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Consider the integral $$U(t)=\int\frac{d^3p}{(2\pi)^3}e^{-ip^2t/2m}e^{i\vec p\cdot\Delta\vec x}$$ for the free non-relativistic propagator. I'm not quite sure about the gritty details of radial Fourier transformation required to turn this into standard Gaussian rubbish on $(0,\infty)$ like the 1d case (it doesn't feel like integrating the angles should be hard, but I'm not sure...) I attempted: $$U(t)=\frac 1{(2\pi)^3}\int_0^\infty dp\,p^2e^{-ip^2t/2m}\int_0^{2\pi}d\phi\int_0^\pi d\theta\,\sin\theta\,e^{ip\Delta x\cos\theta}$$ with the correct jacobian, but integrating $\theta$ then gives a $\sin(p\Delta x)$ which shouldn't turn out in the final expression. Thus, is the above conversion from cartesian to spherical coordinates incorrect? (I ask as this integration technique is essential for a great deal of 3d QM calculations).

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  • $\begingroup$ You are right, integrating the angles shouldn't be too hard. Since you are integrating over all of momentum space, you are free to change the orientation of your p-axes with impunity. What if you aligned your pz axis along Δx⃗ ? How would this make your life easier? $\endgroup$ – ApproximatelyTrue May 13 '15 at 6:54
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In general, you will run into some problems with "functions" like this one. Consider, for example, U(0), which is $$ U(0)=\frac{1}{(2\pi)^3}\int d^3 p e^{i\vec p \cdot \vec {\Delta x}} $$ $$ =\delta(\vec {\Delta x}) $$ I.e., a three dimension "delta function". This is not really a "function" in the conventional sense, but rather a "functional".

So, okay, who cares, how do I do this integral? Well, try this (b.t.w, I'm going to write $\vec x$ instead of $\vec {\Delta x}$, for brevity) $$ U(t)=\frac{1}{(2\pi)^3}\int d^3 p e^{-itp^2/(2m)}e^{i\vec p \cdot \vec {\Delta x}} $$ $$ =\frac{1}{(2\pi)^3}\int d^3p e^{-\frac{it}{2m}(p^2-\vec p\cdot\vec x 2m/t)} $$ Then complete the square... $$ =\frac{1}{(2\pi)^3}\int d^3p e^{-\frac{it}{2m}((\vec p-\vec xm/t)^2-\frac{x^2m^2)}{t^2}} $$ $$ =e^{it\frac{1}{2}m{\frac{x^2}{t^2}}}{\left(\frac{m}{2\pi i t}\right)}^{\frac{3}{2}} $$ ...Ta Da.

...Notice, btw, the appearance of the classical action $(t\frac{1}{2}mv^2)$ in the phase of the result. Very interesting, eh?

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