Vector must be the same length.

2 次查看（过去 30 天）

Abu Zar 2023-2-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1910780-vector-must-be-the-same-length

评论： Dyuman Joshi 2023-2-22

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);

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Dyuman Joshi 2023-2-11

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1910780-vector-must-be-the-same-length#answer_1169380

移动：Image Analyst 2023-2-11

在 MATLAB Online 中打开

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.

Read - Why Variables should not be named Dynamically

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