- #1
Kaguro
- 221
- 57
- TL;DR
- Using the "good" states, which are eigenstates of H0 and V simultaneously, we can avoid the infinite coefficients and find the correct energy shifts. But how to find the correct state shifts? The coefficients now are in 0/0 form instead of infinity.
Given the unperturbed Hamiltonian ##H^0## and a small perturbating potential V. We have solved the original problem and have gotten a set of eigenvectors and eigenvalues of ##H^0##, and, say, two are degenerate:
$$ H^0 \ket a = E^0 \ket a$$
$$ H^0 \ket b = E^0 \ket b$$
Let's make them orthonormal. Here the first order state correction coefficients ## \sum_{m \neq n} \frac{\bra m V \ket n}{E_n - E_m} ## blow up when considering two kets from our degenerate eigenspace. To "solve" this problem, we want the numerator to be 0 as well. This is only possible if they are eigenkets of V as well. So, let's construct two new orthonormal vectors ##\ket c## and ##\ket d## from old ones. Still in the same eigenspace.
## H^0 \ket c = E^0 \ket c##
## H^0 \ket c = E^0 \ket c##
but now,
## V \ket c = c\ket c##
## V \ket d = d\ket d##
Now, ##\bra c V \ket d = \bra c d \ket d = d \bra c \ket d = 0## (since these are orthonormal.)
Now, using new vectors, we form another complete basis, and we can find the first order corrections to energy. But, we STILL can't find the first order corrections to the states, because these new vectors are STILL in that same stinking eigenspace and have the same energy. When we try to find the first order correction of the eigenstate ##\ket c## in the direction of ##\ket d## we're stuck. Maybe the other directions are fine, but not along ##\ket d##. Now we end up with 0/0 form instead of infinity...
How did this ordeal help us if we can't find the new eigenstates of the complete Hamiltonian?
Any help will be very much appreciated.
$$ H^0 \ket a = E^0 \ket a$$
$$ H^0 \ket b = E^0 \ket b$$
Let's make them orthonormal. Here the first order state correction coefficients ## \sum_{m \neq n} \frac{\bra m V \ket n}{E_n - E_m} ## blow up when considering two kets from our degenerate eigenspace. To "solve" this problem, we want the numerator to be 0 as well. This is only possible if they are eigenkets of V as well. So, let's construct two new orthonormal vectors ##\ket c## and ##\ket d## from old ones. Still in the same eigenspace.
## H^0 \ket c = E^0 \ket c##
## H^0 \ket c = E^0 \ket c##
but now,
## V \ket c = c\ket c##
## V \ket d = d\ket d##
Now, ##\bra c V \ket d = \bra c d \ket d = d \bra c \ket d = 0## (since these are orthonormal.)
Now, using new vectors, we form another complete basis, and we can find the first order corrections to energy. But, we STILL can't find the first order corrections to the states, because these new vectors are STILL in that same stinking eigenspace and have the same energy. When we try to find the first order correction of the eigenstate ##\ket c## in the direction of ##\ket d## we're stuck. Maybe the other directions are fine, but not along ##\ket d##. Now we end up with 0/0 form instead of infinity...
How did this ordeal help us if we can't find the new eigenstates of the complete Hamiltonian?
Any help will be very much appreciated.