Consider the initial-value problem for the Schrödinger equation $$\tag{IVP} \begin{cases} i\frac{\partial u}{\partial t}+\Delta u=0 & x\in \mathbb{R}^n,\ t\in \mathbb{R}\setminus \{0\} \\ &\\ u=g & t=0 \end{cases}$$ As I read in Evans' book on PDEs (2nd ed. pag. 188 eqn (20) ), the unique solution of (IVP) can be expressed by means of the following formula: $$\tag{1} u(x, t)=\frac{1}{\left(4\pi i t \right)^{n/2}}\int_{\mathbb{R}^n}e^{i\frac{\lvert x-y\rvert^2}{4t}}g(y)\, dy,\qquad t\ne 0.$$

Question When $n$ is odd, in equation (1) there appears the square root of an imaginary number. Which branch of the square root should be chosen, and why? What happens if we choose the other branch?

Some considerations. The treatment found in Evans' book is purely formal and it does not provide with an answer to the previous question. However, in this previous post, we have already derived formula (1). As we have seen, the square root in (1) comes out from the anti-Fourier transformation of $$e^{-i t \lvert \xi\rvert^2}.$$ Assuming for the spatial dimension $n=1$, this transform can be computed directly by completing the square (in the linked post we used a different technique): $$\int_{-\infty}^\infty e^{-it\xi^2+ix\xi}\,d\xi= e^{i \frac{x^2}{4t}}\int_{\infty}^\infty e^{-\left(\sqrt{it}\xi-\frac{ix}{2\sqrt{it}}\right)^2}\,d\xi.$$ Up to this point the choice of a branch cut for the square root is irrelevant. To conclude we need to evaluate the rightmost integral. If we apply the change of variable $$\sqrt{it}\xi-\frac{i x}{2\sqrt{it}}=\eta, $$ and if we treat the quantity $\sqrt{it}$ as if it was real, we get $$\int_{-\infty}^\infty e^{-\left(\sqrt{it}\xi-\frac{ix}{2\sqrt{it}}\right)^2}\,d\xi=\frac{1}{\sqrt{it}}\int_{-\infty}^\infty e^{-\eta^2}\, d\eta=\sqrt{\frac{\pi}{it}},$$ which leads to the correct result (1). But this reasoning (apart from being not sufficiently justified) does not answer the Question above. To obtain such an answer a more careful treatment of this last integral should be given.

Thank you for reading.

  • 1
    $\begingroup$ See the solution I provided here: math.stackexchange.com/questions/285348/… $\endgroup$
    – Ron Gordon
    Jan 29, 2013 at 2:54
  • $\begingroup$ @rlgordonma: Thank you very much. Following the instructions you gave in the link above I finally understood the present subject. $\endgroup$ Feb 2, 2013 at 2:44

1 Answer 1


Since I feel like I completely understood the question at hand, I'll provide here a detailed answer, for future reference. Everything is based on the very good answer by Ron Gordon linked here.

Convention. Let $b\ge 0$ and $c>0$ (resp. $c<0$). Then there is a unique $\theta_0\in (0, \pi/2)$ (resp. $\theta_0\in (-\pi/2, 0)$ ) such that $b+ic=(b^2+c^2)^{1/2}e^{i\theta_0}$. We denote $$\sqrt{b+ic}=\left(b^2+c^2\right)^{\frac{1}{4}}e^{i\frac{\theta_0}{2}}.$$

In other words, for complex numbers lying in the first or fourth quadrant we are choosing the unique square root lying in the same quadrant. This is the natural extension of the standard square root of nonnegative numbers.

With this convention fixed, we can compute the following Gaussian integrals.

Proposition. Let $b > 0, c \in \mathbb{R}$ and $a\in \mathbb{R}^n$. Then we have $$\int_{\mathbb{R}^n} e^{-(b+ic)\lvert\xi\rvert^2+ia\cdot\xi}\, d\xi= \left(\frac{\sqrt{\pi}}{\sqrt{b+ic}}\right)^n e^{-\frac{\lvert a \rvert^2}{4(b+ic)}}.$$ If $b=0$ the result still holds if the integral is taken in the principal value sense, that is $$ \lim_{R\to \infty}\int_{[-R,R]^n} e^{-ic\lvert\xi\rvert^2+ia\cdot\xi}\, d\xi=\left(\frac{\sqrt{\pi}}{\sqrt{ic}}\right)^n e^{-\frac{\lvert a \rvert^2}{4ic}}.$$

Corollary. The square root of $4\pi t\, i $ in formula (1) in Question above is to be taken in the sense of the Convention we just fixed.

Proof of Proposition. It suffices to treat the case $n=1$. Completing the square in the exponential we get $$\tag{2}\int_{-\infty}^\infty e^{-(b+ic)\xi^2+ia\xi}\, d\xi = e^{-\frac{a^2}{2(b+ic)}}\int_{-\infty}^\infty e^{-\left( \sqrt{b+ic}\xi -i \frac{a}{2\sqrt{b+ic}}\right)^2}\, d\xi.$$ Since the function $e^{-z^2}$ is exponentially decaying at infinity, the integrals $$\int_{-\infty}^\infty e^{-\left( \sqrt{b+ic} \xi -i \alpha\right)^2}\, d\xi$$ are independent of $\alpha$. So we are left with evaluating $$\tag{*}\int_{-\infty}^\infty e^{-(b+ic)\xi^2}\, d\xi.$$ We do this by following Ron Gordon's answer. This is where we need the Convention fixed above. We consider the region of the complex plane encircled by the paths $$ \begin{array}{ccc} \gamma_1(t)=R(b^2+c^2)^\frac14\, t,\ t\in[0, 1]; & \gamma_2(t)=R (b^2+c^2)^\frac14 e^{i \theta},\ \theta\in [0, \theta_0/2]; & \gamma_3(t)=R\sqrt{b+ic}(1-t),\ t\in[0, 1]. \end{array} $$ We have $\int_{\gamma_1+\gamma_2+\gamma_3}e^{-z^2}\, dz=0$. Note that $$\lim_{R\to \infty} \int_{\gamma_2} e^{-z^2}\, dz=0, $$ because the integrand function decays exponentially when $R\to \infty$. So we can conclude that $$\tag{3}\lim_{R\to \infty} \int_{\gamma_1} e^{-z^2}\, dz=\lim_{R\to \infty} -\int_{\gamma_3}e^{-z^2}\, dz.$$ Explicitly, $$ \int_{\gamma_1} e^{-z^2}\, dz = \int_0^{R(b^2+c^2)^\frac14} e^{-\tau^2}\, d\tau \to \frac{\sqrt{\pi}}2.$$ Also $$ \int_{\gamma_3}e^{-z^2}\, dz = -\sqrt{b+ic} \int_0^R e^{-(b+ic) t^2}\, dt.$$ Letting $R\to \infty$ we obtain $$ \int_0^\infty e^{-(b+ic)t^2}\, dt = \frac{\sqrt{\pi}}{2\sqrt{b+ic}},$$ from which we can evaluate (*) immediately. $\square$

If we had chosen the other branch cut of the square root in the Convention above, we would have not gotten the sign change in equation (3), and so our result would have had the wrong sign.


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