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Statistics of multiphoton emission

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m (Created page with "Our last paper with Guillermo Díaz Camacho et ''al.'' on the dynamics of emission of multiphotons, in particular, of photon bundles <wz tip="A photon bundle is basica...")
 
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Our last paper with [[Guillermo Díaz Camacho]] et ''al.'' on the dynamics of emission of multiphotons, in particular, of [[photon bundles]] <wz tip="A photon bundle is basically the spontaneous emission of a photon Fock state.">(?!)</wz> involves a lot of complicated combinatoric formulas, such as the average time of detection for the $k$th photon from a $N$-photon Fock state that is spontaneously emitted with a radiative rate&nbsp;$\gamma_a$ and measured by a detector with bandwidth&nbsp;$\Gamma$:
 
Our last paper with [[Guillermo Díaz Camacho]] et ''al.'' on the dynamics of emission of multiphotons, in particular, of [[photon bundles]] <wz tip="A photon bundle is basically the spontaneous emission of a photon Fock state.">(?!)</wz> involves a lot of complicated combinatoric formulas, such as the average time of detection for the $k$th photon from a $N$-photon Fock state that is spontaneously emitted with a radiative rate&nbsp;$\gamma_a$ and measured by a detector with bandwidth&nbsp;$\Gamma$:
$$\begin{multline}
+
\begin{multline}
\langle t_{k}^{(N)} \rangle = \\\frac{k \gamma_a^N}{P(N,N)} \left( \frac{\Gamma}{\Gamma_-} \right) ^{2N} {N\choose k} \sum_{k_1 + k_2 + k_3\atop = N-k} {N-k\choose k_1, k_2, k_3} \sum_{k_4 + \cdots + k_9\atop = k-1}{k-1 \choose k_4,k_5,k_6,k_7,k_8,k_9}\sum_{k_{10} + k_{11} = 2} {2\choose k_{10},k_{11}} \\ \frac{(-1)^{k_3 + k_6 + k_7 + k_8 + k_{11}}}{\gamma_a^{k_1 +k_4+k_7} \Gamma^{k_2+k_5+k_8} \left( \frac{\Gamma_+}{4} \right)^{k_3 + k_6 + k_9} ( \gamma_a (k_1 + k_7+k_{11}/2) +\Gamma( k_2 +k_8 +k_{10}/2) + \frac{\Gamma_+}{2} (k_3+k_9))^2 }\,.\label{eq:tkN}
+
  \langle t_{k}^{(N)} \rangle = 2 N! \left(\frac{\Gamma_+ \Gamma \gamma_a}{\Gamma_-^2}\right)^N \sum_{k_1 + k_2 + k_3\atop = N-k}\sum_{k_4 + \cdots+ k_9\atop = k-1} \sum_{k_{10} + k_{11}\atop = 2}\prod_{j=1}^{11}{1\over k_j!}\times\\ \frac{(-1)^{k_3 + k_6 + k_7 + k_8 + k_{11}}}{\gamma_a^{k_1 +k_4+k_7} \Gamma^{k_2+k_5+k_8} \left( \frac{\Gamma_+}{4} \right)^{k_3 + k_6 + k_9} \big( \gamma_a (k_1 + k_7+{k_{11}\over2}) +\Gamma( k_2 +k_8 +{k_{10}\over2}) + \frac{\Gamma_+}{2} (k_3+k_9)\big)^2 }\,.
\end{multline}$$
+
\end{multline}
  
 
Here I provide a [[Mathematica]] code to turn this general result into particular and limiting cases, and discuss briefly the whole thing.  
 
Here I provide a [[Mathematica]] code to turn this general result into particular and limiting cases, and discuss briefly the whole thing.  
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The formula is obtained by integration over the joint probability density function $\phi_\Gamma$ for emitting the photons at times $t_i$, which we show in this paper is given by:
 
The formula is obtained by integration over the joint probability density function $\phi_\Gamma$ for emitting the photons at times $t_i$, which we show in this paper is given by:
 
$$\phi_\Gamma(t_1,\dots,t_N)=(-1)^{N}N!\gamma_a^N\left(\frac{\Gamma}{\Gamma_-}\right)^{2N }\prod_{i=1}^{N}\left[{d\over dt_i}({e^{-\gamma_a t_i}\over\gamma_a}+{e^{-\Gamma t_i}\over\Gamma}-4{e^{-\Gamma_+ t_i/2}\over\Gamma_+})\right]{\mathbf{1}}_{[t_{i-1}, t_{i+1}[}(t_{i})$$
 
