# I-epsilon prescription and the definition of mass

1. Jun 9, 2005

### Cinquero

In order to use the path integral formulation for the derivation of the Feynman propagator, we need to introduce a slightly complex mass.

Problem: how do we interpret the circumstance that we cannot interchange integration and taking the epsilon limes?

Is the parameter m in the free Lagrangian the true mass? Or is the physical mass subject to a redefinition that takes the aforementioned problem into account?

2. Jun 9, 2005

### Hans de Vries

This is supposed to be just a mathematical trick to be able to get a decent
limit. Get of the real axis a bit to circumvent the poles. Then afterwards you
get the limit by taking epsilon to zero. I don't think you should take the
epsilon as something which is "added" to either the mass or momentum.
The equivalent procedure for the Klein Gordon equation for instance uses
the mass squared:

$$\frac{1}{p^2-m^2+i\epsilon}$$

Regards, Hans

Last edited: Jun 9, 2005
3. Jun 9, 2005

### Cinquero

I'd rather say it is an unmathematical trick because we make a divergent quantity convergent... effectively resulting in something we didn't have in the first place. To make "my" problem more explicit: couldn't we just use other prescriptions and end up with different results? What is the essential point of that special prescription beyond the statement of making the integral convergent? Isn't it totally arbitrarily chosen and therefore leads to (in principle) arbitrary results which perhaps just by incident (or because of some deeper and hidden truth) coincide with measurements?

4. Jun 9, 2005

### ohwilleke

Lots of singularities which are infinite at a point have a finite volume in 3D or a path integral around the boundary of the singularity that is convergent.

If there exists a limit a epsilon approach zero then the singularity is well behaved, and there is only one result upon which the answer will converge regardless of the particular mathematical trick you use to get there. I recall a theorem in complex analysis to that effect, but can't remember for the life of me what it is called and don't have the book at hand in my office.

5. Jun 9, 2005

### Hans de Vries

No, It's all about "picking up the poles" You might want to review the Residue
Theory here:

http://mathworld.wolfram.com/ResidueTheorem.html

(See formula 8 and 9)

There some contour intergral images in Peskin and Schroeder (page 31)

Regards, Hans

Last edited: Jun 9, 2005
6. Jun 10, 2005

### Haelfix

You might be correct in worrying about this, but in this case its -ok-. If you look at canonical quantization, the reason the i epsilon shows up in the path integral becomes well understood here.

The more problematic, slightly related weirdness is Wick rotation in general field theories. This is -not- always ok, despite the average physicists best attempts to convince us that it is.

Another prescription that is often used, that is really -not- ok is Fadeev Poppov ghosts. Its more or less mathematically forbidden!

But as is often the case in field theory, the derivation of a result is much worse behaved than the actual result itself.

7. Jun 10, 2005

### Cinquero

Well, I know the theory of residues. But that doesn't solve the problem that we calculate a divergent integral. As far as I have understood, we can use residues to calculate integrals which are otherwise hard to solve. It is a nice method to calculate convergent integrals. Nothing more. If it would make a divergent integral convergent, the theory of residues would be a very bad theory...

Next, if you insist on integrating over more dimensions, then why do we integrate over fewer dimensions in the first place? Does not make sense to me.

8. Jun 10, 2005

### dextercioby

In what respect is there something "fishy" regarding the Wick rotation and Faddeev-Popov ghosts...?

Daniel.

9. Jun 10, 2005

### juvenal

I don't understand what the problem is with taking an epsilon to 0 limit. Have you taken a class on complex variables before?

http://mathworld.wolfram.com/Contour.html

Look at the contour on the right. That's for an integral from 0 to infinity with a pole at 0. See the little circle in the center? The radius of that circle is epsilon. At the end, one takes a limit of epsilon -> 0.

Last edited: Jun 10, 2005
10. Jun 10, 2005

### Cinquero

I do not doubt that this limit exists if taking the limit _after_ integration. But that limit is just not the same as taking it inside the integral. So in practice we are calculating something very different than we wanted to calculate in the first place.

I guess this is similar to the fact that we do not use higher than second-order time-derivatives in the Lagrangian because we don't know how to handle them: because the free field Lagrangian gives us a divergent integral, we just look for a somewhat modified and effectively different "Lagrangian" with which we can actually do the integration.

11. Jun 10, 2005

### juvenal

I haven't done this stuff in a while, but I'm not sure why you're saying that you take the limit inside the integral. p. 31 of Peskin and Schroeder seems to imply that the integral is taken first before the limit.

12. Jun 10, 2005

### dextercioby

We know how to handle more than 2 time derivatives in the lagrangian and,guess what,even the Hamiltonian formalism is possible...

Daniel.

13. Jun 10, 2005

### Cinquero

Read Zee, "QFT in a Nutshell".

14. Jun 10, 2005

### dextercioby

I have,renormalizability is an entirely different matter.You can build clasical theories with as many time derivatives you want.Incidentally,Weyl gravity is renormalizable,but it doesn't have a low energy limit consistent with the experiment.

If you can't handle it,stay out of it.

BTW,this is going the wrong way,i.e.off-topic.

Daniel.

Last edited: Jun 10, 2005
15. Jun 10, 2005

### Haelfix

Yes if you look at P&S page 296-297 for instance.. Exp 9.56 --> 9.57 doesn't make sense for the Fadeev Popov method.

Or rather it can only make sense if the series converges, you simply can't pull infinities outside of an integral (shaky measure dependant territory) and then divide out essentially infinite quantities and expect no one to wave their hands.

Moreover field theories are more or less believed to not converge, but rather to be asymptotic. So this prescription is not just shaky, its explicitly mathematically forbiden.

(and then a miracle occurs) and we get the right result =)

For the myriad problems with wick rotation and the Euclidean path integral, just google for a few minutes. Wick rotation can lead to inequivalent quantizations of the same theory, and the analytic continuation need not be well defined (it is in a few cases)

16. Jun 10, 2005

### dextercioby

I see.I'm not familiar and don't want to be familiar with the F-P "trick".I'm accustomed with the rigurous BRST approach,either Hamiltonian (original),or Batalin-Vilkoviski( Lagrangian).

Daniel.