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So, I am working through section 5.2 of Sakurai's book which is "Time Independent Perturbation Theory: The Degenerate Case", and I see a few equations I'm having some trouble reconciling with probably because of notation. These are equations 5.2.3, 5.2.4, 5.2.5 and 5.2.7.

First, we define a projection operator [itex]P_0[/itex] onto the space defined by [itex]\{|m^{(0)}\rangle\}[/itex]. We define [itex]P_1 = 1-P_0[/itex] to be the projection onto the remaining states. There are g different eigenkets with the same unperturbed energy [itex]E_{D}^{(0)}[/itex]. So, we have

[tex](E-E_{D}^{(0)} - \lambda P_0 V)P_0|l\rangle - \lambda P_0 V P_1|l\rangle = 0[/tex]

[tex]-\lambda P_1 V P_0 |l\rangle + (E - H_0 - \lambda P_1 V)P_1 |l\rangle = 0[/tex]

So the second equation supposedly gives equation 5.2.5

[tex]P_1 |l\rangle = P_1 \frac{\lambda}{E-H_0-\lambda P_1 V P_1}P_1 V P_0|l\rangle[/tex]

Question 1: How does one get the extra [itex]P_1[/itex] on the RHS sticking to the left?

Now if we substitute this into the second of the two equations above, we supposedly get

[tex]\left(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\right)P_0|l\rangle = 0[/tex]

Question 2: In the third term, in the denominator, how does one get [itex]\lambda V[/itex] instead of [itex]\lambda P_1 V P_1[itex]?

Finally, using this last expression, Sakurai obtains for [itex]|l^{(0)}\rangle[/itex] the condition

[tex](E-E_{D}^{(0)}-\lambda P_0 V P_0)(P_0 |l^{(0)}\rangle) = 0[/tex]

Question 3: How does one arrive at this condition?The operator in the brackets is written to order [itex]\lambda[/itex], but the third term also has an order [itex]\lambda[/itex] term. Isn't the idea here to expand both the big bracket and the ket as two power series in [itex]\lambda[/itex] and then equate the "coefficients" of each term order by order to the right hand side, which is identically zero?

Any help or hints will be much appreciated!

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# Sakurai Degenerate Perturbation Theory: projection operators

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