Let $\phi\in \mathcal S(\mathbb R)$ and consider the uni-dimensional global Cauchy problem for the Schrödinger equation of a free particle, that is

$$(*)\qquad iu_t+u_{xx}=0\quad \text{and}\quad u(x, 0)=\phi(x),\quad x\in \mathbb R, t>0.$$ I must find a solution and prove the following equality,

$$(**)\qquad\lim_{t\to +\infty}\int_{|x|<t}|u(x,t)|^2\,dx=\int_{|\xi|<1/2}|\hat{\phi}(\xi)|^2\,d\xi.$$

I've already found a solution (using Fourier transform), a fundamental solution for $(*)$ is $$E(x, t)=\frac{1}{(4\pi i t)^{1/2}} e^{-\frac{x^2}{4it}},$$ and so a solution for $(*)$ will be $$u(x, t)=\big( E(-, t)*\phi \big)(x)=\frac{1}{(4\pi i t)^{1/2}} \int_{\mathbb R}e^{\frac{i(x-y)^2}{4t}}\phi(y)\, dy.$$ However, I can't prove $(**)$, can you help me with that? (It must not be very hard, but I don't see a clean way to prove this)

Also, if possible, can you give me a non-technical physical interpretation of $(**)$ ? Regards

  • $\begingroup$ Physical interpretation? I thought physicists said you should shut up and do the math? :P $\endgroup$ – terrace Nov 6 '17 at 3:53
  • 1
    $\begingroup$ Look at the energy $e(t) = \int_{-\infty}^\infty |u(x,t)|^2 dx =\int_{-\infty}^\infty |\hat{u}(\xi,t)|^2 d\xi$. I don't see any good reason to integrate on $|\xi|< 1/2$. $\endgroup$ – reuns Nov 6 '17 at 4:06
  • $\begingroup$ Compute the left hand side and the right hand side of (**) separately and compare. $\endgroup$ – Winther Nov 6 '17 at 11:44
  • $\begingroup$ The constant on the left of (**) is the probability of the particle at any t, normalized to 1. Plugging in your general solution, and doing the x integral first, you get a Dirac delta function w.r.t. the two y, y' in the remaining double integral: It collapses y'=y and the remaining integral is the probability of the particle being anywhere at t=0. So, probability is preserved. As @runs points out, there is no good reason for limiting $\xi$. See here. $\endgroup$ – Cosmas Zachos Dec 11 '18 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.