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A A problem about branch cut in contour integral

  1. Dec 5, 2016 #1

    I have a difficulty to understand the branch cut introduced to solve this integral.

    [tex]\int_{ - \infty }^\infty {dp\left[ {p{e^{ip\left| x \right|}}{e^{ - it\sqrt {{p^2} + {m^2}} }}} \right]} [/tex]

    here p is a magnitude of the 3-dimensional momentum of a particle, x and t are space and time coordinates respectively and m is the particle mass. This integral is showed up in the quantum field theory textbook.

    The textbook tries to solve this equation by converting it a contour integral in the complex plane. So, p is now replaced with a complex variable z.

    [tex]\int_{ - \infty }^\infty {dz\left[ {z{e^{iz\left| x \right|}}{e^{ - it\sqrt {{z^2} + {m^2}} }}} \right]} [/tex]

    The textbook said [tex]{\sqrt {{z^2} + {m^2}} }[/tex] must be restricted to a single branch and the branch points are [tex]z = \pm im[/tex] and the branch cuts are convenienly set as shown in the attached image.


    Actually, I don't understand why branch cuts are set like this. As far as I know, the branch cut is the line that we never cross with any mathematical operation in order to make a single-valued function. So I think these branch cuts are set in order to make [tex]{{e^{ - it\sqrt {{z^2} + {m^2}} }}}[/tex] the single-valued function but I don't know how.

    Could you please tell me why the branch cuts are set like this? How do these branch cuts make the single-valued function?
    Last edited: Dec 5, 2016
  2. jcsd
  3. Dec 5, 2016 #2


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    To make the square root continuous, you can remove all values where its argument is zero or real and negative. That's exactly what is done here.
    The cuts don't make the function single-valued, but they make the function continuous.
  4. Dec 10, 2016 #3
    Hello. I'm sorry to reply you lately as I've studied complex analysis further to understand what you said.

    I'm afraid that I'm still not understanding your point. It looks you said that a domain of w(z) = z1/2 is restricted to make w(z) continuous.

    Could you please give me details of how the domain of w(z) is restricted so it becomes continuous?
  5. Dec 10, 2016 #4


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    The square root is continuous if the argument doesn't get 0 or negative real values.

    The argument of your square root is ##z^2+m^2##. m as particle mass is real and positive, therefore ##z^2+m^2=0## has the solutions ##z=\pm i m## and the only way ##z^2+m^2## can be real and negative is ##z=ic## with real ##|c|>m##. By avoiding all those values for z, the argument of the square root is never a negative real number, and it is never zero. Therefore, you can treat the square root as continous function.
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