Amplitude Damping of Hermite-Polynomial-Field Excited Coherent State

Abstract

We introduce Hermite-polynomial-field excited coherent state (HPFECS) and then investigate analytically its evolution in an amplitude damping channel. We find that it evolves into a Laguerre-polynomial-weighted-chaotic photon field in this process, which turns out to be a new nonclassical state. The Q-function of this novel state is also given.

Appendix A: Derivation of (12)

Using the operator of the Hermite polynomials method, we have

$$\begin{array}{@{}rcl@{}} & &\sum\limits_{l = 0}^{m}\sum\limits_{k = 0}^{n}\binom{m}{l}\binom{n}{k}{\vdots} H_{l,k}\left( a,a^{\dag}\right) {\vdots} f^{l}g^{k}\\ & =&\sum\limits_{l = 0}^{m}\sum\limits_{k = 0}^{n}\binom{m}{l}\binom{n}{k}\colon a^{\dag k} a^{l}\colon f^{l}g^{k}\\ & =&\colon(ga^{\dag}+ 1)^{n}(fa+ 1)^{m}\colon, \end{array} $$

(56)

then we construct

$$\begin{array}{@{}rcl@{}} && \sum\limits_{m,n = 0}^{\infty}\frac{s^{m}t^{n}\colon(ga^{\dag}+ 1)^{n}(fa+ 1)^{m} \colon}{m!n!}\\ & =&e^{t(ga^{\dag}+ 1)}e^{s(fa+ 1)}\\ & =&e^{sf(a+\frac{1}{f})}e^{tg(a^{\dag}+\frac{1}{g})}e^{sftg}={\vdots} e^{^{sf(a+\frac{1}{f})+tg(a^{\dag}+\frac{1}{g})+sftg}}\vdots\\ & =&\sum\limits_{m,n = 0}^{\infty}\frac{(sf)^{m}(gt)^{n}}{m!n!}{\vdots} H_{m,n}\left( a+\frac{1}{f},a^{\dag}+\frac{1}{g}\right) \vdots. \end{array} $$

(57)

From (57), we see

$$ \colon(ga^{\dag}+ 1)^{n}(fa+ 1)^{m}\colon=f^{m}t^{n}{\vdots} H_{m,n}\left( a+\frac{1}{f},a^{\dag}+\frac{1}{g}\right) \vdots $$

(58)

Combining (56) and (58) we lead to

$$ \sum\limits_{l = 0}^{m}\sum\limits_{k = 0}^{n}\binom{m}{l}\binom{n}{k}{\vdots} H_{l,k}\left( a,a^{\dag}\right) {\vdots} f^{l}g^{k}=f^{m}t^{n}{\vdots} H_{m,n}\left( a+\frac{1}{f},a^{\dag}+\frac{1}{g}\right) $$

(59)

Since the both sides of (59) are in antinormal ordering. we restore to

$$ \sum\limits_{l = 0}^{m}\sum\limits_{k = 0}^{n}\binom{m}{l}\binom{n}{k}H_{l,k}\left( x,y\right) f^{l}g^{k}=f^{m}g^{n}H_{m,n}\left( x+\frac{1}{f},y+\frac{1}{g}\right) $$

(60)

1.1 Appendix B: Derivation of (14)

$$\begin{array}{@{}rcl@{}} H_{m,n}\left( \xi,\kappa\right) & =&\sum\limits_{l = 0}^{\min(m,n)}\frac{m!n!\left( -1\right)^{l}}{\left( n-l\right) !\left( m-l\right) !}\frac{\xi^{m-l}\kappa^{n-l}}{l!}\\ & =&\left( -1\right)^{n}\xi^{m-n}\sum\limits_{l = 0}^{n}\frac{m!n!}{l!\left( m-l\right) !}\frac{\left( -\xi\kappa\right)^{n-l}}{\left( n-l\right) !}\\ & =&n!\left( -1\right)^{n}\xi^{m-n}\sum\limits_{k = 0}^{n}\frac{m!}{\left( n-k\right) !\left( m-n+k\right) !}\frac{\left( -\xi\kappa\right)^{k}} {k!}\\ & =&n!\left( -1\right)^{n}\xi^{m-n}L_{n}^{m-n}\left( \xi\kappa\right) \end{array} $$

(61)

1.2 Appendix C: Derivation of (35)

Using basis functions in complete space, we have

$$\begin{array}{@{}rcl@{}} & &\int\frac{d^{2}z}{\pi}(z+i\sigma)^{m}(z^{\ast}+i\lambda)^{n}e^{-\left\vert z\right\vert^{2}}\\ & =&\sum\limits_{l = 0}^{+\infty}\sum\limits_{k = 0}^{+\infty}\left( \begin{array} [c]{c} n\\ l \end{array} \right) \left( \begin{array} [c]{c} m\\ k \end{array} \right) (i\lambda)^{n-l}(i\sigma)^{m-k}\int\frac{d^{2}z}{\pi}z^{k}z^{\ast l}e^{-\left\vert z\right\vert^{2}}\\ & =&\sum\limits_{l = 0}^{+\infty}\sum\limits_{k = 0}^{+\infty}\left( \begin{array} [c]{c} n\\ l \end{array} \right) \left( \begin{array} [c]{c} m\\ k \end{array} \right) (i\lambda)^{n-l}(i\sigma)^{m-k}\sqrt{l!k!}\delta_{l,k}, \end{array} $$

