Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Functional integration

  1. Jan 8, 2009 #1
    In free-field theory, the functional integral

    [itex]\int \mathcal{D}\varphi \exp\left(i \frac{1}{2} \int d^4 x (\partial_\mu \varphi \partial^\mu \varphi - m^2 \varphi^2)\right)[/itex]

    can be done exactly (see e.g., Peskin and Schroeder p. 285).

    I'm tyring to understand the step in their derivation where they change integration variables from the field [itex]d\varphi(x)[/itex], to the real and imaginary parts [itex]d\Re[\varphi(x)],d\Im[\varphi(x)][/itex]. They claim that since the transformation is unitary, they have

    [itex]\prod_i d\varphi(x_i) = \prod_i d\Re[\varphi(x_i)]d\Im[\varphi(x_i)][/itex].

    I don't understand this claim. Suppose the unitary xfm relating [itex]x_i[/itex] to [itex]X_i[/itex] is [itex]U[/itex]. Then inEinstein notation,

    [itex]dx_i = U_{ij} dX_j [/itex].


    [itex]\prod_i dx_i = \prod_i U_{ij} dX_j = (U_{1i}U_{2j}U_{3k}\cdots)(dX_i dX_j dX_k \cdots)[/itex].

    Thus P&S's claim amounts to the assertion that

    [itex]\prod_{n=1} U_{n ,i_n} = \prod_{n=1}\delta_{n, i_n}[/itex].

    I don't understand this?

    Any help would be appreciated.
  2. jcsd
  3. Jan 9, 2009 #2


    User Avatar
    Science Advisor

    First of all, it's just a definition. The measure for ordinary integration over a complex variable [itex]z=x+iy[/itex] is defined to be [itex]dx\,dy[/itex].

    More generally, a change of variable involves the determinant of the jacobian matrix of the transformation.
  4. Jan 10, 2009 #3
    That's interesting, I desperately need to take a course in complex analysis.

    I also forgot that the change of variables involves the Jacobian determinant, which is unity for a unitary matrix.

    I still don't understand why P&S go through a long argument involving integrating only over the wavevectors k such that [itex]k^0 > 0[/itex]?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook