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

[stevendaryl]

The fact you can do that change of variables implies analytic continuation anyway. Its not legitimate to substitute a complex value into an equation of real variables although its often done

Suppose that you're in the following situation (well, maybe not you, but somebody else):

• You are trying to compute some physically meaningful quantity F(t)
  
• Your theoretical analysis gives you a mathematical expression, \sum_n e^{-i \frac{E_n t}{\hbar}}
• This expression does not converge for t real.
• Therefore, it cannot literally be the case that F(t) = \sum_n e^{-i \frac{E_n t}{\hbar}}, since the left-hand side is something meaningful, and the right-hand side is nonsense.
  
• However, the similar expression \sum_n e^{-\beta E_n} does converge, when \beta is real and positive, to a function \tilde{F}(\beta).
• The function \tilde{F}(\beta) can be analytically continued to the region where \beta is purely imaginary.

So you hypothesize that the F(t) = \tilde{F}(\frac{i t}{\hbar})

I'm not sure which step you're saying is not legitimate. There is no claim that the original sum converges to \tilde{F}(\frac{it}{\hbar}). It certainly doesn't. The claim is that the physically meaningfully value F(t) is equal to \tilde{F}(\frac{it}{\hbar}). That's a hypothesis, not a conclusion, so the notion of "legitimate" versus "illegitimate" doesn't come up. The only issue is whether that way of computing F(t) agrees with observation.

 

[atyy]

I looked at the Hida distributions approach that bhobba linked to, and they are closer to what he is saying. I think stevendaryl is thinking more of the Wick rotation which changes the problem from the hand-wavy derivation of the real time path integral to a statistical mechanics problem via imaginary time, which is invalid from the point of view of the hand-wavy derivation, yet can be shown to rigourously define relativistic quantum field theory under some conditions.

 

[bhobba]

I'm not sure which step you're saying is not legitimate.

It not legitimate to substitute a complex variable for a real t unless the equation is extended by analytic continuation.

But that is beside the point. Both Wiener processes and path integrals have convergence issues that are solved by Hida Distributions.

Thanks
Bill

 

[stevendaryl]

I looked at the Hida distributions approach that bhobba linked to, and they are closer to what he is saying. I think stevendaryl is thinking more of the Wick rotation which changes the problem from the hand-wavy derivation of the real time path integral to a statistical mechanics problem via imaginary time, which is invalid from the point of view of the hand-wavy derivation, yet can be shown to rigourously define relativistic quantum field theory under some conditions.

Yes, I'm talking about path integrals. They don't literally converge to anything. The way that people get meaningful results out of them is by analytically continuing similar expressions that do converge.

Even for the simplest case of the free one-dimensional, nonrelativistic, massive spin-zero particle, evaluating the path integrals involve expressions such as:

\int e^{(i k x - i t \frac{\hbar k^2}{2m})} dk

That expression doesn't converge. However, the related expression

\int e^{(k u - \beta \frac{\hbar^2 k^2}{2m})} dk

does converge, to F(u, \beta) = \frac{2m \pi}{\beta \hbar^2} e^{- \frac{mu}{2\beta\hbar^2}} (if I did that right). So the assumption is that the free propagator is actually obtained by replacing u by ix and \beta by i \frac{t}{\hbar} in F(u,\beta).

 

[stevendaryl]

It not legitimate to substitute a complex variable for a real t unless the equation is extended by analytic continuation.

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.

 

[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 \beta and u are real, the integral \int e^{k u - \beta \frac{\hbar k^2}{2m}} dk converges to \sqrt{\frac{2 \pi m}{\beta \hbar}} e^{\frac{-m u^2}{2 \beta \hbar}}
2. The function F(u, \beta) = \sqrt{\frac{2 \pi m}{\beta \hbar}} e^{\frac{-m u^2}{2 \beta \hbar}} can be analytically extended to the case where u and \beta 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 (0,0) to the point (x,t) is given by F(i x, -i \frac{t}{\hbar})

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.