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Sakurai Degenerate Perturbation Theory: projection operators

  1. May 4, 2014 #1
    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 [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!
     
  2. jcsd
  3. May 5, 2014 #2

    Fredrik

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    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.
     
    Last edited: May 5, 2014
  4. May 5, 2014 #3

    Bill_K

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    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.
     
  5. May 5, 2014 #4

    Bill_K

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    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.
     
    Last edited: May 5, 2014
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