- #1
frustrationboltzmann
- 5
- 2
Hello, I have a little problem understanding the quantum mechanics of a hydrogen atom.
Im troubled with the following question: before i measure the state of a (simplified: without fine-, hyperfinestructure) hydrogen atom, which is the right probability density of finding the electron? is it the density of the groud state (s) or the 2s, 3s...?
my assumption is the following:
the probability of a state to be occupied is given by the fermi dirac distribution for a certain temperature. so in order to get the probability density of the location of the electron before it is measured, I multiply each (normalized) probability density with the occupation probability of every eigenstate and sum over all those products. is this right or complete nonsense?
also if its right, how would I treat degenerated quantum states. my assumption. multiply every degenerated states by the probability of the fermi dirac distribution and divide by the number of degenerated states, since I assume that they all have the same probability.
this would be a mixed state, no?
I appreciate any help!
thank you very much
Im troubled with the following question: before i measure the state of a (simplified: without fine-, hyperfinestructure) hydrogen atom, which is the right probability density of finding the electron? is it the density of the groud state (s) or the 2s, 3s...?
my assumption is the following:
the probability of a state to be occupied is given by the fermi dirac distribution for a certain temperature. so in order to get the probability density of the location of the electron before it is measured, I multiply each (normalized) probability density with the occupation probability of every eigenstate and sum over all those products. is this right or complete nonsense?
also if its right, how would I treat degenerated quantum states. my assumption. multiply every degenerated states by the probability of the fermi dirac distribution and divide by the number of degenerated states, since I assume that they all have the same probability.
this would be a mixed state, no?
I appreciate any help!
thank you very much