# A Feeling uncomfortable about Wick-rotations

1. Nov 17, 2017

### haushofer

Dear all,

I'm writing a (semi)popular science book about fundamental physics (in Dutch). I want to include a section about Wick rotations in the path integral (in the context of Hawkings no-boundary proposal, but let's keep quantum gravity out for the moment and stick to ordinary QFT). I've read many blogs and papers about a justification (like Lubos Motl's blog and a dozen of QFT books like "Gifted amateur", Peskin&Schroeder, Srednicki, Zee, etc) and I just can't see how I would explain it conceptually to a laymen (or graduate student, for all that matters). besides a "shut up and calculate" kind of answer.

I'd like to compare Wick rotation to the renormalization hypothesis, which says that divergencies in Feynmandiagrams are encountered due to the use of naked parameters instead of physical ones. Is there any similar "hypothesis" about why we can use Wick rotations? Up to now I have the impression that the reasoning is like "well, the path integral is ill-defined anyway so let's try an analytical continuation, because the correspondence between QFT and thermodynamics in equilibrium looks so nice. This analytic continuation is unique, so if it gives a sensible answer we can compare with experiments and only wonder about a possible physical deaper meaning of this trick."

I guess something similar goes for zeta-function regularization in e.g. the Casimir effect or string theory, in which one analytically continues the zeta function to obtain a finite answer to the sum of all integers.

I also see that for the propagator a Wick rotation is equivalent to the usual $i\epsilon$-prescription, which regulates the Green's function by imposing an infinitesimal damping and as a bonus includes boundary conditions. But this does not automatically implies full equivalence with the Wick rotation of the path integral, right?

I hope my question is clear. If someone could make me more comfortable with this Wick-rotation business and give me a conceptual explanation of why we are allowed to do it/a decent philosophy behind it, I'd be happy. And if someone can supplement it with simple algebraic examples, I'd be even more happy.

2. Nov 17, 2017

### vanhees71

Hm, I think it's very difficult to explain this quite formal math to lay people, and I also don't see the merit of it. Physically it's of course related to time ordering and analytic continuation from Euclidean (imaginary-time) to Minkowski space.

3. Nov 17, 2017

### haushofer

We can "repair" the divergencies of Feynman diagrams by the philosophy that we are just too dumb to start out with physical parameters. How would you phrase the philosophy behind "repairing the divergence of the path integral by wick rotating"?

I see that it works. And I'm not intended to explain how it works mathematically. I want to explain the philosophy behind it, the same way as I understand renormalization. Wick rotation seems to me now like renormalization in the pre-Wilsonian era.

Last edited: Nov 17, 2017
4. Nov 17, 2017

### MathematicalPhysicist

Well, QFT looks like an ill-defined mathematically theory;

For example instead of saying that for the Casimir effect we don't use the diverging sum itself, we are looking at a resummation method for this sum and not the sum itself which diverges.

For the Wick trick, it's just the transformation $t=i\tau$.

Nothing fancy with regards to maths.

I wouldn't bother writing a popular book, only technical books... :-D
If I had the time, I planned on writing in Latex solution manuals to some Logic books and then abandon it to other matters.

5. Nov 17, 2017

### rubi

The Wick-rotation can be made rigorous by the Osterwalder-Schrader reconstruction theorem. It says that given an Euclidean (Wick-rotated) quantum field theory, i.e. a certain path integral measure that satisfies certain axioms, one can reconstruct a perfectly Minkowskian Wightman QFT from it. Hence, the Wick-rotated path integral is just a means to write down a Minkowskian QFT.

6. Nov 18, 2017

### haushofer

So in writing down the usual path integral in minkowski spacetime, noting that it diverges and analytically continuing by wick rotating were just too dumb to write down the Euclidean theory in the first place?

With renormalization I see that the naked parameters are just mathematical devices which we don't measure. But what does it mean that Euclidean theories are better behaved than Minkowski-ones?

And how are we assured that e.g. interference is kept in this analytical continuation? It sure seems to disappear.

Is there any conceptual way to understand this without having to dig myself into all the formalities of axiomatic QFT?

7. Nov 18, 2017

### haushofer

Well, I want to understand what I'm doing and why. Not just noting that it works. Like I said, renormalization can be understood in a Wilsonian way with effective field theories. I'm suprised that apparently such a conceptual understanding of Wick rotations lacks.

And what suprises me even more is that so many textbooks don't spend any words on this, as far as i can see.

