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Time integrals of free particle propagator

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    As part of an assignment on matter wave diffraction I'm to calculate the following integrals
    [tex]
    I_1=\int_0^{\infty}G(\vec r_2,\vec r_1;\tau)e^{i\omega\tau}d\tau,\quad
    I_2=\int_0^{\infty}G(\vec r_2,\vec r_1;\tau)e^{i\omega\tau}\frac{d\tau}{\tau}
    [/tex]


    2. Relevant equations
    To do so, introduce the following integral
    [tex]
    J(u)=\int_0^{\infty}G(\vec r_2,\vec r_1;\tau)e^{(i\omega-u)\tau}d\tau
    [/tex]
    It is easy to see that
    [tex]
    I_1=J(u=0),\quad I_2=\int_0^{\infty}J(u)du
    [/tex]


    3. The attempt at a solution
    We have that
    [tex]
    J(u)=\int_0^{\infty}d\tau \langle \vec r_2|e^{-iH\tau/\hbar}e^{(i\omega -u)\tau}|\vec r_1\rangle
    [/tex]

    If we switch to the momentum representation, we can exploit the fact the momentum eigenstates are eigenstates of the Hamiltonian and hence get (constants suppressed)
    [tex]
    \int d^3p_1 d^3p_2 e^{i\vec p_2\cdot\vec r_2/\hbar}e^{-i\vec p_1\cdot\vec r_1\hbar}\frac{i\hbar}{2m\hbar(\omega+iu)-p_1^2}\delta^3(\vec p_2-p_1)
    [/tex]

    We can integrate over p2, and noticing that [itex]\hbar\omega=p^2/2m[/itex] we have
    [tex]
    J(u)=-\frac{i\hbar 2m}{(2\pi\hbar)^3}\int d^3p_1 e^{i\vec p_1\cdot(\vec r_2-\vec r_1)/\hbar}\frac{1}{-i\hbar 2mu-p^2+p_1^2}
    [/tex]

    Switch to spherical coordinates aligned in such a way that [itex]\vec p_1\cdot(\vec r_2-\vec r_1)=|\vec p_1||\vec r_2-\vec r_1|\cos\theta[/itex]. Integrate over the angles to get
    [tex]
    J(u)=-\frac{(i\hbar2m)(2\pi\hbar)}{i(2\pi\hbar)^3r}\int_{-\infty}^{\infty}dp_1 p_1 \frac{e^{ip_1r/\hbar}}{p_1^2-p^2-i\hbar 2mu}
    [/tex]

    We can see there are two simple poles at [itex]p_1^{\pm}=\pm\sqrt{p^2+i\hbar 2mu}[/itex]. We consider a semi-circle contour in the upper half plane according to Jordan's lemma. And so the residue is just [itex]\frac{1}{2}e^{ir\sqrt{p^2+i\hbar 2mu}/\hbar}[/itex]

    Then altogether we have that
    [tex]
    J(u)=-\frac{(i\hbar 2m)(2\pi)(i\pi)\hbar}{i(2\pi\hbar)^3r}e^{ir\sqrt{p^2+i\hbar 2mu}/\hbar}
    [/tex]

    Now, I am given the expressions for [itex]I_1,I_2[/itex] and they are
    [tex]
    I_1=\frac{m}{2i\pi\hbar}\frac{1}{r}e^{ipr/\hbar} ,\quad
    I_2=\frac{m}{2i\pi\hbar}\frac{1}{r^2}(p+\frac{i \hbar}{r})e^{ipr/\hbar}
    [/tex]

    Now from the expression I calculated for J(u) I can see that J(u=0)=I_1 as found, my problem is that I don't know how to calculate
    [tex]
    I_2=\frac{m}{2i\pi\hbar}\frac{1}{r}\int_0^{\infty} e^{ir\sqrt{p^2+i\hbar 2mu}/\hbar}du
    [/tex]

    I feel like there's a clever change of variables that I'm just not seeing. Any help is appreciated, thanks for reading.
     
    Last edited: Feb 20, 2012
  2. jcsd
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