CDL
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Hi, I have a question about calculating probabilities in situations where a particle experiences a sudden change in potential, in the case where both potentials are time independent.
For example, a tritium atom undergoing spontaneous beta decay, and turning into a Helium-3 ion. The orbital electron is initially in the ground state. The main point about this example is that the orbital electron essentially experiences a sudden change in Coulomb potential, with atomic number Z = 1 to Z = 2. Suppose we want to calculate the probability that after this beta decay the electron is left in the ground state of the new potential. In order to do this, we just take the inner product of the wave-function for the ground state state with Z = 1 with the 1s state, Z = 2, and take the modulus squared. There is our probability. Something like $$\mathbb{P}(\text{1s initial to 1s final}) = |\langle \psi_{1sf} | \psi_{1si} \rangle|^2$$ Where ## \psi_{1si}## and ##\psi_{1sf}## are the wave functions of the ground states in the initial and final potentials.
From what I have seen, this is what is generally done in these situations where the potential suddenly changes, but I'm not entirely comfortable with why this is done. Is this just another way of writing $$\mathbb{P}(\text{transition from state a to state b}) = |\langle \psi_{b} | \psi_{a} \rangle|^2$$
Could someone please explain why this is done, and why we can do this even though the potential changes?
Is it the fact that the wave-function doesn't have time to react to such a change in potential, and so the probability is a measure of 'how much of the proposed state b is present in the state a' ?
For example, a tritium atom undergoing spontaneous beta decay, and turning into a Helium-3 ion. The orbital electron is initially in the ground state. The main point about this example is that the orbital electron essentially experiences a sudden change in Coulomb potential, with atomic number Z = 1 to Z = 2. Suppose we want to calculate the probability that after this beta decay the electron is left in the ground state of the new potential. In order to do this, we just take the inner product of the wave-function for the ground state state with Z = 1 with the 1s state, Z = 2, and take the modulus squared. There is our probability. Something like $$\mathbb{P}(\text{1s initial to 1s final}) = |\langle \psi_{1sf} | \psi_{1si} \rangle|^2$$ Where ## \psi_{1si}## and ##\psi_{1sf}## are the wave functions of the ground states in the initial and final potentials.
From what I have seen, this is what is generally done in these situations where the potential suddenly changes, but I'm not entirely comfortable with why this is done. Is this just another way of writing $$\mathbb{P}(\text{transition from state a to state b}) = |\langle \psi_{b} | \psi_{a} \rangle|^2$$
Could someone please explain why this is done, and why we can do this even though the potential changes?
Is it the fact that the wave-function doesn't have time to react to such a change in potential, and so the probability is a measure of 'how much of the proposed state b is present in the state a' ?