I was trying to prove that Quantum Mechanics violates causality.

To do that, I started by computing the transition amplitude between the fixed position $x_0$ and an arbitrary position $x$, during a certain time interval $t$:

$$A(t) = \langle x |\hat{U}(t,0)|x_0 \rangle $$

Using the momentum basis and integrating the angular contribution, one can obtain

$$A(t) = \frac{1}{(2\pi)^2 |x-x_0|i} \int_{-\infty}^{\infty}p e^{i(-t\sqrt{p^2+m^2} + p |x-x_0|)} \ dp$$

This integral can be evaluated using complex analysis. To do so, I started by noting that the region of the imaginary axis where $|\mathrm{Im}(p)| > m $ does not contribute to the integral, since it is physically irrelevant. So, I choose the following path: enter image description here

It can be easily proven that, when $R \to \infty$ the semicircular regions contribute $0$ to the integral. My problem is in evaluating the vertical regions. In doing so, I used (for the leftmost region)

$$ z_l = - \epsilon + iy$$

and, for the rightmost region

$$ z_r = \epsilon + iy$$

where $\epsilon > 0 $.

Therefore, I need to evaluate

$$ I_1 = - \lim_{\epsilon \to 0}\int_{m}^{\infty} e^{i f(z_l)} z_l i\ dy $$


$$ I_2 = \lim_{\epsilon \to 0}\int_{m}^{\infty} e^{i f(z_r)} z_r i\ dy $$

where $$f(k) = -t \sqrt{k^2+m^2}+k \lvert x-x_0 \rvert $$

In doing so, I get that the sum of the integrals yields $0$, whereas I am supposed to get

$$ I = \int_{m}^{\infty} \ e^{- p \mid x-x_0 \mid} \sinh \left( t\sqrt{p^2-m^2}\right)p\ dp $$

What am I doing wrong?

  • $\begingroup$ What branches of the square root are you using? $\endgroup$ – mike stone Feb 16 at 12:43
  • $\begingroup$ @Andrew if proven, the result states that a particle could propagate faster than the speed of light, in contraction with the causality principles imposed by the Theory of Relativity. $\endgroup$ – miniplanck Feb 16 at 13:49
  • $\begingroup$ @mikestone What do you mean by branch in this case? $\endgroup$ – miniplanck Feb 16 at 13:53
  • $\begingroup$ @Andrew it's the standard proof that the propagator is non-zero outside the future light-cone in naive RQM. $\endgroup$ – Nihar Karve Feb 16 at 15:19
  • $\begingroup$ I means that the root has opposite signs on the two sides of the branch cut --- as explained by Nihar Karve in his answer. $\endgroup$ – mike stone Feb 16 at 17:32

Here, $x-x_0$ is a spatial interval and $p =|\mathbf{p}|$. Start with $$\langle x|\hat{U}(t,0)|x_0\rangle = \frac{1}{(2\pi)^2 |x-x_0|i} \int_{-\infty}^{\infty}\mathrm{d}p\ p e^{i(-t\sqrt{p^2+m^2} + p |x-x_0|)} $$ You must restrict the square root $\sqrt{p^2+m^2}$ to a single branch in order for this to make any sense. As $\sqrt{p^2+m^2} = 0$ for $p=\pm im$, a convenient choice for the branch cut in the $p$-plane is from $im$ upwards and $-im$ downwards, along the imaginary axis. Since we're looking at amplitudes outside the lightcone, $|x-x_0|>t$, so the $e^{ip|x-x_0|}$ factor ensures that the integrand does not blow up in the upper half plane since it damps any exponential contribution from $e^{-it\sqrt{p^2+m^2}}$.

Therefore you can adjoin a semicircle in the upper half plane to the real line, and we take its radius to infinity - but we deform the contour so that it dips below the upper branch cut. By the residue theorem, the contour integral is zero, and so the integral over the real line is equal to minus the integral around the branch cut (the integral over the semicircle vanishes for the reasons mentioned above).

enter image description here

[Peskin and Schroeder, Introduction to Quantum Field Theory, Figure 2.3]

Then, substituting $p\rightarrow iz-\varepsilon$ for the left side and $p\rightarrow iz+\varepsilon$ for the right side, we get $$ \langle x|\hat{U}(t,0)|x_0\rangle = \frac{1}{(2\pi)^2 |x-x_0|i}\times\lim_{\varepsilon\to0}\left(g(z+i\varepsilon) - g(z-i\varepsilon)\right) $$ where $$ g(z) = \int_{m}^{\infty}\mathrm{d}z\ z e^{i(-t\sqrt{z^2+m^2} + z|x-x_0|)} $$ After some cancellation, you'll see that the only place the $\varepsilon$ actually makes a difference is inside the square root, so $$ \langle x|\hat{U}(t,0)|x_0\rangle = \frac{1}{(2\pi)^2 |x-x_0|i} \times\lim_{\varepsilon\to0}\int_{m}^{\infty}\mathrm{d}(iz)\ iz \left(e^{-it\sqrt{(iz+\varepsilon)^2+m^2}}-e^{-it\sqrt{(iz-\varepsilon)^2+m^2}}\right)e^{-z|x-x_0|} $$ The square root on the right of the branch cut has a positive imaginary part, while that to the left of the branch cut has a negative imaginary part - this is because the branch cut determines the phase of the function on either side of it - there will be a discontinuous phase jump as you move across the branch cut, for $p>m$. Thus we pull $i$ out of the first square root and $-i$ out of the second square root, and take the limit $\varepsilon\to 0$. $$ \langle x|\hat{U}(t,0)|x_0\rangle = \frac{1}{(2\pi)^2 |x-x_0|i} \int_{m}^{\infty}\mathrm{d}z\ (-1)z \left(e^{t\sqrt{z^2-m^2}}-e^{-t\sqrt{z^2-m^2}}\right)e^{-z|x-x_0|} $$

Finally, using that the $\sinh z$ is the odd part of $e^z$ and renaming $z\rightarrow p$ $$ \langle x|\hat{U}(t,0)|x_0\rangle \propto \int_{m}^{\infty} \mathrm{d}p \ p\sinh \left( t\sqrt{p^2-m^2}\right) e^{- p|x-x_0|} $$ which is the desired result.

  • $\begingroup$ Why can we state that 'The square root on the right of the branch cut has a positive imaginary part, while that to the left of the branch cut has a negative imaginary part'? I think that's what is missing in my proof. $\endgroup$ – miniplanck Feb 16 at 15:13
  • $\begingroup$ @miniplanck hint: what does the branch cut do? $\endgroup$ – Nihar Karve Feb 16 at 16:34
  • $\begingroup$ As far as I know, it enables us to define a range of values in order to obtain 'unique points' (I'm talking about the case of the log function). However, I don't think it's the same concept you're referring to. $\endgroup$ – miniplanck Feb 16 at 16:41
  • 1
    $\begingroup$ @miniplanck it is indeed the same concept. The branch cut turns a multi valued map into a function, and in doing so, determines the phase of the resultant on either side of it. I believe there are some good math.stackexchange.com posts on this, though I can't find one off hand. I'll make sure to notify you when I have time. $\endgroup$ – Nihar Karve Feb 16 at 17:29
  • $\begingroup$ I'll search into it. Thank you very much! $\endgroup$ – miniplanck Feb 16 at 17:56

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