# A Degenerate perturbation theory -- Sakurai

1. Apr 13, 2016

### ShayanJ

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

2. Apr 13, 2016

### blue_leaf77

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.

3. Apr 17, 2016

### ShayanJ

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. Apr 19, 2016

### secur

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.