# Sudden Perturbation Approximation Question

## Homework Statement

In a beta decay H3 -> He3+, use the sudden perturbation approximation to determine the probability of that an electron initially in the 1s state of H3 will end up in the |n=16,l=3,m=0> state of He3+

|<n'l'm'|nlm>|^2

## The Attempt at a Solution

I actually know the answer to this but I am not clear as to why and I am wondering if there is an easier way to determine the solution.

The answer comes out to be 0. When integrating the wavefunctions |n'l'm'> and |nlm> in spherical coordinates to calculate the inner product (i.e. the probability amplitude), the integral over $$d\theta$$ returns 0. Is there an easier way to see this other than going through the calculations?

How is this result interpreted?

Thanks,

jsc

nrqed
Homework Helper
Gold Member

## Homework Statement

In a beta decay H3 -> He3+, use the sudden perturbation approximation to determine the probability of that an electron initially in the 1s state of H3 will end up in the |n=16,l=3,m=0> state of He3+

|<n'l'm'|nlm>|^2

## The Attempt at a Solution

I actually know the answer to this but I am not clear as to why and I am wondering if there is an easier way to determine the solution.

The answer comes out to be 0. When integrating the wavefunctions |n'l'm'> and |nlm> in spherical coordinates to calculate the inner product (i.e. the probability amplitude), the integral over $$d\theta$$ returns 0. Is there an easier way to see this other than going through the calculations?

How is this result interpreted?

Thanks,

jsc

The spherical harmonics are orthonormal so <l' m' | l m> is zero unless l=l' and m= m'. Clearly here the result is zero since |3,0> is orthogonal to |0,0>

Physically, it simply says that a particle in an s state has no angular momentum so the probability of it being observed with l=3 is zero. The sudden approximation simply assumes that the transition was so quick that the orbital angular momentum of the electron remained unchanged.

Last edited:
Thanks nrged.

That makes it clear.