Degenerate perturbation theory -- Sakurai

  • A
  • Thread starter ShayanJ
  • Start date
  • #1
2,793
594
I'm reading section 5.2 "Time-Independent Perturbation Theory: The Degenerate Case" of the book "Modern Quantum Mechanics" by Sakurai and Napolitano and I have trouble with some parts of the calculations.
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
 

Answers and Replies

  • #2
blue_leaf77
Science Advisor
Homework Helper
2,629
784
I will provide a way which I think rather heuristic because it will involve the use of approximation ##E\approx E_D^{(0)}##. Moreover, since ##\lambda## is assumed to be quite small for the perturbation method to work, the operator ##\frac{\lambda}{E-H_0-\lambda P_1 V P_1}## is approximated to be ##\frac{\lambda}{E-H_0}##. If you want, you can approach this issue from a different perspective. Consider the power expansion of the denominator of ##\frac{\lambda}{E-H_0-\lambda P_1 V P_1}## in ##\lambda##. Since we want to be working with the terms linear in ##\lambda##, one should only pick the zeroth term out of the expansion because there is already ##\lambda## in the numerator.
We will only care about the terms which contain ##\lambda##, therefore in the left hand side of
$$
P_1|l\rangle=P_1 \frac{\lambda}{E-H_0-\lambda P_1 V P_1}P_1 V P_0 |l\rangle
$$
substitute ##|l\rangle = \lambda | l^{(1)}\rangle## and in the right hand side of the above equation, substitute ##|l\rangle = |l^{(0)}\rangle## (this is so because we can be sure that the higher order terms in the expansion of ##|l\rangle## will only contribute to the terms nonlinear in ##\lambda##). Performing this step as well as incorporating the two approximations mentioned in the beginning, the above equation transforms to
$$
P_1|l^{(1)}\rangle=P_1 \frac{\lambda}{E_D^{(0)}-H_0}P_1 V P_0 |l^{(0)}\rangle
$$
The projection ##P_1## in the left most of the right hand side can be brought past ##\frac{\lambda}{E_D^{(0)}-H_0}## because they commute giving you ##P_1^2=P_1##. Using the fact that ##P_0 |l^{(0)}\rangle = |l^{(0)}\rangle## and ##P_1 = \sum_{k \notin D} |k^{(0)} \rangle \langle k^{(0)}|##, it should be pretty straightforward to see that the above equation will lead to the equation you wanted to prove.
 
  • Like
Likes ShayanJ
  • #3
2,793
594
I have another equation.
After the above calculations, the formula below is derived:
## (E-E_D^{(0)}-\lambda P_0 V P_0-\lambda^2 P_0 V P_1 \frac{1}{E-H_0-\lambda V} P_1 V P_0 )P_0 |l\rangle=0 ##.

Then its said that although there is a term of order ## \lambda^2 ## in the above formula, it actually produces a term of order ## \lambda ## in the state ## P_0 |l\rangle ##. I don't understand how this is possible!
 
  • #4
381
118
I don't have the book and am not familiar with these perturbation calculations but ...

Evidently when applying P0 the perturbed energy E approaches H0 quickly as λ goes to 0. So, in that fourth term, divide numerator and denominator by λ. Then ignoring the P-V-P terms the numerator becomes λ, denominator becomes (E-H0)/λ - V. As λ -> 0, (E-H0)/λ must remain finite, so overall the term is O(λ), as the book says.

I guess.
 
  • Like
Likes ShayanJ

Related Threads on Degenerate perturbation theory -- Sakurai

Replies
9
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
760
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
757
Replies
13
Views
9K
Replies
10
Views
7K
  • Last Post
Replies
12
Views
3K
Top