A question about the Hamiltonian of the fission of Uranium

In summary, the lecturer discussed the concept of dispersion in a quantum mechanics course, using the example of uranium fission by alpha rays. He explained that a detector must be placed to detect the particles, but it can only detect particles with specific energies and angles. He then introduced the Hamiltonian as H=H0+V, with V considered as a perturbation. The H0 contains both continuous and discreet states, similar to the nuclear shell model used to describe the atomic model. The lecturer used V to represent a free particle Hamiltonian for the alpha particle subsystem.
  • #1
raeed
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so I am taking a quantum mechanics course, we started taking about dispersion.
so he the lecturer gave us an example about the fission of uranium by alpha ray... he said that we should place a detector in order to detect the alpha particlee , but the detector can only detect particlees with specific energies and and also specific angles. so he said the Hamiltonian will be H=H0+V, and we will look at V as a perturbation, and thus V=∫([p[2]2m) |p)(p| [d][3]p
he said that its true because the normal hamiltonian H0 contains both continuous states and discreet states.
so my question is how did he get to V and why the H0 contiains continuous and discreet states?
 
  • #3
The H0 is the (probably simplified) Hamiltonian of the nucleus. Usually one needs a nuclear model: http://en.wikipedia.org/wiki/Atomic_nucleus#Nuclear_models to write it down. This model: http://en.wikipedia.org/wiki/Nuclear_shell_model resembles the atomic model (for the electron shells and subshells). Atomic hamiltonians are notorious to have also a continuous part of the spectrum which would correspond to the scattering states of the free electrons. His V looks like a free particle Hamiltonian (probably for the alpha particle seen as a different subsystem (the other subsystem would be the nucleus/disintegrated nucleus)).
 

1. What is the Hamiltonian of the fission of Uranium?

The Hamiltonian of the fission of Uranium is a mathematical operator used in quantum mechanics to describe the total energy of a system undergoing nuclear fission. It takes into account the kinetic energy of the fragments, the Coulomb interaction between them, and the potential energy of the nuclear forces.

2. How is the Hamiltonian of the fission of Uranium calculated?

The Hamiltonian of the fission of Uranium is calculated by solving the Schrödinger equation, which describes the behavior of quantum systems. This involves using a combination of theoretical models and experimental data to determine the various terms in the Hamiltonian, such as the masses and charges of the fragments.

3. Why is the Hamiltonian of the fission of Uranium important?

The Hamiltonian of the fission of Uranium is important because it allows us to understand and predict the behavior of nuclear fission, which is a crucial process in nuclear power and weapons. It also provides insight into the fundamental forces and interactions that govern the behavior of particles at the nuclear level.

4. How does the Hamiltonian of the fission of Uranium relate to other nuclear reactions?

The Hamiltonian of the fission of Uranium is specific to the process of nuclear fission, which involves the splitting of a heavy nucleus into smaller fragments. It is different from the Hamiltonian of other nuclear reactions, such as fusion, which involves the merging of two lighter nuclei to form a heavier one.

5. Can the Hamiltonian of the fission of Uranium be altered or controlled?

Yes, the Hamiltonian of the fission of Uranium can be altered or controlled through various means, such as changing the shape or size of the Uranium nucleus, or bombarding it with particles to induce fission. This allows for the manipulation of nuclear reactions and the production of energy in nuclear power plants.

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