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

  1. Oct 31, 2012 #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 [itex]L^4[/itex]. The measure is
    \mathcal{D}\phi = \prod_i d\phi(x_i)

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

    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.
  2. jcsd
  3. Nov 2, 2012 #2
    Bear in mind that this is a functional measure. That means that what they're trying to do is consider every possible value that [itex]\phi(k)[/itex] could take on at position [itex]k_i[/itex]. Since the Fourier transform requires [itex]\phi[/itex] to be complex, we need a way to parameterize all of the complex plane. We can do this by defining [itex]\phi[/itex] in terms of two real numbers [itex]a[/itex] and [itex]b[/itex], by setting [itex]\phi(k) = a(k) + i b(k)[/itex], and integrating both of them over the entire real line, leading to an integration measure of [itex]da\:db[/itex]. Writing [itex]d Re\phi\:d Im\phi[/itex] is just another way of saying the same thing.
    Last edited: Nov 2, 2012
  4. Nov 3, 2012 #3
    I think I understand that part now, thanks. I do have a problem still with the change of variables from [itex]\phi(x_i)\to\phi(k_n)[/itex]. I might be missing something, but there would presumably be a factor of [itex]V^{1/n}[/itex] from the 1/V factor in the Fourier series expansion. And then when I transform from [itex]\phi(k_n)\phi^*(k_n)\to Re \phi(k_n)Im \phi(k_n)[/itex] I get a Jacobian that's not equal to 1.
  5. Nov 5, 2012 #4
    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 [itex]\phi = \frac{a + ib}{\sqrt{2}}[/itex] or something like that in order for the Jacobian to work out correctly.
  6. Nov 7, 2012 #5
    Well the reason I thought there would be a factor is because, unless I'm doing something wrong, I thought the jacobian for [itex]\phi(x_i)\to\phi(k_n)[/itex] is [itex]\frac{1}{V}e^{-ik_n\cdot x_i}[/itex]. 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.
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