basically, as far as I know we can derive 1/2mv(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}from

∫F⋅ds=1/2mv^{2}=(p^{2})/2m

for wave equation we use Hamiltonian H=P^{2}/2m+V where P and V are both operators

However, I wonder how we can say that P^{2}/2m is the term for kinetic energy because

∫F⋅ds=∫(dp/dt)⋅ds=1/2mv^{2}is saying that knowing F and path s, we can determine (p^{2})/2m. and given with an appropriate boundary condition, vice versa.

Then, we can say that P^{2}/2m is a functional F[p,s]=∫(dp/dt)⋅ds in which we have to determine p and s separately?

Then KE is dependent on both s and p, and this seems a bit ambiguous to me because in Hamiltonian KE is dependent on momentum and potential energy is dependent on position. Dimension-wise, it would agree with energy but I wonder how we took (p^{2})/2m from ∫F⋅ds=∫(dp/dt)⋅ds=1/2mv^{2}

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# B Question about Hamiltonian and kinetic energy

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