- #1

ShayanJ

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At firsts he explains that there is a g-dimensional subspace(which he calls D) of degenerate energy eigenstates of the unperturbed Hamiltonian which he calls ## \{|m^{0}\rangle\}##. Then he says that the perturbation will remove the degeneracy and these states will split into g states with different energies ## \{|l \rangle \} ## but if you let the perturbation parameter ## \lambda ## go to zero, you may end up with different states than ## \{|m^{0}\rangle\}##, let's call them ## \{|l^0 \rangle \} ##. Then he introduces the projection operators ## P_0 ## and ## P_1=1-P_0 ## where ## P_0 ## projects on the subspace spanned by ## \{|m^{0}\rangle\}##. Then he splits the time-independent Schrodinger equation for the ## \{|l \rangle \} ## states as below:

## 0=( E-H_0-\lambda V) |l\rangle=(E-E_D^{(0)}-\lambda V)P_0|l\rangle+( E-H_0-\lambda V)P_1 |l\rangle ##

Then he projects the above equation on the left by ## P_1 ## to get:

## -\lambda P_1 V P_0 |l\rangle+(E-H_0-\lambda P_1V)P_1|l\rangle=0 ##

Which after some manipulations, gives:

## P_1|l\rangle=P_1 \frac{\lambda}{E-H_0-\lambda P_1 V P_1}P_1 V P_0 |l\rangle ##.

Then he substitutes the expansion ## |l\rangle=|l^{(0)}\rangle+\lambda |l^{(1)}\rangle+ \dots ## into the above equation to get the equation below to order ## \lambda ##:

##\displaystyle P_1|l^{(1)}\rangle=\sum_{k \notin D} \frac{|k^{(0)} \rangle V_{kl}}{E_D^{(0)}-E_k^{(0)}} ##

But I don't understand how he got this. I just can't see what should be done!

One of the things that confuses me is that there is a ## \lambda ## in the denominator, so can we still say the order of each term just by counting the number of ## \lambda##s in the nominator? How should we account for the presence of ## \lambda ## in the denominator?

Any hint would be appreciated.

Thanks