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why MATLAB is busy for this code?

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Abu Zar
Abu Zar el 15 de En. de 2023
Comentada: Abu Zar el 15 de En. de 2023
1.What is the result?
creation=circshift(diag(sqrt(0:1:14)),-1);
annihilation=creation';
syms n
alpha = sym(0.03);
ketalpha1 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14);
syms n
alpha = sym(0.06);
ketalpha2 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14);
syms n
alpha = sym(0.09);
ketalpha3 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14);
syms n
alpha = sym(0.12);
ketalpha4 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14);
syms n
alpha = sym(0.15);
ketalpha5 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14);
syms n
alpha = sym(0.18);
ketalpha6 = symsum(exp(-abs(alpha)^2 * alpha^n / sqrt(factorial(n)) ), n, 0, 14);
V1=annihilation*ketalpha1-0.03*ketalpha1;
V2=annihilation*ketalpha2-0.06*ketalpha2;
V3=annihilation*ketalpha3-0.09*ketalpha3;
V4=annihilation*ketalpha4-0.12*ketalpha4;
V5=annihilation*ketalpha5-0.15*ketalpha5;
V6=annihilation*ketalpha5-0.18*ketalpha5;
Vnorm1=V1'*V1;
Vnorm2=V2'*V2;
Vnorm3=V3'*V3;
Vnorm4=V4'*V4;
Vnorm5=V5'*V5;
Vnorm6=V6'*V6;
alpha=[0.03;0.06;0.09;0.12;0.15;0.18];
Vnorm=[Vnorm1;Vnorm2;Vnorm3;Vnorm4;Vnorm5;Vnorm6];
size(alpha)
ans = 1×2
6 1
size(Vnorm)
ans = 1×2
90 15
plot(alpha,Vnorm)
Error using plot
Vectors must be the same length.
xlabel('alpha','FontSize',12,'FontWeight','bold')
ylabel('Vnorm','FontSize',12,'FontWeight','bold')
  2 comentarios
Torsten
Torsten el 15 de En. de 2023
As you can see, alpha has dimension 6x1 and Vnorm has dimension 90x15.
How to plot Vnorm against alpha ?
Abu Zar
Abu Zar el 15 de En. de 2023
Thank you 🌹

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Respuesta aceptada

Walter Roberson
Walter Roberson el 15 de En. de 2023
My tests show it executing quickly but failing the plot.
annihilation is 15x15 so your Vnorm variables are 15x15. You [] 6 of those together to get 90 x 15 but expect to be able to plot that against 6 alpha values.
annihilation is created by a diag() so it is not going to be a vector.

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