My MSc. Project
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The document describes simulation and experimentation of pulsed light interacting with multilevel quantum systems. It presents two coupled differential equations modeling the interaction and applies the rotating wave approximation to neglect fast oscillating terms. It also only considers the case of exact resonance between the light frequency and the system energy levels.
![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]);
}
}](https://image.slidesharecdn.com/poster2-120729141917-phpapp01/85/My-MSc-Project-12-320.jpg)
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.
- 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 !!!