8. Nov 18, 2017

### rubi

It's not related to renormalization. It's the same problem that we have with standard integrals. For example, the Gaussian integral $\int_{-\infty}^{\infty} e^{-\alpha x^2}\,\mathrm d x=\sqrt{\frac\pi\alpha}$ is finite, but the integral $\int_{-\infty}^{\infty} e^{-i\alpha x^2}\,\mathrm d x=\int_{-\infty}^{\infty} \cos(\alpha x^2)\,\mathrm d x-i \int_{-\infty}^{\infty} \sin(\alpha x^2)\,\mathrm d x$ doesn't exist, because the integrand is oscillating rather than approaching $0$ at $\infty$. For a Lebesgue integral to exist, the absolute value of the function must be integrable, so there is no chance for the negative parts to cancel out the positive parts (no conditional convergence). However, you can assign a value to the integral by chosing a contour in the complex plane and taking an appropriate limit (see Fresnel integral). You get the result $\sqrt{\frac\pi{i\alpha}}$, which happens to be the analytically continued Gaussian integral. However, you wouldn't call this the Lebesgue integral of the function, because it is not strictly a Lebesgue integral, but rather a limit of Lebesgue integrals.

The same problems occurs in path integrals. Roughly speaking, the Minkowskian integrands are just not well-behaved enough for the integral to exist, so it doesn't make sense to speak of a path integral. However, we can assign a value to it by performing the Euclideanized integral and then analytically continuing. It's just a mathematical trick that lets us escape the limitations of standard integration theory and the justification for the trick is the Osterwalder-Schrader reconstruction theorem.

Edit: In other words: The function $e^{-i\alpha x^2}$ is not Lebesgue integrable, but we can formally write down $\int e^{-i\alpha x^2}=\sqrt{\frac\pi{i\alpha}}$ and pretend that is is the integral of $e^{-i\alpha x^2}$, when really it is not. In the same sense, the function $e^{i S[\Psi]}$ is not path integrable, but we can formally write down $\int\mathcal D\Psi\, e^{i S[\Psi]}$ and assign the analytical continuation of the integral $\int\mathcal D\Psi\, e^{-S_E[\Psi]}$ to it, which hopefully exists. $\int\mathcal D\Psi\, e^{i S[\Psi]}$ isn't a path integral either, but we can pretend it is.

Last edited: Nov 18, 2017
9. Nov 18, 2017

### haushofer

But how can we justify the trick if our expression for the path integral is ill-defined in the first place? What determines the "right expression"?

10. Nov 18, 2017

### rubi

Well, $\int_{-\infty}^{\infty} e^{-i\alpha x^2}\,\mathrm d x$ is also ill-defined in the first place. We just assign the value $\sqrt{\frac\pi{i\alpha}}$ to it, because we believe that this would be a reasonable assignment to the purely formal expression $\int_{-\infty}^{\infty} e^{-i\alpha x^2}\,\mathrm d x$. It's not justified by integration theory. In the same sense, the expression $\int\mathcal D\Psi \,e^{i S[\Psi]}$ is ill-defined, but $\int\mathcal D\Psi \,e^{-S_E[\Psi]}$ isn't. Euclidean path integrals are in principle mathematically well-defined objects. The OS theorem tells us that if we succeed in defining an Euclidean path integral that satisfies certain axioms, then we get a Minkowskian QFT for free. Whether this Minkowskian QFT matches experimental observations is of course a matter of further research that needs to be decided on a case-by-case basis.

11. Nov 18, 2017

### haushofer

Of course, observations ultimately decide :P

I understand that wick-rotation is regularization, and differs from renormalization. But can I summarize all this by "we simply don't understand QFT mathematically well enough, so we cross our fingers and hope that an analytical continuation of time gives us answers coinciding with experiments?"

As I said, I have the feeling that the philosophy behind renormalization is much more solide and well understood than Wick rotations. Am I right?

12. Nov 18, 2017

### rubi

No, all of it is mathematically well-understood at the highest level of rigour. It's just that there is a one-to-one correspondence between Minkowskian QFT's and Euclidean QFT's and the relation is given by Wick-rotation. It's not a regularization, just an equivalent formulation. Path integrals are not a necessity for quantum theories. One can perfectly do quantum theory without path integrals, but if one wants to have them, one must formulate them on the Euclidean side, because of convergence issues on the Minkowskian side. So if we want to include path integrals in our toolset, we need to deal with Euclidean QFT's and that's the only purpose of Euclidean QFT's in the first place. If we are willing to do without path integrals, we can also dispense with Euclidean QFT's and Wick-rotation.

