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If we only took ``half'' of the commutator, above, we would get an expression that looks like: \begin{eqnarray} \langle 0 \vert \phi(x)\phi(y) \vert 0\rangle &=& \int_0^\infty \frac{dk}{(2\pi)^2}\frac{k}{2\omega_k}\frac{i}{\xi}e^{-i\omega_k \tau}\left(e^{ik\xi}-e^{-ik\xi}\right) \end{eqnarray} Now this integral is even $k$, since we can rewrite the two exponential terms as: \begin{eqnarray} &=& \int_0^\infty \frac{dk}{(2\pi)^2}\frac{1}{2\omega_k}\frac{i}{\xi}e^{-i\omega_k \tau}\left(ke^{ik\xi}-ke^{-ik\xi}\right) \end{eqnarray} We therefore extend integration to $-\infty \to \infty$, and write: \begin{eqnarray} \langle 0 \vert \phi(x)\phi(y) \vert 0\rangle &=& \frac{i}{\xi} \int_{-\infty}^\infty \frac{dk}{(2\pi)^2}\frac{k}{2\omega_k}e^{-i\omega_k \tau+ik\xi} \end{eqnarray} And now we have this strange integral to deal with. The first thing to do is choose a case: either $\xi=0$ or $\tau=0$, since we can always find a proper lorentz boost to accomplish this. If we look at the space-like separated case, we have \begin{eqnarray} \langle 0 \vert \phi(x)\phi(y) \vert 0\rangle &=& \frac{i}{\xi} \int_{-\infty}^\infty \frac{dk}{(2\pi)^2}\frac{k}{2 \sqrt{k^2+m^2}}e^{ik\xi} \end{eqnarray} which has a pesky square root function, and therefore a branch cut, at either $im$ or $-im$. We choose our branch cut to be in the negative imaginary regime, and make a transformation $k^\prime=-ik$, such that: \begin{eqnarray} \langle 0 \vert \phi(x)\phi(y) \vert 0\rangle &=& \frac{i}{\xi} \int_{i\infty}^{-i\infty} \frac{dk}{(2\pi)^2}\frac{-k}{2 \sqrt{m^2-k^2}}e^{-k\xi}\\ &=& \frac{i}{\xi} \int_{-i\infty}^{i\infty} \frac{dk}{(2\pi)^2}\frac{k}{2 \sqrt{m^2-k^2}}e^{-k\xi} \end{eqnarray} Now we notice that if we make a ``keyhole contour'' that goes along the imaginary axis in this rotated space, goes out along $k\to \infty$ and then wraps around the branch cut, then we can say that such a closed contour integral will yield zero, since it contains no singularities. \begin{eqnarray} \oint_{C_r + C_i + C_b} \end{eqnarray} Where the $C$'s stand for rounded arcs, the imaginary and branch axis contours, respectively. We can claim immediately that the arcs will yield zero as we go out to infinity, since the real part of $k$ damps our integrand exponentially. The next step is to notice that our branch cut function can be written as: \begin{eqnarray} \sqrt{m^2-k^2}&=& e^{\frac{1}{2}\log(m^2-k^2)+\frac{i}{2}\phi }\\ &=& \sqrt{m^2-k^2}e^{i\phi/2} \end{eqnarray} Where the angle $\phi$, in this rotated space, takes on the values $0<\phi<2\pi$. This means that above the branch cut, on the top side of the real axis, we will get a positive sign in our branched function. But below the real axis we will get a negative sign, due to the phase $e^{i \frac{2\pi}{2}}$ (with some delta in there so that we're not exactly on the branch cut!). So we can equate the imaginary axis integration and the branch cut integration: \begin{eqnarray} \frac{i}{\xi} \int_{-i\infty}^{i\infty} \frac{dk}{(2\pi)^2}\frac{k}{2 \sqrt{m^2-k^2}}e^{-k\xi} &=& - \int_m^\infty \frac{dk}{(2\pi)^2}\frac{k}{\sqrt{m^2-k^2}}e^{-k\xi} \end{eqnarray} So we may write \begin{eqnarray} \langle 0\vert \phi(x)\phi(y) \vert 0\rangle &=& - \int_m^\infty \frac{dk}{(2\pi)^2}\frac{k}{\sqrt{m^2-k^2}}e^{-k\xi} \end{eqnarray} Up to some possibly dropped negative sign. The important thing is that now, we may take the $\xi \to \infty$ limit, and note that the asymptotic contribution to this integral comes from the minimum value of $k$, and so we get \begin{eqnarray} \lim_{\vert x-y \vert \to \infty}\langle 0\vert \phi(x)\phi(y) \vert 0\rangle &\sim & e^{-m\vert \vec{x}-\vec{y}\vert} \end{eqnarray} Similarly, for the time-like separation case, we can do the same yakkety-yak, rotating our complex integration space and then wrapping around the a fitting branch cut, in order to write: \begin{eqnarray} \langle 0\vert \phi(x)\phi(y) \vert 0\rangle &=& - \int_m^\infty \frac{dk}{(2\pi)^2}\frac{k}{\sqrt{m^2-k^2}}e^{i\sqrt{m^2-k^2}\tau} \end{eqnarray} Except in this case we have not really helped ourselves, as this integral needs some sort of regularization, or infinitesimal offset of the parameter $m$ in order to converge, since, $k=0$ corresponds to the point of stationary phase in the asymptotic $\tau \to \infty$ limit. We could have just taylor expand our function $\sqrt{m^2-k^2}$ about $0$, without all the branch and rotation nonsense, to write: \begin{eqnarray} \langle 0 \vert \phi(x)\phi(y) \vert 0\rangle &=& \int_{-\infty}^\infty \frac{dk}{(2\pi)^2}\frac{k^2}{2 \sqrt{k^2+m^2}}e^{-i\sqrt{k^2+m^2} \tau}\\ &\approx & e^{-im\tau} \times \dots \end{eqnarray} The rest of the expression can be written as a series of expectation values of a 1-D gaussian integral, with variance $\sigma^2 = \frac{m}{\tau}$. So we find that the naive scalar propagator does not vanish for space-like separation, but it's commutator does. We also find that for time-like separation we get an oscillatory result, which not coincidentally, for the commutator, will result in something like \begin{eqnarray} \langle 0 \vert \phi(x)\phi(y) \vert 0\rangle &=& e^{-im\tau}-e^{im\tau} \end{eqnarray} an interference between the two measurements.