(62)

Introducing a special function for two-mode hermite polynomials, then we can get

$$\begin{array}{@{}rcl@{}} && \int\frac{d^{2}z}{\pi}(z+i\sigma)^{m}(z^{\ast}+i\lambda)^{n}e^{-\left\vert z\right\vert^{2}}\\ & =&i^{m+n}\sum\limits_{l = 0}^{+\infty}\frac{(-1)^{l}n!m!}{l!(m-l)!(n-l)!}\sigma^{m-l}\lambda^{n-l}=i^{m+n}H_{m,n}(\sigma,\lambda). \end{array} $$

(63)

The (35) is derived by simple subsitution.

1.3 Appendix D: Derivation of (37)

In the spirit of the operator of Hermite polynomial method, we consider

$$ \sum\limits_{l = 0}^{n}\left( \begin{array} [c]{c} n\\ l \end{array} \right) z^{n-l}q^{l}{\vdots} H_{l,m}(a,a^{\dag})\vdots=\sum\limits_{l = 0}^{n}\left( \begin{array} [c]{c} n\\ l \end{array} \right) z^{n-l}q^{l}\colon a^{\dag m}a^{l}\colon=\colon a^{\dag m} (z+qa)^{n}\colon, $$

(64)

Then we make the sum

$$\begin{array}{@{}rcl@{}} \sum\limits_{m,n = 0}^{\infty}\frac{s^{m}t^{n}}{m!n!}\colon a^{\dag m}(z+qa)^{n} \colon&=&e^{sa^{\dag}}e^{(z+qa)t}={\vdots} e^{(\frac{z}{q}+a)tq}e^{sa^{\dag}} e^{-stq}\vdots\\ &=&\sum\limits_{m,n = 0}^{\infty}\frac{s^{m}(tq)^{n}}{m!n!}{\vdots} H_{n,m}(\frac{z}{q}+a,a^{\dag})\vdots. \end{array} $$

(65)

Combining (64) and (65) we lead to

$$ \sum\limits_{l = 0}^{n}\left( \begin{array} [c]{c} n\\ l \end{array} \right) z^{n-l}q^{l}{\vdots} H_{l,m}(a,a^{\dag})\vdots=q^{n}{\vdots} H_{n,m}(\frac{z}{q}+a,a^{\dag})\vdots $$

(66)

Since the both sides of (66) are in antinormal ordering, we restore to

$$ \sum\limits_{l = 0}^{n}\left( \begin{array} [c]{c} n\\ l \end{array} \right) z^{n-l}q^{l}H_{l,m}(x,y)=q^{n}H_{n,m}(\frac{z}{q}+x,y) $$

(67)

1.4 Appendix E: Derivation of (46)

Using the operator of the Hermite polynomials method, we have

$$\begin{array}{@{}rcl@{}} & &\sum\limits_{n,m = 0}^{\infty}\frac{t^{n}s^{m}}{n!m!}H_{m,n}(x,y){\vdots} H_{m,n}(a^{\dag},a)\vdots\\ & =&\sum\limits_{n,m = 0}^{\infty}\frac{\colon s^{m}a^{\dag m}t^{n}a^{n}\colon} {n!m!}H_{m,n}(x,y)\\ & =&\colon\exp\left( -sta^{\dag}a+sa^{\dag}x+tay\right) \colon, \end{array} $$

(68)

where the symbol ⋮ ⋮ denotes the anti-normally ordered product of operators a and a^‡.

Converting normal ordering to antinormal ordering, we lead to

$$ (E1)=\frac{1}{1-st}\vdots\frac{\exp\left( tay+a^{\dag}sx-sxty-sta^{\dag} a\right)} {1-st}\vdots. $$

(69)

Letting $a^{\dag }\rightarrow x^{\prime },a\rightarrow y^{\prime },$ we obtain

$$ \sum\limits_{n,m = 0}^{\infty}\frac{t^{n}s^{m}}{n!m!}H_{m,n}(x,y)H_{m,n}(x^{^{\prime}} ,y^{^{\prime}})=\frac{1}{1-st}\frac{\exp\left( ty^{\prime}y+sx^{\prime} x-sxty-stx^{\prime}y^{\prime}\right)} {1-st}. $$

(70)

Using the formula

$$ \sum\limits_{m = 0}^{\infty}\frac{s^{m}}{m!}H_{m,n}(x,y)=\left( y-s\right)^{n}e^{sx} $$

(71)

we have

$$ \sum\limits _{m = 0}^{\infty}\frac{s^{m}}{m!}H_{m,n}(x,y)H_{m,n}(x^{^{\prime}} ,y^{^{\prime}})=(-s)^{n}H_{n,n}[i(\frac{y}{s}-x^{^{\prime}}),i(y^{^{\prime}} -sx)]e^{sx^{^{\prime}}x}. $$

(72)