Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A A problem about branch cut in contour integral

  1. Dec 5, 2016 #1

    goodphy

    User Avatar
    Gold Member

    Hello.

    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.

    upload_2016-12-6_1-5-52.png

    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

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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

    goodphy

    User Avatar
    Gold Member

    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

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A problem about branch cut in contour integral
  1. Branch cut integration (Replies: 2)

Loading...