$$\phi_\Gamma(t_1,\dots,t_N)=(-1)^{N}N!\gamma_a^N\left(\frac{\Gamma}{\Gamma_-}\right)^{2N }\prod_{i=1}^{N}\left[{d\over dt_i}({e^{-\gamma_a t_i}\over\gamma_a}+{e^{-\Gamma t_i}\over\Gamma}-4{e^{-\Gamma_+ t_i/2}\over\Gamma_+})\right]{\mathbf{1}}_{[t_{i-1}, t_{i+1}[}(t_{i})$$
where $\Gamma_\pm\equiv\Gamma\pm\gamma_a$. When tracing out all the photons to keep only the $k$th of interest, through some clever math hackery, we arrive to the computation of $\int_0^\infty\left({e^{-\gamma_a t}\over\gamma_a}+{e^{-\Gamma t}\over\Gamma}-4{e^{-\Gamma_+ t/2}\over\Gamma_+}\right)^N\,dt$ which is easy to do thanks to the multinomial theorem, which is where all the mad coefficients come from. That's how one gets down to something like Eq.&nbsp;\ref{eq:tkN}
+
where $\Gamma_\pm\equiv\Gamma\pm\gamma_a$. When tracing out all the photons to keep only the $k$th of interest, through some clever math hackery, we arrive to the computation of $\int_0^\infty\left({e^{-\gamma_a t}\over\gamma_a}+{e^{-\Gamma t}\over\Gamma}-4{e^{-\Gamma_+ t/2}\over\Gamma_+}\right)^N\,dt$ which is easy to do with the multinomial theorem, which is where all the mad coefficients come from. The nice thing, of course, is that it covers all cases with a single formula, i.e., it can tell you about the 1st photon from a biphoton or the 7th photon from a 15-photon bundle. Now the direct calculation of these particular cases is a bit tedious. We can ask a formal computer language, like Mathematica, to do it. A first thing is to
 +
 
 +
<syntaxhighlight lang="python">
 +
tkN[k1_, k2_, k3_, k4_, k5_, k6_, k7_, k8_, k9_, k10_, k11_] :=
 +
2 \[CapitalNu]! (((\[CapitalGamma] + \[Gamma]a) \[CapitalGamma] \
 +
\[Gamma]a)/(\[CapitalGamma] - \[Gamma]a)^2)^\[CapitalNu] 1/(
 +
  k1! k2! k3! k4! k5! k6! k7! k8! k9! k10! k11!) (-1)^(
 +
  k3 + k6 + k7 + k8 + k11)/(\[Gamma]a^(k1 + k4 + k7) \[CapitalGamma]^(
 +
  k2 + k5 + k8) ((\[CapitalGamma] + \[Gamma]a)/4)^(
 +
  k3 + k6 + k9) (\[Gamma]a (k1 + k7 + k11/2) + \[CapitalGamma] (k2 + k8 + k10/
 +
        2) + (\[CapitalGamma] + \[Gamma]a)/2 (k3 + k9))^2)
 +
</syntaxhighlight>
  
 
This is the formula for the average of the squared time of the $k$th photon from a $N$-photon bundle:
 
This is the formula for the average of the squared time of the $k$th photon from a $N$-photon bundle:
  
$$\begin{multline}
+
\begin{multline}
 
   \langle \big(t_k^{(N)}\big)^2\rangle  =  
 
   \langle \big(t_k^{(N)}\big)^2\rangle  =  
 
   \frac{2\gamma_a^N}{P(N,N)}\left( \frac{\Gamma}{\Gamma_-} \right)^{2N}k {N\choose k} \sum_{k_1+k_2+k_3=N-k}{N-k\choose k_1, k_2, k_3}\\{}\times \sum_{k_4+\cdots+k_9=k-1}{k-1\choose k_4, k_5, k_6, k_7,k_8,k_9} \sum_{k_{10}+k_{11}=2}{2\choose k_{10}, k_{11}}\\{}\times \frac{(-1)^{k_3 + k_6 + k_7 + k_{8}+k_{11}}}{\gamma_a^{k_1+k_4+k_7} \Gamma^{k_2+k_5+k_8} \left( \frac{\Gamma_+}{4} \right)^{k_3 + k_6 +k_9} ( \gamma_a (k_1+k_7+k_{11}/2) +\Gamma(k_2+k_8 +k_{10}/2) + \frac{\Gamma_+}{2} (k_3+k_9))^3}
 
   \frac{2\gamma_a^N}{P(N,N)}\left( \frac{\Gamma}{\Gamma_-} \right)^{2N}k {N\choose k} \sum_{k_1+k_2+k_3=N-k}{N-k\choose k_1, k_2, k_3}\\{}\times \sum_{k_4+\cdots+k_9=k-1}{k-1\choose k_4, k_5, k_6, k_7,k_8,k_9} \sum_{k_{10}+k_{11}=2}{2\choose k_{10}, k_{11}}\\{}\times \frac{(-1)^{k_3 + k_6 + k_7 + k_{8}+k_{11}}}{\gamma_a^{k_1+k_4+k_7} \Gamma^{k_2+k_5+k_8} \left( \frac{\Gamma_+}{4} \right)^{k_3 + k_6 +k_9} ( \gamma_a (k_1+k_7+k_{11}/2) +\Gamma(k_2+k_8 +k_{10}/2) + \frac{\Gamma_+}{2} (k_3+k_9))^3}
\end{multline}$$
+
\end{multline}

