- #1
vega12
- 11
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Right now, I'm self-studying from J. J. Sakurai's book Modern Quantum Mechanics. In section 7.3, Optical Theorem, there is one step in the proof that he uses that escapes me. His proof involves using the transition operator T defined as:
[itex]
V \mid\psi^{(+)} \rangle = T \mid\phi \rangle \\
[/itex]
where [tex]\mid\phi \rangle[/tex] is the free particle state and [tex]\mid\psi^{(+)} \rangle[/tex] is the scattered state. In the proof for the optical theorem, he says "Now we use the well-known relation":
[itex]
\frac{1}{E - H_0 - i\epsilon} = Pr. \left( \frac{1}{E-H_0}\right) + i \pi \delta(E-H_0)
[/itex]
But I am at a loss as to what "Pr." means. I can follow the rest of the proof except for this one part, so a bit of clarification would be much appreciated. Thanks!
[itex]
V \mid\psi^{(+)} \rangle = T \mid\phi \rangle \\
[/itex]
where [tex]\mid\phi \rangle[/tex] is the free particle state and [tex]\mid\psi^{(+)} \rangle[/tex] is the scattered state. In the proof for the optical theorem, he says "Now we use the well-known relation":
[itex]
\frac{1}{E - H_0 - i\epsilon} = Pr. \left( \frac{1}{E-H_0}\right) + i \pi \delta(E-H_0)
[/itex]
But I am at a loss as to what "Pr." means. I can follow the rest of the proof except for this one part, so a bit of clarification would be much appreciated. Thanks!