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February 21, 2007

Quantization and Cohomology (Week 15)

Posted by John Baez

This week in our course on Quantization and Cohomology, we finished off the path-integral quantization of the free particle:

  • Week 15 (Feb. 20) - The free particle on a line (part 2). Showing the path-integral approach agrees with the Hamiltonian approach. Fourier transforms and Gaussian integrals.

Last week’s notes are here; next week’s notes are here.

Avoiding an unpleasant direct proof that the stationary phase approximation is exact for the free particle on a line, we instead took an indirect route which covered all the basic facts about Fourier transforms and Gaussians. The Wick rotation was cleverly disguised as a little maneuver to compute the integral

e iu 2/2du\int_{-\infty}^\infty e^{-i u^2/2} d u

which, you will note, fails to converge absolutely!

Posted at February 21, 2007 1:23 AM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1172

5 Comments & 2 Trackbacks

Re: Quantization and Cohomology (Week 15)

I am currently teaching a similar subject and thus are naturally very interested in how you introduce these subjects.

However I was a bit disappointed when I realised you were cheating: Your analytic continuation looks very clever but of course a priori there is no reason why you can interchange taking the integral and doing the limit k->i. You could have argued similarly and taken the limit k->-i and obtained a mess. There is another difference: As it is unitary, you can compute the quantum mechnical propagator for both positive and negative t but the heat kernel (the euclidean version) only exists for t>0. You could see this as a manifestation of QM being time reversal invariant but thermodynamics being not (because of the second law).

Posted by: Robert on February 21, 2007 7:10 AM | Permalink | Reply to this

Re: Quantization and Cohomology (Week 15)

I certainly don’t feel I found the best way to explain this stuff. It’s the first time I’ve tried to explain path integrals.

But, I said a lot of stuff that didn’t find its way into the notes, since I didn’t write it on the board.

Among other things, I said that path integrals are a bit mysterious — they really go beyond our standard textbook ideas on integration theory, which is why most quantum field theories haven’t been made rigorous. I said that in this simple problem, we see just a tiny taste of this: none of the integrals I wrote down converge absolutely! So the usual Lebesgue theory of integration does not apply. I explained that this is why we get, at the very end, an integral

e iu 2/2du \int_{-\infty}^\infty e^{-i u^2 / 2} d u

that does not converge absolutely. Like all real-time path integrals (as opposed to Wick-rotated path integrals), it involves an ‘infinite cancellation of phases’.

I mentioned that we could compute this integral using a step-function cutoff:

lim c+ c ce iu 2/2du \lim_{c \to +\infty} \int_{-c}^c e^{-i u^2 / 2} d u

but said that it was easier to compute it using a Gaussian cutoff. That amounts to computing

e au 2/2du \int_{-\infty}^\infty e^{- a u^2 / 2} d u

for aa with Re(a)>0Re(a) > 0, and then taking the limit as aia \to i.

Your analytic continuation looks very clever but of course a priori there is no reason why you can interchange taking the integral and doing the limit kik \to i.

Just to be painfully clear: the issue of interchanging the limit and the integral works as follows. If we first do the integral and then the limit we get

lim ai e au 2/2du=2π/i lim_{a \to i} \int_{-\infty}^\infty e^{- a u^2 / 2} d u = \sqrt{2 \pi / i}

where of course I mean the limit as aa approaches ii while keeping Re(a)>0Re(a) &gt; 0, since the integral diverges wildly for Re(a)<0Re(a) &lt; 0.

By the way: we can even decide which branch of the square root function is the right one!

If on the other hand we try to do the limit first, we get

lim aie au 2/2du= e iu 2/2du \int_{-\infty}^\infty \lim_{a \to i} e^{- a u^2 / 2} d u = \int_{-\infty}^\infty e^{-i u^2/2} d u

and now we’re trying to integrate a function that’s not Lebesgue integrable. In other words, this integral is not absolutely convergent. So, we need some other method of defining the integral. Any sensible method of defining the integral will give the answer

e iu 2/2du=2π/i \int_{-\infty}^\infty e^{-i u^2 / 2} d u = \sqrt{2 \pi / i}

but Lebesgue integration is not one of these methods.

This issue afflicts all the path integrals in this week’s notes.

Posted by: John Baez on February 21, 2007 9:57 PM | Permalink | Reply to this

Re: Quantization and Cohomology (Week 15)

I agree that the oscillatory integral is not absolutely convergent. However (without remembering any good argument for it) I think it is still Lebesque integrable. Thus this one dimensional integral as it is (without giving a real part to a) can be made sense of but this is not the origin of imaginary path integrals not having a mathematical founding (as opposed to Wiener integrals) but again, I have no argument or reference at hand.

