# Transition probabilities (basic concepts no math)

1. Jan 23, 2005

### RedX

Does it make sense to speak of the probability of finding a system which was once in the ground state in a higher state after a certain time? Since the Hamiltonian depends on time, once you collapse the wavefunction at that time, the energy you get can't be one of the values of the unperturbed system, but rather the allowed values for the Hamiltonian at the moment you collapse the wavefunction.

For example, take beta decay, where the nucleus gets an extra positive charge and an electron flies out. Because this process happens so quick, the state of a orbital electron in the ground state remains there, but that doesn't mean it has that energy, but instead one has to write the ground state of the unperturbed system as a linear combination of the energy eigenstates for a postive ion nucleus?

2. Jan 24, 2005

### dextercioby

Well,not really.Once a perturbation is applied,the system's quantum states will undergo a change and it will not make any sense to speak about the probability of finding the system in one of the previous states.The number given by (time-dependent) perturbation theory will not be the amplitude of probability of finding the system in the quantum state $|\psi\rangle$,but the amplitude of probability of TRANSITION from the quantum state $|\psi_{0}\rangle$ to the $|\psi\rangle$

Daniel.

P.S.We cannot fing the eigenstates of the perturbed Hamiltonian...

3. Jan 25, 2005

### RedX

Say the system starts in the ground state at time zero. For adiabatic perturbations, where the Hamiltonian changes slowly over a long period of time, then the system will be found in the ground state of the Hamiltonian at that later time, the ground state of H(t). I guess this has something to do with the time energy uncertainty relation, which I don't understand, as it has nothing to do with the general uncertainty relations for two Hermitian operators. But that's fine I asked a question about the energy-time relationship long ago and didn't understand the answer. We can always find eigenvectors of any operator, whether time is included and not, and the Hilbert is still spanned by these vectors, so any state can be expressed as a linear combo of these vectors. Whether the state collapses into these vectors is physics not maths and I guess depends on the postulates, so that with a time-Hamiltonian you can't find the eigenstates must be something in the postulates.