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Berezin's correspondance of (anti-)symmetric function with functional

  1. Jun 9, 2012 #1
    Hi all, I have recently been reading the book ``The Method of Second Quantization'' by Felix Berezin but I got trapped on just page 4, where the concept of generating functionals is introduced. It seems to be assigning each (anti-) symmetric function of N variables with a functional of a function of just the degrees of freedom of one of the particles. And in the last sentence of the page, it is commented that ``Knowing the functional \Phi(a^*) and \tilde{A}(a^*, a), one can obviously construct the vector \Phi and the operator \tilde{A}''. But even after a serious amont of thinking, I am still not able to be the obviousness here. Google search did not seem to have yielded some clear answer. Even the book seems to have been highly cited, but I really cannot find a detailed explanation to it. Could someone here give me some guidance? Thank you so much!
  2. jcsd
  3. Jun 9, 2012 #2


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    That book is a difficult study, as Berezin assumes quite a lot of the reader.

    The stuff on page 4 you mentioned is part of an "introduction". A more detailed explanation follows in chapter 1 (though, as I look through it, I can't help thinking there must be more helpful ways of presenting this stuff).

    Given the ##\Phi(a^*)## in Berezin's eq(0.10), i.e.,
    \Phi(a^*) ~:=~ \sum_n \frac{1}{\sqrt{n!}} \int K_n(x_1,\cdots,x_n) \;
    a^*(x_1) \cdots a^*(x_n) \; d^n x ~,
    we want to extract the components of the vector ##\hat\Phi## in eq(0.1) which consists of the various ##K_n(\dots)## functions in the integrand. This is usually done with the help of a functional derivative. E.g., to extract ##K_1(y)##, use a single functional derivative like this:
    K_1(y) ~=~ \frac{\delta \Phi(a^*)}{\delta a^*(y)} \; \Big|_{a^*=0}
    This uses
    \frac{\delta a^*(x)}{\delta a^*(y)} ~=~ \delta(x-y)
    (i.e., a Dirac delta on the right hand side). This extracts one term from the sum of integrals, and all the others vanish after applying ##a^*=0## as the last step.

    For higher order ##K_n(\dots)## we use higher order functional derivatives, apply the Leibniz product rule carefully when differentiating the integrands (which results in a factor of ##n!##, iirc), and possibly introduce an extra factor of ##\sqrt{n!}## somewhere to compensate.

    I hope that's enough to give you the basic idea. Such use of functional differentiation is very common when working with generating functionals.
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