How to prove that an operater is hamiltonian in nature?
Answers
Answer:
quantum mechanics states that the hamiltonian, defined as
H=iℏ∂∂t
is a hermitian operator. But i don't really see how I have to interpret this. First of all: from which to which space is this operator working? They are defining a vectorspace called the "wavefunctionspace F" which contains all square-integrable functions that are continious and infinite differentiable (and everywhere defined). But it looks to me, that if the hamiltonian acts on this space, it's not necessary true that the image of a random vector of F is again in F.
I think in fact, that there are some vectors of F so that the hamiltonian of those vectors is not an element of F (so that it's not an endomorphism on F). And if the hamiltonian has to be hermitian, it has to be an endomorphism on some space.
If we define instead the vectorspace V, which is the same space as F but where functions don't have to be square-integrable, the hamiltonian will be an endomorphism (so at first I thought this was the solution). But now the inner product on functions
<f,g>=∫∞−∞dx f∗g
which was defined well on F because the integral will always exist if f and g are function of F, is no longer properly defined.
Explanation:
Answer:
In quantum mechanics, a Hamiltonian is an operator corresponding to the sum of the kinetic energies plus the potential energies for all the particles in the system (this addition is the total energy of the system in most of the cases under analysis). It is usually denoted by , also or .
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