Recent content by alex23

The first thing I have to consider is that, since ##\Delta t \rightarrow 0##, the potential ##U## is not going to contribute and we can consider it to be ##0##. Next thing I did was calculate ##\left<xt|x_1t_1\right>## directly with the definition considering what I said before, and I got...

You are absolutely right, I had a sign error in the Lagrangian. After fixing that and doing the integration by parts as you suggested I managed to get ##\delta S=\frac{1}{2}\displaystyle\int_{0}^{T}{\dot\alpha^2(t)dt}## as I as supposed to. Thanks!

Using the Lagrangian of the system I reached that ##x(t)=\frac{1}{2} gt^2+ut ## is the real trajectory of the particle. After that, I consider different trajectories: ##x(\alpha,t) = x(t) + \alpha(t)## with ##\alpha(t)## being an arbitrary function of t expect for the conditions...