Is the Placement of Branch Cuts in ln(-s^2-iε) Physically Significant?

[RedX]

In the expression ln(-s^2-i\epsilon), s^2 and \epsilon are positive (this expression can result from for example a loop diagram where s^2 is a Mandelstam variable). In mathematics, the branch cut of ln() is usually taken to be the negative real axis, so that the value above the negative axis differs from the value below the negative axis by 2\pi i.

But shouldn't the physical result be independent of where you place the branch cut? If you place it on the positive real axis, then ln(-s^2-i\epsilon) has the same value as ln(-s^2+i\epsilon).

 

[Ben Niehoff]

In the expression ln(-s^2-i\epsilon), s^2 and \epsilon are positive (this expression can result from for example a loop diagram where s^2 is a Mandelstam variable). In mathematics, the branch cut of ln() is usually taken to be the negative real axis, so that the value above the negative axis differs from the value below the negative axis by 2\pi i.

But shouldn't the physical result be independent of where you place the branch cut? If you place it on the positive real axis, then ln(-s^2-i\epsilon) has the same value as ln(-s^2+i\epsilon).

The i\varepsilon part is telling you which branch to use. Yes, you can move the branch cut to the real axis, but you still have to choose the correct branch (i.e., do you have \ln e = 1 or \ln e = 1 + 2\pi i, etc.).

 

[RedX]

The i\varepsilon part is telling you which branch to use. Yes, you can move the branch cut to the real axis, but you still have to choose the correct branch (i.e., do you have \ln e = 1 or \ln e = 1 + 2\pi i, etc.).

How are you supposed to know which branch of the function to choose? Physically, does it matter if argument is \theta or \theta+2\pi? Separating a function into branches seems to me a mathematical convenience, to get the function to be single-valued.