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Page 1: My MSc. Project

Simulation and Experiment of the Interaction of Pulsed Light with

Multilevel Systems

Abhijit MondalY6010

Department of ChemistryIndian Institute of Technology, Kanpur

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∂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.

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NOT Gate

Hadamard Gate

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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,"%f\n",q[i]); }

}

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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

c2= a2b0 + a1b1 + a0b2 b2= (c2- a1b1- a2b0) / a0

.

. cn-1= ∑i=0

n-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

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Thank You for noticing me !!!