A free particle Hamiltonian, at time=0, the particle is located at origin, a delta function.


Then t>0 solution should be the Green function:

$\Psi(r,t)=G(r,r'=0,t)=\sqrt{\frac{m}{2 \pi i \hbar t}} e^{ i \frac{ m}{2 \hbar}\frac{r^2}{t}}$

However, the probability distribution does not depend on position $r$ anymore, and it is not normalizable.

$P(r,t)=|\Psi(r,t)|^2=\frac{m}{2 \pi \hbar t} $

What's the problem here? should the initial condition be



$\Psi(r,t)\neq G(r,r'=0,t) $

It seems that a lot of textbooks are wrong, what do you think then?

  • $\begingroup$ More on the square root of the Dirac delta distribution: physics.stackexchange.com/q/135569/2451 and mathoverflow.net/q/235827/13917 $\endgroup$ – Qmechanic Aug 29 '16 at 22:08
  • $\begingroup$ Why are you saying that the probability distribution does not depend on position? Judging from your second expression, it seems to me that it does. $\endgroup$ – flippiefanus Sep 1 '16 at 9:40
  • $\begingroup$ If you take the absolute square, the dependence of position on imaginary exponential will be a constant one. $\endgroup$ – wwwjjj Sep 1 '16 at 21:34

The problem is that you're using a position eigenstate as your initial state. If you use a state with finite tunable width you can see what's happening. Examples to try: $$\begin{align} \Psi(\mathbf{x},0) & = \frac{1}{\left(s \sqrt{2\pi}\right)^{d/2}} \operatorname{e}^{-\frac{(\mathbf{x}-\mathbf{x}_0)^2 }{ 4s^2} + i\mathbf{k}\cdot \mathbf{x}}, \ \mathrm{or}\\ & = \prod_{n=1}^d \frac{2 \sin\left(\frac{k_w (x_n-x_0)}{2}\right)}{k_w (x_n-x_0)} \operatorname{e}^{i k_n (x_n- x_0)}\sqrt{\frac{k_w}{2\pi}}, \end{align}$$ where $d$ is the number of spatial dimensions. The momentum space versions (Fourier transforms) of these are:$$\begin{align}\Psi(\mathbf{p},0) & = \left(s\sqrt{\frac{2}{\pi}}\right)^{d/2}\operatorname{e}^{i \frac{(\hbar \mathbf{k} - \mathbf{p})\cdot\mathbf{x}_0}{\hbar} - \frac{s^2}{\hbar^2} \left(\mathbf{p} - \hbar\mathbf{k}\right)^2},\ \mathrm{and} \\ & = \prod_{n=1}^d \frac{\operatorname{e}^{-ik_n x_0}}{\sqrt{\hbar k_w}} \Theta\left(\frac{p_n}{\hbar} - k_n + \frac{k_w}{2}\right) \Theta\left(k_n + \frac{k_w}{2} - \frac{p_n}{\hbar}\right).\end{align}$$ Notice how as the $\mathbf{x}$-space widths go to $0$ ($s\rightarrow 0$ or $k_w \rightarrow \infty$) the $\mathbf{p}$-space widths behave in the opposite fashion. So, the reason you get a non-normalizable and flat state is not just because $\delta$ functions are not square integrable, but also because the momentum wave function has equal probabilities of all momenta, including infinity. So it's no surprise when the state instantly transitions into one that has equal probability at all positions.


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