Effacer les filtres
Effacer les filtres

Vector must be the same length.

2 vues (au cours des 30 derniers jours)
Abu Zar
Abu Zar le 11 Fév 2023
Commenté : Dyuman Joshi le 22 Fév 2023
I created the m.file to plot alpha vs Vnorm but the vector is not the same length ,its request to hellp me at this stage
D=15;
tmpI=eye(D);
ket0=tmpI(:,1); %|0>
ket1=tmpI(:,2); %|1>
ket2=tmpI(:,3); %|2>
ket3=tmpI(:,4); %|3>
ket4=tmpI(:,5); %|4>
ket5=tmpI(:,6); %|5>
ket6=tmpI(:,7); %|6>
ket7=tmpI(:,8); %|7>
ket8=tmpI(:,9); %|8>
ket9=tmpI(:,10); %|9>
ket10=tmpI(:,11); %|10>
ket11=tmpI(:,12); %|11>
ket12=tmpI(:,13); %|12>
ket13=tmpI(:,14); %|13>
ket14=tmpI(:,15); %|14>
sym n
alpha1 =0.03;
ketalpha1 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14)*(ket1+ket2+ket3+ket4+ket5+ket6+ket7+ket8+ket9+ket10+ket11+ket12+ket13+ket14);
sym n
alpha2 =0.06;
ketalpha2 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14)*(ket1+ket2+ket3+ket4+ket5+ket6+ket7+ket8+ket9+ket10+ket11+ket12+ket13+ket14);
sym n
alpha3 =0.09;
ketalpha3 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14)*(ket1+ket2+ket3+ket4+ket5+ket6+ket7+ket8+ket9+ket10+ket11+ket12+ket13+ket14);
sym n
alpha4 =0.12;
ketalpha4 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14)*(ket1+ket2+ket3+ket4+ket5+ket6+ket7+ket8+ket9+ket10+ket11+ket12+ket13+ket14);
sym n
alpha5 =0.15;
ketalpha5 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14)*(ket1+ket2+ket3+ket4+ket5+ket6+ket7+ket8+ket9+ket10+ket11+ket12+ket13+ket14);
sym n
alpha6 =0.18;
ketalpha6 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14)*(ket1+ket2+ket3+ket4+ket5+ket6+ket7+ket8+ket9+ket10+ket11+ket12+ket13+ket14);
creation=circshift(diag(sqrt(0:1:14)),-1);
annihilation=creation';
V1=annihilation*ketalpha1-alpha1*ketalpha1;
V2=annihilation*ketalpha2-alpha2*ketalpha2;
V3=annihilation*ketalpha3-alpha3*ketalpha3;
V4=annihilation*ketalpha4-alpha4*ketalpha4;
V5=annihilation*ketalpha5-alpha5*ketalpha5;
V6=annihilation*ketalpha6-alpha6*ketalpha6;
Vnorm1=V1*V1';
Vnorm2=V2*V2';
Vnorm3=V3*V3';
Vnorm4=V4*V4';
Vnorm5=V5*V5';
Vnorm6=V6*V6';
plot(Vnorm1,alpha1,Vnorm2,alpha2,Vnorm3,alpha3,Vnorm4,alpha4,Vnorm5,alpha5,Vnorm6,alpha6);

Réponse acceptée

Dyuman Joshi
Dyuman Joshi le 11 Fév 2023
Déplacé(e) : Image Analyst le 11 Fév 2023
I have tidied up your code. Avoid using dynamically named variables as much as you can, it is not recommended, Indexing is much simpler and efficient.
Coming onto the issue, alpha1 is a scalar and Vnorm1 is a matrix. How exactly do you plan to plot a scalar against a matrix (or vice-versa) ?
D = 15;
tmpI = eye(D);
%sum of ket1 to ket14 is
ket = [0;ones(14,1)];
syms n
ct = 6;
creation = circshift(diag(sqrt(0:1:14)),-1);
annihilation = creation';
%pre-allocation
[ketalpha, V, Vnorm] = deal(cell(1,ct));
for k=1:ct
alpha = 0.03*k;
temp0 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14)*ket;
ketalpha{k} = temp0;
temp1 = annihilation*temp0 - alpha*temp0;
V{k} = temp1;
Vnorm{k} = temp1*temp1';
end
  19 commentaires
Abu Zar
Abu Zar le 22 Fév 2023
hello sir you know my codes ,i check values of every variables but i did not got the values of n, i put the values of n from 0 to 14.
Dyuman Joshi
Dyuman Joshi le 22 Fév 2023
"but i did not got the values of n"
Because you have not assigned any value to n.
"i put the values of n from 0 to 14."
Yes, to find the value of a sum, with respect to a variable. But the value of n has not been explicitly defined.

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