Peskin & Schroeder p. 285, change of variables integration measure

Click For Summary
SUMMARY

The discussion focuses on the integration measure for path integrals following a unitary change of variables in a 4D spacetime lattice with volume L4. The measure is expressed as \mathcal{D}\phi = \prod_i d\phi(x_i), and the field variables are expanded using a Fourier series. The participants clarify that the real and imaginary parts of \phi(k_n) are treated as independent variables to parameterize the complex plane, leading to the measure \mathcal{D}\phi(x)=\prod_{k_n^0>0}dRe\phi(k_n)dIm \phi(k_n). They also discuss the implications of the Jacobian in the change of variables and the normalization factor from the Fourier expansion.

PREREQUISITES
  • Understanding of path integrals in quantum field theory
  • Familiarity with Fourier series and transforms
  • Knowledge of complex variables and their parameterization
  • Basic concepts of Jacobians in variable transformations
NEXT STEPS
  • Study the derivation of path integral measures in quantum field theory
  • Learn about the implications of unitary transformations on integration measures
  • Explore the role of Jacobians in variable transformations in functional integrals
  • Investigate the properties of Fourier transforms in multi-dimensional spaces
USEFUL FOR

Quantum physicists, theoretical physicists, and students studying quantum field theory who seek to deepen their understanding of integration measures and transformations in path integrals.

naele
Messages
199
Reaction score
1
I had a question about about the integration measure for the path integral after a unitary change of variables. First they consider a 4D spacetime lattice with volume L^4. The measure is
<br /> \mathcal{D}\phi = \prod_i d\phi(x_i)<br />

They expand the field variables in a Fourier series \phi(x_i)=\frac{1}{V}\sum_n e^{-ik_n\cdot x_i}\phi(k_n). My questions are as follows:
1) Why do they consider the real and imaginary parts of \phi(k_n) as independent variables?
2) Why do they re-write the measure as
<br /> \mathcal{D}\phi(x)=\prod_{k_n^0&gt;0}dRe\phi(k_n)dIm \phi(k_n)<br />

I've never seen a measure re-written like that, I was wondering what allows them to do so.

There's already a thread about this here but I wasn't comfortable bumping a three year old thread, and the response didn't clear up my confusion.

I appreciate any help.
 
Physics news on Phys.org
Bear in mind that this is a functional measure. That means that what they're trying to do is consider every possible value that \phi(k) could take on at position k_i. Since the Fourier transform requires \phi to be complex, we need a way to parameterize all of the complex plane. We can do this by defining \phi in terms of two real numbers a and b, by setting \phi(k) = a(k) + i b(k), and integrating both of them over the entire real line, leading to an integration measure of da\:db. Writing d Re\phi\:d Im\phi is just another way of saying the same thing.
 
Last edited:
I think I understand that part now, thanks. I do have a problem still with the change of variables from \phi(x_i)\to\phi(k_n). I might be missing something, but there would presumably be a factor of V^{1/n} from the 1/V factor in the Fourier series expansion. And then when I transform from \phi(k_n)\phi^*(k_n)\to Re \phi(k_n)Im \phi(k_n) I get a Jacobian that's not equal to 1.
 
For your first question, I think the answer is that we're just dealing with the measure, not the full integral. So the 1/V will probably show up in the full integral expression.

As for the Jacobian of the measure, that one may have been my fault--I think you need to define the parameterization as \phi = \frac{a + ib}{\sqrt{2}} or something like that in order for the Jacobian to work out correctly.
 
Well the reason I thought there would be a factor is because, unless I'm doing something wrong, I thought the jacobian for \phi(x_i)\to\phi(k_n) is \frac{1}{V}e^{-ik_n\cdot x_i}. Although now that I think about it, they do say that it is a unitary transformation, so presumably the Jacobian would have unit modulus, but I'm having difficulties checking that.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 10 ·
Replies
10
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
10K