Where does the 'i' come from in QFT path integral?

[bhobba]

What do you mean by "legitimate"? As I said, there is no mathematical claim being made. It's a physical claim that a particular physically measurable quantity is equal to a particular mathematical expression.

There is a mathematical claim being made.

When you take beta as real that in effect means you are taking t as pure imaginary.

You can only do that if the function has been extended to the complex plane ie t is complex.

Thanks
Bill

 

[stevendaryl]

There is a mathematical claim being made.

The mathematical claims that are being made (for example):

1. When [itex]\beta[/itex] and [itex]u[/itex] are real, the integral [itex]\int e^{k u - \beta \frac{\hbar k^2}{2m}} dk[/itex] converges to [itex]\sqrt{\frac{2 \pi m}{\beta \hbar}} e^{\frac{-m u^2}{2 \beta \hbar}}[/itex]
2. The function [itex]F(u, \beta) = \sqrt{\frac{2 \pi m}{\beta \hbar}} e^{\frac{-m u^2}{2 \beta \hbar}}[/itex] can be analytically extended to the case where [itex]u[/itex] and [itex]\beta[/itex] are pure imaginary.

Those are mathematical claims, but they are just true. They are provable. And those are the only mathematical claims being made. (In this case, anyway).

Then there is a physical claim being made--a hypothesis--which is that the amplitude for a spin-zero massive nonrelativistic particle to travel from the point [itex](0,0)[/itex] to the point [itex](x,t)[/itex] is given by [itex]F(i x, -i \frac{t}{\hbar})[/itex]

Now, in cases that are more complicated than a free particle, the analogs of 1. and 2. are not provable, so I would agree that they are additional, mathematical assumptions. But they aren't the assumptions that you say are being made. Neither of those assumptions involves assuming that you can replace a real integration parameter by an imaginary one.