% numerics for problem 3
x = -5:0.01:5;
psi = (exp(-(x-1).^2) + i*exp(-(x+1).^2))/sqrt(2);
area = trapz(x,abs(psi).^2);
C2 = 1/sqrt(area);
psi = C2*psi; % now psi is normalized
expvalx = trapz(x,conj(psi).*x.*psi);
expvalx2 = trapz(x,conj(psi).*(x.^2).*psi);
% There is a really small imaginary part,
% in my code of order 10^-20; however, here
% we take the real part.
expvalx = real(expvalx);
expvalx2 = real(expvalx2);
dx = x(2) - x(1);
dpsidx = diff(psi)/dx;
% These are at the midpoint values because the derivatives are
% evaluated at the midpoints.
psimid = 0.5*(psi(1:(length(psi) -1)) + psi(2:length(psi)));
xmid = 0.5*(x(1:(length(psi) -1)) + x(2:length(psi)));
expvalp = -i*trapz(xmid,conj(psimid).*dpsidx); % with hbar = 1
% Again the imaginary part is very small.
expvalp = real(expvalp);
N = length(x);
d2psidx2 = diff(dpsidx)/dx;
x = x(2:(N-1));
psi = psi(2:(N-1));
expvalp2 = - trapz(x,conj(psi).*d2psidx2); % with hbar = 1
% The imaginary part is much smaller than the real part.
expvalp2 = real(expvalp2);
Deltax = sqrt(expvalx2 - expvalx^2);
Deltap = sqrt(expvalp2 - expvalp^2);
Deltax*Deltap
% This number should be greater or equal to 1/2, which it is.