Revision as of 17:48, 21 October 2021

Our last paper with Guillermo Díaz Camacho et al. on the dynamics of emission of multiphotons, in particular, of photon bundles (?!) involves a lot of complicated combinatoric formulas, such as the average time of detection for the $k$th photon from a $N$-photon Fock state that is spontaneously emitted with a radiative rate $\gamma_a$ and measured by a detector with bandwidth $\Gamma$: \begin{multline} \langle t_{k}^{(N)} \rangle = 2 N! \left(\frac{\Gamma_+ \Gamma \gamma_a}{\Gamma_-^2}\right)^N \sum_{k_1 + k_2 + k_3\atop = N-k}\sum_{k_4 + \cdots+ k_9\atop = k-1} \sum_{k_{10} + k_{11}\atop = 2}\prod_{j=1}^{11}{1\over k_j!}\times\\ \frac{(-1)^{k_3 + k_6 + k_7 + k_8 + k_{11}}}{\gamma_a^{k_1 +k_4+k_7} \Gamma^{k_2+k_5+k_8} \left( \frac{\Gamma_+}{4} \right)^{k_3 + k_6 + k_9} \big( \gamma_a (k_1 + k_7+{k_{11}\over2}) +\Gamma( k_2 +k_8 +{k_{10}\over2}) + \frac{\Gamma_+}{2} (k_3+k_9)\big)^2 }\,. \end{multline}

Here I provide a Mathematica code to turn this general result into particular and limiting cases, and discuss briefly the whole thing. The formula is obtained by integration over the joint probability density function $\phi_\Gamma$ for emitting the photons at times $t_i$, which we show in this paper is given by: $$\phi_\Gamma(t_1,\dots,t_N)=(-1)^{N}N!\gamma_a^N\left(\frac{\Gamma}{\Gamma_-}\right)^{2N }\prod_{i=1}^{N}\left[{d\over dt_i}({e^{-\gamma_a t_i}\over\gamma_a}+{e^{-\Gamma t_i}\over\Gamma}-4{e^{-\Gamma_+ t_i/2}\over\Gamma_+})\right]{\mathbf{1}}_{[t_{i-1}, t_{i+1}[}(t_{i})$$ where $\Gamma_\pm\equiv\Gamma\pm\gamma_a$. When tracing out all the photons to keep only the $k$th of interest, through some clever math hackery, we arrive to the computation of $\int_0^\infty\left({e^{-\gamma_a t}\over\gamma_a}+{e^{-\Gamma t}\over\Gamma}-4{e^{-\Gamma_+ t/2}\over\Gamma_+}\right)^N\,dt$ which is easy to do with the multinomial theorem, which is where all the mad coefficients come from. The nice thing, of course, is that it covers all cases with a single formula, i.e., it can tell you about the 1st photon from a biphoton or the 7th photon from a 15-photon bundle. Now the direct calculation of these particular cases is a bit tedious. We can ask a formal computer language, like Mathematica, to do it. A first thing is to

tkN[k1_, k2_, k3_, k4_, k5_, k6_, k7_, k8_, k9_, k10_, k11_] := 
 2 \[CapitalNu]! (((\[CapitalGamma] + \[Gamma]a) \[CapitalGamma] \
\[Gamma]a)/(\[CapitalGamma] - \[Gamma]a)^2)^\[CapitalNu] 1/(
  k1! k2! k3! k4! k5! k6! k7! k8! k9! k10! k11!) (-1)^(
  k3 + k6 + k7 + k8 + k11)/(\[Gamma]a^(k1 + k4 + k7) \[CapitalGamma]^(
   k2 + k5 + k8) ((\[CapitalGamma] + \[Gamma]a)/4)^(
   k3 + k6 + k9) (\[Gamma]a (k1 + k7 + k11/2) + \[CapitalGamma] (k2 + k8 + k10/
        2) + (\[CapitalGamma] + \[Gamma]a)/2 (k3 + k9))^2)

This is the formula for the average of the squared time of the $k$th photon from a $N$-photon bundle:

\begin{multline} \langle \big(t_k^{(N)}\big)^2\rangle = \frac{2\gamma_a^N}{P(N,N)}\left( \frac{\Gamma}{\Gamma_-} \right)^{2N}k {N\choose k} \sum_{k_1+k_2+k_3=N-k}{N-k\choose k_1, k_2, k_3}\\{}\times \sum_{k_4+\cdots+k_9=k-1}{k-1\choose k_4, k_5, k_6, k_7,k_8,k_9} \sum_{k_{10}+k_{11}=2}{2\choose k_{10}, k_{11}}\\{}\times \frac{(-1)^{k_3 + k_6 + k_7 + k_{8}+k_{11}}}{\gamma_a^{k_1+k_4+k_7} \Gamma^{k_2+k_5+k_8} \left( \frac{\Gamma_+}{4} \right)^{k_3 + k_6 +k_9} ( \gamma_a (k_1+k_7+k_{11}/2) +\Gamma(k_2+k_8 +k_{10}/2) + \frac{\Gamma_+}{2} (k_3+k_9))^3} \end{multline}