Plotting a Piecewise function Dg?

Abdelkader Hd

2023 4 28

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2 次查看（30 天）

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Hello community I look to get the figure of Dg, please verifiy if the definition of Piecewise function Dg is correct and Thank you

clc 
clear all
x=[]; y=[]; z=[];
for n=1:1001
    x1=0.01*(n-1);
 
    x2=4;
    B=[0,x1,4,x2;
        x1,3,x2,x2;
        x1,0,5,x2;
        0,5,x2,x1];
  
    %{
    Pd=eig(B);
    if max(real(Pd))<0
        disp('fail'); 
        disp(n);
    end
    %}
    B11=[B(1,1),B(1,2);B(2,1),B(2,2)];
    B22=[B(3,3),B(3,4);B(4,3),B(4,4)];
    B12=[B(1,3),B(1,4);B(2,3),B(2,4)];
    V1_2=[B11,B12;B12',B22];
A1=det(B11);
B1=det(B22);
C1=det(B12);
D1=det(V1_2);
    Sum=det(B11)+det(B22)+2.*det(B12);
    nup=sqrt(Sum+sqrt(Sum.^2-4.*det(V1_2)))./sqrt(2);
    nun=sqrt(Sum-sqrt(Sum.^2-4.*det(V1_2)))./sqrt(2);
 
    %defin quantum discord
    
fmp=(nup+1)/2*log((nup+1)/2)-(nup-1)/2*log((nup-1)/2);
fmn=(nun+1)/2*log((nun+1)/2)-(nun-1)/2*log((nun-1)/2);
fB1=(sqrt(B1)+1)/2*log((sqrt(B1)+1)/2)-(sqrt(B1)-1)/2*log((sqrt(B1)-1)/2);
Dg1=fB1-fmp-fmn;
%inf epsilon
G1=(2*C1^2+(B1-1)*(D1-A1)+2*abs(C1)*sqrt(C1^2+(-1+B1)*(-A1+D1)))/(B1-1)^2;%val1
G2=(A1*B1-C1^2+D1-sqrt(C1^4+(-A1*B1+D1)^2-2*C1*(A1*B1+D1)))/2*B1;%val2
G3=(D1-A1*B1)^2-(1+B1)*C1^2*(A1+D1);%<=0 condition
DgG1=(sqrt(G1)+1)/2*log((sqrt(G1)+1)/2)-(sqrt(G1)-1)/2*log((sqrt(G1)-1)/2); %val if %G3<=0 condition
DgG2=(sqrt(G2)+1)/2*log((sqrt(G2)+1)/2)-(sqrt(G2)-1)/2*log((sqrt(G2)-1)/2); %else G3>0
if G3 <= 0
       Dg = real(DgG1+Dg1); 
  else     
       Dg = real(DgG2+Dg1);
 end
    
  x(n)=x1; y(n)=Dg; 
    n=n+1;
    
end
plot(x,y)

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Abdelkader Hd 2023-4-28

I need to define this function which is given in the code as G1 and G2 and the condition as G3. I want to plot this function as a function of other parameters such as temperature and coupling, since the variables A, B, C, and D are indirectly dependent on the mentioned parameters.@John D'Errico