If you compute the probability amplitude of a free 1D non-relativistic particle with mass $m$, located at position $x_0$ at time $t_0$, for beeing detected at some other point $x_N$ at time $t_N$ you will find it to be given by $$ \mathcal{M} = \left\langle x_N \right| \text{e}^{-\frac{\text{i}}{\hbar} \frac{P^2}{2m}(t_N-t_0)}\left| x_0\right\rangle =\left(\frac{m}{2\pi\text{i}\hbar\ (t_N-t_0)}\right)^{1/2} \text{e}^{\frac{\text{i}}{\hbar}\frac{m}{2}\frac{(x_N-x_0)^2}{t_N-t_0}} $$ Now, if I compute the corresponding probability(density) according to $$ P = \left|\mathcal{M}\right|^2 = \mathcal{M} \mathcal{M}^* = \frac{m}{2\pi\hbar\ (t_N-t_0)} $$ it somehow strikes me that it does not depend on the distance $(x_N-x_0)$ at all. Does this mean, that the probability of detecting the particle is the SAME everywhere? I expected something like the initial (i.e. $t_N \rightarrow t_0$) delta function "melting away" like a Gaussian wave packet... Can anyone tell me what the correct interpretation of $P$ should be?
Answer
Does this mean, that the probability of detecting the particle it the SAME everywhere?
No, it does not. This is quite a common mistake, stemming from the idea that the Green function $\mathcal{M}$ can be used in the role of the $\psi$ function of free particle with the Born interpretation of $|\psi|^2$ as probability density. But that is not possible, since $\mathcal{M}$ is not normalizable.
The quantity $\mathcal{M}$ is simply the Green function of the time-dependent Schroedinger equation for free particle. It can be used to express $\psi$ function of the particle at time $t$ as $$ \psi(x,t) = \int \mathcal{M}(x,t;x_0, t_0) \psi_0(x_0,t_0) dx_0 $$ where $\psi_0(x_0,t_0)$ is normalized initial $\psi$ function at time $t_0$.
No comments:
Post a Comment