Sakurai Degenerate Perturbation Theory: projection operators

[maverick280857]

Hi,

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

$$(E-E_{D}^{(0)} - \lambda P_0 V)P_0|l\rangle - \lambda P_0 V P_1|l\rangle = 0$$
$$-\lambda P_1 V P_0 |l\rangle + (E - H_0 - \lambda P_1 V)P_1 |l\rangle = 0$$

So the second equation supposedly gives equation 5.2.5

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

Question 1: How does one get the extra $P_1$ on the RHS sticking to the left?

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

$$\left(E-E_{D}^{(0)} - \lambda P_0 V P_0 - \lambda^2 P_0 V P_1 \frac{1}{E-H_0-\lambda V}P_1 V P_0\right)P_0|l\rangle = 0$$

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

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

$$(E-E_{D}^{(0)}-\lambda P_0 V P_0)(P_0 |l^{(0)}\rangle) = 0$$

Question 3: How does one arrive at this condition? The operator in the brackets is written to order $\lambda$, but the third term also has an order $\lambda$ term. Isn't the idea here to expand both the big bracket and the ket as two power series in $\lambda$ 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!

 

[Fredrik]

Question 1: How does one get the extra $P_1$ on the RHS sticking to the left?

I only have time to look at this one right now. You multiply both sides of the second equation by this:
\begin{align}
(E-H_0-\lambda P_1 V)^{-1} &=P_1P_1^{~-1}(E-H_0-\lambda P_1 V)^{-1} =P_1\big((E-H_0-\lambda P_1 V)P_1\big)^{-1}\\
&=P_1\big((E-H_0)P_1-\lambda P_1VP_1\big)^{-1}
\end{align} I guess we must have $(E-H_0)P_1=E-H_0$ somehow. Hm...I think that this is not true, but we can pretend that it is as long as the operator above only acts on states of the form $P_1|\text{something}\rangle$. To put it differently, the restriction of $E-H_0$ to the subspace $P_1(\mathcal H)$ (where $\mathcal H$ is the Hilbert space) is equal to the restriction of $(E-H_0)P_1$ to that subspace.

Edit: D'oh, this doesn't make sense. Projection operators aren't invertible. Unfortunately I have to leave the computer for a couple of hours now.

 

[Bill_K]

Question 1: How does one get the extra $P_1$ on the RHS sticking to the left?

I believe Sakurai gives the answer to this. E - H₀ - λP₁VP₁ is singular and can't be inverted. But it is nonsingular in the P₁ subspace. So we invert its projection, P₁(E - H₀ - λP₁VP₁)P₁, gettting P₁(E - H₀ - λP₁VP₁)^-1 P₁.

 

[Bill_K]

Question 3: How does one arrive at this condition? The operator in the brackets is written to order $\lambda$, but the third term also has an order $\lambda$ term.

No, the third term already has a λ² in front, and if you expand (E - H₀ - λV)^-1 in a power series for small λ, it will contribute only more positive powers of λ.