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注册日期: 2007-07-07
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自洽求解薛定谔方程的系统势阱图
下面是非自洽方法的程序 clear; clc; s0=-250:1:250; v=0; j=1; m=1; t1=zeros(1,1000000); t2=zeros(1,1000000); for E=25:.0001:30 s1=0:1:500;s1(1)=0;s1(2)=1; for k=2:1:500 if k>=201 & k<=300 v=0; else v=167; end s1(k+1)=(1.9294e-5*(v-E)+2)*s1(k)-s1(k-1); end t1(m)=s1(501);t2(m)=E; m=m+1; end for i=2:999999 if abs(t1(i))<abs(t1(i-1))&abs(t1(i))<abs(t1(i+1)) E=t2(i); s1=0:1:500;s1(1)=0;s1(2)=1; for k=2:1:500 if k>=201 & k<=300 v=0; else v=167; end s1(k+1)=(1.9294e-5*(v-E)+2)*s1(k)-s1(k-1); end h=0;a=0; for m=1:501 h=h+s1(m)^2; if h~=0 a=(1/h)^0.5; end end b=a*s1(501); s1=a*s1; %s1=s1+.005*E; subplot(2,2,1) plot(s0,s1); e(j)=E;j=j+1; else continue end end for i=1:501 if i>=201 & i<=300 r1(i)=-(2e16*(s1(i).^2)-2e14)*1e-14; else r1(i)=-2e16*(s1(i).^2)*1e-14; end end subplot(2,2,2); plot(s0,r1); r2=0:500; %r2(1)=-0.904*sum(r1(2:501))/13; for i=1:1:500 r2(i)=.904*(sum(r1(1:i))-sum(r1((i+1):501)))/13; end r2(501)=.904*sum(r1(1:501))/13; subplot(2,2,3); plot(s0,r2); %r2=r2-1.808; r3=0:500;r3(1)=.1*r2(1); for j=2:501 r3(j)=.1*sum(r2(1:j)); end subplot(2,2,4); plot(s0,r3); %end |
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