A free particle has a continuous energy spectrum. So, like other observables like position and momentum, which are also not discrete, we must be able to calculate the probability that out of so many values of energies, what is the probability that the particle will have energies in a specific range say $E_1$ to $E_2$. I think that, maybe we have to integrate some "wave function in Energy-space", in the given range since for getting position probability we needed to integrate over wave function in position space, and for momentum probability, in momentum space. But, I haven't heard of a Energy space, so maybe I am not on the right track, and maybe what I am seeking is ill-defined (as other properties of quantum mechanical free particle). How to approach this?
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2$\begingroup$ A free particle has only kinetic energy, so $E=p^2/(2m)$. $\endgroup$– TurbotantenCommented Apr 12, 2020 at 16:16
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1 Answer
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You can label the eigenstates of the Hamiltonian by their energy, i.e:
$$\mathcal{H}|E\rangle=E|E\rangle$$
If the energy spectrum is not discrete, the probability amplitude will be given by an integral with respect to $E$ rather than a sum over eigenstates. Then, the probability between $E_1$ and $E_2$ is:
$$\int_{E_1}^{E_2}dE|\langle E|\psi\rangle |^2$$
