# Homework Help: Time integrals of free particle propagator

1. Feb 20, 2012

### naele

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
$$I_1=\int_0^{\infty}G(\vec r_2,\vec r_1;\tau)e^{i\omega\tau}d\tau,\quad I_2=\int_0^{\infty}G(\vec r_2,\vec r_1;\tau)e^{i\omega\tau}\frac{d\tau}{\tau}$$

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

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

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)
$$\int d^3p_1 d^3p_2 e^{i\vec p_2\cdot\vec r_2/\hbar}e^{-i\vec p_1\cdot\vec r_1\hbar}\frac{i\hbar}{2m\hbar(\omega+iu)-p_1^2}\delta^3(\vec p_2-p_1)$$

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

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

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

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

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

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
$$I_2=\frac{m}{2i\pi\hbar}\frac{1}{r}\int_0^{\infty} e^{ir\sqrt{p^2+i\hbar 2mu}/\hbar}du$$

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