Posted by: Robert on February 23, 2007 7:01 AM | Permalink | Reply to this

Re: Quantization and Cohomology (Week 15)

Robert wrote:

I agree that the oscillatory integral is not absolutely convergent. However (without remembering any good argument for it) I think it is still Lebesque integrable.

It’s not.

In Lebesgue integration theory, first we define the integral for nonnegative measurable functions f:X[0,)f : X \to [0,\infty). This integral always exists, but it can equal ++\infty. We say that a nonnegative measurable function ff is integrable if f<\int f &lt; \infty.

Then, for a general function f:Xf: X \to \mathbb{R}, we separate it into its positive and negative parts:

f=f +f f = f_+ - f_-

with f +,f 0f_+, f_- \ge 0. We say ff is integrable if both f +f_+ and f f_- are integrable, and then we define

f=f +f \int f = \int f_+ - \int f_-

Similarly, we say a complex-valued function f:Xf : X \to \mathbb{C} is integrable if Re(f)Re(f) and Im(f)Im(f) are integrable, and in this case we again define the integral in the obvious way:

f=Re(f)+iIm(f) \int f = \int Re(f) + i \int Im(f)

So, any function ff for which |f||f| is not integrable cannot itself be integrable, in the Lebesgue theory!

The point is that the integral we’re trying to do:

e iu 2/2du \int_\mathbb{R} e^{-i u^2/2} d u

has a sensible value if we ‘start integrating it near the middle of the real line and work our way out’, e.g by doing

lim c+ c ce iu 2/2du, \lim_{c \to + \infty} \int_{-c}^c e^{-i u^2/2} d u,

or using a Gaussian cutoff, or any sort of sensible cutoff function that ‘approaches 1 starting from the middle of the real line and working its way out’.

But, this notion relies on the domain \mathbb{R} having structure beyond a mere measure space! It’s just like doing this sort of sum

11/2+1/31/4+1 - 1/2 + 1/3 - 1/4 + \cdots

If you sum from left to right, you get a well-defined answer, ln(2)ln(2). But if you allow arbitrary rearrangements, you can get any answer you want. The Lebesgue theory of integration applies to functions f:Xf: X \to \mathbb{R} defined on a mere measure space XX, so its answers, when well-defined, must be invariant under arbitrary ‘rearrangements’: measure-preserving transformations of XX. So, any sort of integral that fails to converge absolutely is not Lebesgue integrable.

So, we need some other concept of integration to tackle even the simplest real-time path integrals! The whole point of these integrals is ‘infinite cancellation of phases’, and Lebesgue integration is blind to this.

Posted by: John Baez on February 23, 2007 6:34 PM | Permalink | Reply to this

Re: Quantization and Cohomology (Week 15)

Robert suspected:

I agree that the oscillatory integral is not absolutely convergent. However (without remembering any good argument for it) I think it is still Lebes[g]ue integrable.

Lebesgue integration is only about absolutely convergent integrals. (In fact, a sequence n: N → an: R defines a Lebesgue-integrable function x: R+ → a[x], where [x] is the floor of x, if and only if it defines an absolutely convergent series.)

You may be thinking about (or want to look into) the Henstock integral (also known by many other names). Like the improper Riemann integral, it deals with integration over infinite intervals by working from the inside out; like the Lebesgue integral, it can handle all bounded measurable functions. (In fact, a function is Henstock integrable if it’s either Riemann integrable or Lebesgue integrable.)

A function is Lebesgue integrable if and only if it absolutely Henstock integrable. Thus Lebesgue : Henstock :: asbolutely convergent : convergent.

Of course, even the Henstock integral can’t handle series that require tricks like Abel summation. Is there a theory of Abel–Henstock integration? I don’t know.

Posted by: Toby Bartels on February 23, 2007 7:33 PM | Permalink | Reply to this
Read the post Quantization and Cohomology (Week 16)
Weblog: The n-Category Café
Excerpt: More examples of path-integral quantization: the particle in a potential.
Tracked: February 27, 2007 11:34 PM
Read the post QFT of Charged n-Particle: The Canonical 1-Particle
Weblog: The n-Category Café
Excerpt: On the category of paths whose canonical Leinster measure reproduces the path integral measure appearing in the quantization of the charged particle.
Tracked: March 19, 2007 9:03 PM

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