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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.

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See the solution I provided here:… – Ron Gordon Jan 29 '13 at 2:54
@rlgordonma: Thank you very much. Following the instructions you gave in the link above I finally understood the present subject. – Giuseppe Negro Feb 2 '13 at 2:44
up vote 2 down vote accepted

Since I feel like I completely understood the question at hand, I'll try to provide here a detailed answer, for future reference. Everything is based on the very good answer by rlgordonma 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 third quadrant we are choosing the unique square root lying in the same quadrant. 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 $$\int_{\infty}^\infty e^{-(b+ic)\xi^2}\, d\xi.$$ This is exactly the procedure rlgordonma follows in the linked post and it is the place where we will need the Convention fixed above. Indeed, we consider the region of the complex plane encircled by the paths $$ \begin{array}{ccc} \gamma_1(t)=t,\ t\in[0, R]; & \gamma_2(t)=R e^{i \theta},\ \theta\in [0, \theta_0/2]; & \gamma_3(t)=\sqrt{b+ic}(R-t),\ t\in[0, R]. \end{array} $$ We have $\int_{\gamma_1+\gamma_2+\gamma_3}e^{-z^2}\, dz=0$, but we have also that $$\lim_{R\to \infty} \int_{\gamma_2} e^{-z^2}\, dz=0, $$ so we can conclude that $$\tag{3}\lim_{R\to \infty} \int_{\gamma_1} e^{-z^2}\, dz=\lim_{R\to \infty} -\int_{\gamma_3}e^{-z^2}\, dz.$$ This is exactly what claimed. $\square$

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 changed by a sign.

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