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Simulation and Experiment of the
 Interaction of Pulsed Light with
       Multilevel Systems

             Abhijit Mondal
                  Y6010
        Department of Chemistry
 Indian Institute of Technology, Kanpur
∂c1/∂t = i(E0 μ12/2ħ) (ei(ω−ω0)t + e−i(ω+ω0 )t ) c2 (t)
     ∂c2/∂t = i(E0 μ12/2ħ) (e-i(ω−ω0)t + ei(ω+ω0 )t ) c1 (t)

      where E0 is the amplitude of the light wave,
         light wave of angular frequency ω and
                    ω0 = (E2 − E1 )/ħ.
                Rabi frequency defined by:
                     ΩR = |μ12 E0 /ħ |.
 We apply the rotating wave approximation to neglect the
           terms that oscillate at ±(ω + ω0 ),
Second, we only consider the case of exact resonance with
                         δω = 0.
NOT Gate




Hadamard
Gate
int main(int argc,char *argv[])
{

    FILE *fp1,*fp2,*fp3,*fp4; fp1=fopen("1.txt","r"); fp2=fopen("gr 5.txt","r");
    fp3=fopen("quotient.txt","w"); fp4=fopen("remainder.txt","w");
    char *buffer1=new char[20]; char *buffer2=new char[20]; double *f=new double[2000];
    double *g=new double[2000];

    int i=0;
    while(!feof(fp1)) {fseek(fp1,7,SEEK_CUR); fgets(buffer2,8,fp1); double y=atof(buffer2); f[i]=y;i+
    +;
    }

    i=0;
    while(!feof(fp2)) {fseek(fp2,7,SEEK_CUR); fgets(buffer2,8,fp2); double y=atof(buffer2); g[i]=y;i+
    +;
    }

    int deg=i;
    for(i=0;i<deg;i++) {
    int j=1;double sum=0.0;
    while(j<=i) {sum+=f[j]*q[i-j]; j++; }


    q[i]=(g[i]-sum)/f[0]; fprintf(fp3,"%fn",q[i]);
    }

}
Let us suppose that the original response signal be denoted
 as g(x), the h(x) is the convolution of f(x) and g(x). Then

                     c0= a0              b0= c0 / a0
           c1= a1b0 + a0b1               b1= (c1- a1b0) / a0
    c 2= a2b 0 + a 1b 1 + a 0b 2         b2= (c2- a1b1- a2b0) / a0
                                     .
                                     .
   cn-1= ∑i=0n-1 ai bn-1-i         bn-1= (cn-1 - ∑i=1n-1 ai bn-1-i) / a0

Hence we use a dynamic programming approach to solve for
                    bi for each i as:
                bi = (ci - ∑j=1i aj bi-j) / a0
Thank You for noticing me !!!

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My MSc. Project

  • 1. Simulation and Experiment of the Interaction of Pulsed Light with Multilevel Systems Abhijit Mondal Y6010 Department of Chemistry Indian Institute of Technology, Kanpur
  • 2. ∂c1/∂t = i(E0 μ12/2ħ) (ei(ω−ω0)t + e−i(ω+ω0 )t ) c2 (t) ∂c2/∂t = i(E0 μ12/2ħ) (e-i(ω−ω0)t + ei(ω+ω0 )t ) c1 (t) where E0 is the amplitude of the light wave, light wave of angular frequency ω and ω0 = (E2 − E1 )/ħ. Rabi frequency defined by: ΩR = |μ12 E0 /ħ |. We apply the rotating wave approximation to neglect the terms that oscillate at ±(ω + ω0 ), Second, we only consider the case of exact resonance with δω = 0.
  • 3.
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12. int main(int argc,char *argv[]) { FILE *fp1,*fp2,*fp3,*fp4; fp1=fopen("1.txt","r"); fp2=fopen("gr 5.txt","r"); fp3=fopen("quotient.txt","w"); fp4=fopen("remainder.txt","w"); char *buffer1=new char[20]; char *buffer2=new char[20]; double *f=new double[2000]; double *g=new double[2000]; int i=0; while(!feof(fp1)) {fseek(fp1,7,SEEK_CUR); fgets(buffer2,8,fp1); double y=atof(buffer2); f[i]=y;i+ +; } i=0; while(!feof(fp2)) {fseek(fp2,7,SEEK_CUR); fgets(buffer2,8,fp2); double y=atof(buffer2); g[i]=y;i+ +; } int deg=i; for(i=0;i<deg;i++) { int j=1;double sum=0.0; while(j<=i) {sum+=f[j]*q[i-j]; j++; } q[i]=(g[i]-sum)/f[0]; fprintf(fp3,"%fn",q[i]); } }
  • 13. Let us suppose that the original response signal be denoted as g(x), the h(x) is the convolution of f(x) and g(x). Then c0= a0 b0= c0 / a0 c1= a1b0 + a0b1 b1= (c1- a1b0) / a0 c 2= a2b 0 + a 1b 1 + a 0b 2 b2= (c2- a1b1- a2b0) / a0 . . cn-1= ∑i=0n-1 ai bn-1-i bn-1= (cn-1 - ∑i=1n-1 ai bn-1-i) / a0 Hence we use a dynamic programming approach to solve for bi for each i as: bi = (ci - ∑j=1i aj bi-j) / a0
  • 14. Thank You for noticing me !!!