+a_{2}a_{2}^{\dagger })\sin \theta +i(a_{2}^{\dagger }a_{1}+a_{1}^{\dagger }a_{2})\cos \theta ] \\ &&\times \exp [(a_{1}a_{1}^{\dagger }+a_{2}a_{2}^{\dagger })(\cos \theta -1)-i(a_{2}^{\dagger }a_{1}+a_{1}^{\dagger }a_{2})\sin \theta ]: \\ &=&-a_{1}^{\dagger }B(\theta )(a_{1}\sin \theta +ia_{2}\cos \theta )-a_{2}^{\dagger }B(\theta )(a_{2}\sin \theta +ia_{1}\cos \theta ). \end{array} $$

(A1)

Inserting 1 = B(θ)B^†(θ) on the left side of (A1) and using the following transformation formulas

$$ B^{\dagger }(\theta )a_{1}^{\dagger }B(\theta )=a_{1}^{\dagger }\cos \theta +a_{2}^{\dagger }i\sin \theta \text{ and } B^{\dagger }(\theta )a_{2}^{\dagger }B(\theta )=a_{1}^{\dagger }i\sin \theta +a_{2}^{\dagger }\cos \theta $$

we can rewrite (A1) as

$$ \frac{dB(\theta )}{d\theta }=-iB(\theta )(a_{1}^{\dagger }a_{2}+a_{2}^{\dagger }a_{1}). $$

(A2)

In fact, (A1) can also represented as

$$ \frac{dB(\theta )}{d\theta }=-(a_{1}^{\dagger }\sin \theta +a_{2}^{\dagger }i\cos \theta )B(\theta )a_{1}-(a_{1}^{\dagger }i\cos \theta +a_{2}^{\dagger }\sin \theta )B(\theta )a_{2}. $$

(A3)

Inserting 1 = B^†(θ)B(θ) on the right side of (A3) and using the transform relations

$B(\theta )a_{1}B^{\dagger }(\theta )=a_{1}{\cos \limits } \theta +a_{2}i{\sin \limits } \theta$ and $B^{\dagger }(\theta )a_{2}B(\theta )=a_{1}i{\sin \limits } \theta +a_{2}{\cos \limits } \theta$

yields directly that

$$ \frac{dB(\theta )}{d\theta }=-i(a_{1}^{\dagger }a_{2}+a_{2}^{\dagger }a_{1})B(\theta ). $$

(A4)

The (A2) and (A4) guarantee that

$$ \frac{dB(\theta )}{B(\theta )}=-i(a_{1}^{\dagger }a_{2}+a_{2}^{\dagger }a_{1})d\theta . $$

(A5)

The (A5) together with B(0) = 1 leads to $B(\theta )=\exp [-i\theta (a_{1}^{\dagger }a_{2}+a_{2}^{\dagger }a_{1})]$.

In a similar way, assuming a real rotation matrix $M=\left (\begin {array}{cc} {\cos \limits } \theta & {\sin \limits } \theta \\ -{\sin \limits } \theta & {\cos \limits } \theta \end {array} \right )$, we can also get

$$ B(\theta )=\exp [\theta (a_{2}^{\dagger }a_{1}-a_{1}^{\dagger }a_{2})], $$

(A6)

which is just the (12b). □