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?
2 Answers
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 $t_0$.
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The Feynman propagator/kernel/amplitude is, as OP writes,
$$ \langle x_f ,t_f | x_i ,t_i \rangle~=~ \sqrt{\frac{m}{2\pi i\hbar} \frac{1}{\Delta t}} \exp\left[ \frac{im}{2\hbar}\frac{(\Delta x)^2}{\Delta t}\right], \tag{1}$$
where $\Delta x:=x_f-x_i$ and $\Delta t:=t_f-t_i > 0$. It is implicitly understood in eq. (1), that one should perform the Feynman $i\epsilon$ prescription. More precisely one should substitute $\Delta t\to\Delta t-i\epsilon$, i.e. the $\Delta t\in\mathbb{C}$ in eq. (1) is actually situated just below the real axis in the complex $\Delta t$ plane. This $i\epsilon$ prescription ensures that the propagator (1) becomes a Dirac delta distribution in the short time limit:
$$ \langle x_f ,t_f | x_i ,t_i \rangle ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^{+}.\tag{2}$$
As explained in Ref. 1 there is no absolute notion of probability since the position space is non-compact, but there is a relative notion of probability. The relative probability distribution in position space becomes uniform
$$\begin{align} P(x_f,t_f;x_i,t_i)~=~&|\langle x_f ,t_f | x_i ,t_i \rangle|^2\cr ~=~&\frac{m}{2\pi \hbar} \frac{1}{\sqrt{(\Delta t)^2+\epsilon^2}}\exp\left[ -\frac{m\epsilon}{\hbar}\left(\frac{\Delta x}{\Delta t}\right)^2\right]\cr \longrightarrow&~ \frac{m}{2\pi \hbar} \frac{1}{\Delta t}\quad \text{for} \quad \epsilon \to 0^{+}.\end{align}\tag{3}$$
Physically, this may be understood as that the corresponding relative probability distribution in momentum space is uniform as well. Phrased differently, a position eigenstate is a superposition of all momentum eigenstates, and it turns out that the particle can be near or far away with the same probability density.
For more on normalization of the path integral, see e.g. this Phys.SE post and links therein.
References:
- R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965, Problem 3.1.
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