It's just about extending our toolset to include path integrals. We can either define the Minkowskian QFT directly (without path integrals) or we define a Euclidean QFT using by a path integral and analytically continue it to a Minkowskian QFT (and this procedure is rigorously underpinned by the OS theorem). In both cases, we obtain a bona-fide Minkowskian QFT. It's just that the path integral approach might be easier in certain cases, because one needn't deal with as many functional analytic difficulties, even though we need to perform the additional step of analytical continuation.

The hope comes in when we have to decide which QFT (Euclidean of Minkowskian) to write down in the first place. Nobody tells us what the correct model is, neither in the Euclidean formulation nor in the Minkowskian formulation, so we just have to guess a (Euclidean or Minkowskian) QFT and compare its predictions to experiments.

I would say it's the other way around. Wick-rotation is supported by rigorous mathematics, while renormalization is still an ad-hoc procedure with lots of ambiguities and no firm mathematical basis. Different regularization schemes may lead to different physics. Physicists claim they don't, but there is no mathematical proof, not even a sloppy one.

13. Nov 18, 2017

### haushofer

This really helps, especially your comment that path integrals are just one of the tools we use for QFT's. I didn't appreciate that fact untill now.

I have to think more carefully about how the rigor of renormalization compares to Wick rotations. Many thanks, your comments have been very helpful.

14. Nov 18, 2017

### haushofer

May I summarize my conclusion as follows?

"We can define QFT's either by a path integral or by canonical quantization. The path integral is a nice intuïtive picture wich however lacks mathematical rigour: it doesn't convergence. One can however show that a Wick rotation analytically continuates the path integral, making it convergence, and that if we rotate back, the QFT obtained coincides with the one we would obtain by canonical quantization. This last result is ensured by the Osterwalder-Schrader theorem."

15. Nov 18, 2017

### rubi

That's fine, except that I would replace "canonical quantization" by "Hilbert space quantization" to be more general and include interacting QFT's as well. Also, you don't really Wick-rotate the non-existent Lorentzian path integral, but rather you rotate the Lorentzian theory into a Euclidean one, which then happens to admit a path integral formulation.

16. Nov 18, 2017

### haushofer

Ok. I think most of my discomfort has been taken away. You've been very helpful. Thanks again!

I'm still puzzled why many textbooks on QFT (the ones I've (partly) read) don't really explain all this stuff at least at the conceptual level. But then again, many of these textbooks don't mention Wightman-axioms or Osterwalder-Schrader reconstruction. This would make a great Insights-article on this forum; it would take away a lot of magic, which I'm sure a lot of students are experiencing. I'm not intending to understand axiomatic QFT, but I just want to have some assurance that the math one applies to make expressions well-defined has some justification.

17. Nov 18, 2017

### rubi

You're welcome!

Unfortunately, the textbook situation in QFT is very bad. Many textbooks are more or less just recipes for calculating Feynman diagrams. They don't go into depth when it comes to the conceptual framework.

I probably don't have enough time to write an Insight article at the moment. Especially since it would have to be a longer series, but let's see.

The Wightman axioms are just a list of features that we expect from a reasonable QFT. They just say that a QFT is a quantum theory whose observables are field operators that transform like fields under Poincare transformations. Furthermore, there should be a Poincare invariant vacuum state such that all other states arise as excitations of the vacuum. Also, field operators localized in spacelike separated regions should commute. All of these axioms seem very reasonable from a physical point of view. The rest of rigorous QFT is just proving theorem from these axioms.

18. Nov 18, 2017

### haushofer

Ha, I didn't mean you for the writing, but if you find the time sometime, I'll be very happy to read it :D The probleme of time is familiar, though ;)

Urs is also writing insights about qft right now, I'll give them a try soon.

19. Dec 9, 2017

### Iliody

There is a way of defining as a distribution the oscillatory not-wick rotated path-integral, restricting the test functions that you use as observables (Sergio Albeverio's work on infinite-dimensional oscillatory integrals, you can search for the references to his papers and books in scholarpedia). Explaining with 'there are physical arguments that tells you that there is one and only one natural extension to the plane of the function in time that you would integrate, and that it's known that the integral in the real line can be thinked as the limit of the integrals on half circles, and that integral that you can deform the contour and its integral doesn't change if it's well defined, and because it's easier to make that kind of integral than the other physicists integrate in the rotated plane, and that is wick rotating' (we can see if this is useful if we see how to make every affirmation that I used more pedagical?)

Sorry for my poor redaction.