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A Projection trick to obtain time-ordered correlator

  1. Oct 7, 2016 #1
    To compute the vacuum expectation value

    ##\langle \Omega | T\{q(t_{1})\cdots q(t_{n})\}|\Omega\rangle##

    in the path integral formalism, we start with the time-ordered product in the path integral representation

    ##\langle q_{f},t_{f}|\ T\{q(t_{1})\cdots q(t_{n})\}\ |q_{i},t_{i}\rangle=\int\ \mathcal{D}\phi\ e^{iS}\ q(t_{1})\cdots q(t_{n}),##

    use the fact that

    ##|\psi\rangle = \int\ dq_{i}\ |q_{i},t_{i}\rangle\ \langle q_{i},t_{i}|\psi\rangle##

    and the projection trick

    ##\lim\limits_{T \to\infty}e^{-iHT(1-i\epsilon)}|\Omega\rangle = \sum\limits_{n}e^{-iE_{n}T(1-i\epsilon)}|n\rangle\langle n|\Omega\rangle##

    to project out all the states with ##n \neq 0## and obtain

    ##\langle \Omega | T\{q(t_{1})\cdots q(t_{n})\}|\Omega\rangle = \frac{\int\mathcal{D}q(t)\ e^{iS[q]}q(t_{1})\cdots q(t_{n})}{\int \mathcal{D}q(t)\ e^{iS[q]}}.##

    How can we be sure that the projection trick is a legitimate step in the calculation and not some sleight of hand? Is the projection trick performed in the limit that ##\epsilon \rightarrow 0##?

    I would also like some help in deriving
    ##\langle \Omega | T\{q(t_{1})\cdots q(t_{n})\}|\Omega\rangle## using the above hints. Do I substitute for ##|\Omega\rangle## using the expansion of ##|\psi\rangle## I have from above?
     
  2. jcsd
  3. Oct 8, 2016 #2
    The limit ##T \rightarrow \infty## kills the exponential, though of course we need ##\epsilon \rightarrow 0^{+}##. It's true that this somewhat feels a bit like a sleight-of-hand, but the introduction of this extra imaginary term works. Similar tricks are used to ensure convergence of various other integrals, although I must admit I'm not entirely sure how to justify the tricks rigorously.
    If you do want to treat the problem with mathematical rigor though (i.e. without this small imaginary time trick), then you might want to look at the Riemann-Lebesgue lemma, which states that
    [tex]\lim_{\mu \rightarrow \infty} \int^{b}_{a} f(x) e^{i \mu x} dx = 0[/tex]
     
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