Index a matrix multiplied by a time dependent function

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Hi there,
I'm trying to index a matrix that's multiplied by a time dependent function, but keep getting the error "inner dimensions must agree" in the U expression line. I need to know the exact matrix values for each interaction related to the function E(t) in the time, i.e., U(1) related to E(1) and so on. Thanks in advance!
Here's the code
%% Quantum Control
format long
clear all
clc
%% Variables
fs=10;
ft = -20000:1/fs:20000; % Time
alpha = 0.001; % Time parameter
beta = 0.001; % Time parameter
ai = 0.2; % Pop |1>
ai2 = 0.8; % Pop |2>
af = 1; % Pop Final |1>
w0 = 0.02; % w0 = (E2 - E1)/h
mi = 6; % Dipole
phii = pi/24; % Relative phase
phif = pi/4; % Relative phase final
E1 = 1; % Energy
E2 = 2; % Energy 2
H = [0 1;1 0]; % Interaction hamiltonian
H0 = [E1 0;0 E2]; % Matriz do Hamiltoniano sem Interação
%% Control function
g = 1./(1+exp(-alpha.*ft));
f = ai*(1-g)+af*g;
p = 1./(1+exp(-beta.*ft));
h = phii*(1-p)+phif*p;
%% Electric field
E = alpha.*(af-ai).*exp(alpha.*ft).*sin(w0.*ft+h)./(mi.*(1+exp(alpha.*ft)).*sqrt((1-ai+(1-af).*exp(alpha.*ft)).*(ai+af.*exp(alpha.*ft))))+2.*beta.*(phif-phii).*exp(beta.*ft).*sqrt(f.*(1-f)).*cos(w0.*ft+h)/(mi.*(1-2.*f).*(1+exp(beta.*ft).^2));
%% Time evolution operator
[V, D]=eig(H);
Z = expm(H0);
Z1 = expm(D);
U = exp(-1i.*ft./2).*Z.*exp(1i.*mi.*E.*ft).*V.*Z1.*inv(V).*exp(-1i.*ft./2).*Z;
%% End
  2 Comments
Walter Roberson
Walter Roberson on 10 Jul 2019
V.*Z1.*inv(V) is probably a mistake. Any multiplication of a matrix and its inverse should be using * instead of .*
You should likely also be coding using Z1/V instead of Z1.*inv(V)
In your long expression you have a / that should probably be ./
guilherme stahlberg
guilherme stahlberg on 10 Jul 2019
I've tried either * or .* but keep getting the same error "inner dimensions must agree". Same regarding the / and./

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Accepted Answer

KSSV
KSSV on 10 Jul 2019
U = zeros(2,2,length(ft)) ;
for i = 1:length(ft)
U(:,:,i) = exp(-1i.*ft(i)./2).*Z.*exp(1i.*mi.*E(i).*ft(i)).*V.*Z1.*inv(V).*exp(-1i.*ft(i)./2).*Z;
end
  1 Comment
Walter Roberson
Walter Roberson on 12 Jul 2019
This is likely to be mathematical nonsense though. V.*Z1.*inv(V) is much more likely to need V*Z1*inv(V) or similar.

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