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I am trying to get my head around the Wick rotation technique. I have tried to play around with some elementary examples. Let us imagine I need to solve on the real line $$ y’ = \cos (x)$$ the prime denoting differentiation. I can view the function on the r.h.s as the real part of a complex function and re-formulate the differential equation, now on the imaginary line ( I suppose this is the "rotation"), as $$ y’ = e^{ix} $$ Whose solution is $$y = \frac{1}{i} e^{ix}$$ Now to get my solution on the real line I have to re-rotate back, by multiplying by $$-i$$, and then take the imaginary part of the ensuing expression to get my solution $ y = - \sin (x)$. Is this a correct,formally rigorous, application of the Wick rotation technique? I would also be the most grateful if somebody could show me an elementary example of analytical continuation by the Wick rotation. Thanks

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$\newcommand{\Cpx}{\mathbf{C}}$Not sure I've seen a formal definition of "Wick rotation", but I'd probably call your example "complexification" rather than "Wick rotation". "Wick rotation", in my experience, is more along the lines of replacing the Lorentz metric $dx^{2} - dt^{2}$ on a $1$-dimensional spacetime with the Euclidean metric $dx^{2} - (i\, dy)^{2} = dx^{2} + dy^{2}$ by introducing $y = -it$.

In your situation, this might be more analogous to replacing $y' = \cos(x)$ with $y' = i\cos(iu) = i\cosh(u)$, with $x = iu$ and the factor of $i$ outside the $\cosh$ coming from the chain rule. Then you get (up to an additive constant) $$ y = i\sinh(u) = i\sinh(-ix) = \sin(x), $$ as expected. (That said, I've never seen this type of computation called "Wick rotation".)

According to Goethe (in free paraphrase), "Mathematicians...translate everything you say into their own language, whereupon it is something entirely different." At risk of doing that to your question, it seems you're "really" asking about analytic continuation (though I'm not sure there's a "precise" question in the sense of the Math.SE guidelines). In that spirit, here are some (helpful, I hope) facts, examples, and exercises:

  • "Permanence of analytic identities": If $f$ is a real-analytic function satisfying a functional identity, such as $$ f(x + y) = f(x) \cdot f(y) \quad\text{for all real $x$ and $y$,} $$ and if $F$ is an entire (complex-analytic on the whole complex plane) function such that $F(x) = f(x)$ for all real $x$, then $F$ satisfies the same identity, here $$ F(w + z) = F(z) \cdot F(w) \quad\text{for all complex $z$ and $w$.} $$ The idea is straightforward mathematical bootstrapping: An entire function that vanishes on the real axis is well-known to vanish identically on the complex plane. Using the preceding example, fix a real number $w$, and consider the entire function $G$ defined by $$ G(z) = F(z + w) - F(z) \cdot F(w). $$ Since $G$ vanishes on the real axis, $G(z) = 0$ for all complex $z$. Since $w$ was an arbitrary real number, $F(z + w) - F(z) \cdot F(w) = 0$ for all complex $z$ and all real $w$. That is, if we fix an arbitrary complex $z$ and define $H(w) = F(z + w) - F(z) \cdot F(w)$, then $H$ is entire and vanishes on the real axis, so $H(w) = 0$ for all _complex $w$; since $w$ was arbitrary, $F(z + w) - F(z) \cdot F(w) = 0$ for all complex $w$ and $z$.

  • If $f(x) = \sum_{k} a_{k} (x - x_{0})^{k}$ is a real-analytic function on some open interval, there is a unique analytic continuation $F(z) = \sum_{k} a_{k} (z - x_{0})^{k}$ defined in some disk centered at $x_{0}$. For example, the extensions of $\exp$, $\cos$, and $\sin$ are, writing $z = x + iy$ with $x$ and $y$ real, \begin{align*} \exp(z) &= e^{x} e^{iy} = e^{x} \cos y + ie^{x} \sin y, \\ \cos(z) &= \cos(x) \cos(iy) - \sin(x) \sin(iy) = \cos(x) \cosh(y) - i\sin(x) \sinh(y), \\ \sin(z) &= \sin(x) \cos(iy) + \cos(x) \sin(iy) = \sin(x) \cosh(y) + i\cos(x) \sinh(y). \end{align*} The expansions on the right come from permanence of analytic identities together with the algebraic identities $$ \left. \begin{alignedat}{2} \cos(iy) - i\sin(iy) &= e^{y} &&= \cosh(y) + \sinh(y) \\ \cos(iy) + i\sin(iy) &= e^{-y} &&= \cosh(y) - \sinh(y) \end{alignedat}\right\}\quad \text{for all real $y$,} $$ which immediately imply $\cos(iy) = \cosh(y)$ and $\sin(iy) = i\sinh(y)$ for all real $y$.

In the following examples, let $D = \Cpx \setminus(-\infty, 0]$ be the "slit plane", and write the general element of $D$ uniquely as $z = re^{i\theta}$ with $r > 0$ and $-\pi < \theta < \pi$.

  • The function $f(x) = \sqrt{x}$, defined for all positive real $x$, has unique analytic continuation $F:D \to \Cpx$ defined by $$ F(z) = F(re^{i\theta}) = \sqrt{r} e^{i\theta/2}. $$ It's easy to check $F(z)^{2} = z$ for all $z$ in $D$, and $F(x) = f(x)$ for all positive real $x$. Further, $F$ has no continuous extension over the negative real axis, since for all $r > 0$, $$ \lim_{\theta \nearrow \pi} F(r e^{i\theta}) = -\lim_{\theta \searrow -\pi} F(r e^{i\theta}) = i\sqrt{r} \neq 0. $$

    1. Exercise: Show that the "double-valued function" $z \mapsto \pm F(z)$ extends holomorphically over the negative reals in the following sense: If $D' = \Cpx \setminus [0, \infty)$ and we write the general element of $D'$ uniquely as $z = re^{i\theta}$ with $r > 0$ and $0 < \theta < 2\pi$, then the function $$ G(z) = G(re^{i\theta}) = \sqrt{r} e^{i\theta/2} $$ is analytic on $D'$, and for all $z$ in $D \cap D'$ we have $\{\pm F(z)\} = \{\pm G(z)\}$ as sets of complex numbers. (Note that $G = F$ in the open upper half-plane, and $G = -F$ in the lower half-plane. This is the analytic version of the famous construction of the Riemann surface of the square root function by "taking two copies of the slit plane and cross-joining the edges of the slits".)

    2. Exercise: If $h(x) = \sqrt[3]{x}$ for $x > 0$, write out the unique analytic continuation $H:D \to \Cpx$ and analyze its limit behavior along the negative real axis; how do the resulting values compare with the real cube root of a negative real number?

  • The function $\ell(x) = \log(x)$ has unique analytic continuation $L:D \to \Cpx$ defined by $$ L(z) = \ell(r) + i\theta. $$ It's easy to check $\exp\bigl(L(z)\bigr) = z$ for all $z$ in $D$, and $L(x) = \ell(x)$ for all positive real $x$. Further, $L$ has no continuous extension over the negative real axis, since for all $r > 0$, $$ \lim_{\theta \nearrow \pi} L(r e^{i\theta}) = \lim_{\theta \searrow -\pi} L(r e^{i\theta}) + 2\pi i. $$

These are among the simplest non-trivial examples of analytic continuation. In each case, there is no analytic continuation over a "larger" open subset of the plane than $D$, though there do exist analytic continuations over different open sets than $D$.

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    $\begingroup$ Many thanks for this clear and thorough explanation, much appreciated, I am on the exercises now. $\endgroup$ May 29, 2015 at 20:23

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