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Branch points and branch cuts

  1. Jan 12, 2012 #1

    The following function [itex]\Omega = [x^2 + y^2 + (z - ia)^2]^{1/2}[/itex] has the following branch points:
    [itex] z_1 = i (a + \lambda), z_2 = i (a - \lambda) [/itex] where [itex] \lambda^2 = x^2 + y^2 [/itex].
    Now I do not understand the theory about branch cuts. It is said in the paper I read: "We make cuts on the imaginary z axis from [itex] z_1 [/itex] up to [itex] z = i\infty [/itex] and from [itex] z_2 [/itex] down to [itex] z = -i\infty [/itex]". OK. Now, let's consider the case [itex] \lambda < a [/itex]. Then the lower branch line crosses the real z axis. What I do not understand is that
    [itex]\Omega^- = -\Omega^+ , z=0 [/itex]
    where [itex]\Omega^- [/itex] denotes the value of the radical for [itex] z = 0^- [/itex] and [itex]\Omega^+ [/itex] for [itex] z = 0^+ [/itex]. (By the way, what is 'radical'?) They further wrote that
    [itex] \Omega^+ = [\lambda^2 + (-ia)^2]^{1/2} = -i(a^2-\lambda^2)^{1/2} [/itex]. How they calculated that? When I tried to calculate that I got +i instead of -i for the prefactor.
    I thought that that lower branch cut means that the angle is from the interval [itex] (-\pi/2,3/2\pi) [/itex] and so
    [itex] \Omega^+ = [\lambda^2 + (\epsilon -ia)^2]^{1/2} = [\lambda^2 -a^2 + \epsilon^2 - 2\epsilon ai]^{1/2} = [a^2 - \lambda^2]^{1/2} e^{i\pi/2} = [a^2 - \lambda^2]^{1/2} i[/itex] which is wrong.

    Any help much appreciated.
  2. jcsd
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