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?
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1You can describe Gaussian wave-packet centered at $r_0$ with position variance $\sigma$, if you begin with initial condition $\psi_0(x_0,t_0) = \frac{1}{\sqrt{\sqrt{2\pi}\sigma}} e^{-\frac{(x_0-r_0)^2}{4\sigma^2}}$. You can then find its time evolution by calculating the integral below. – Ján Lalinský Dec 18 '13 at 04:28
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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Is it then also invalid to take $\psi$ in the integral to be the position operator eigenfunction? – BMS Dec 18 '13 at 00:57
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Position operator does not have normalizable eigenfunction. Delta distribution is often called its eigenfunction, but it is not normalizable. We can plug it in the integral, but the resulting $\psi$ won't be normalizable either. – Ján Lalinský Dec 18 '13 at 01:30
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Does that imply, that it is really IMPOSSIBLE for a particle to be in a position operator eigenstate? – André Dec 18 '13 at 09:08
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It is impossible if we work with the Born interpretation of $|\psi|^2$ as probability distribution.If we dropped this interpretation and understood $\psi$ in some other way, it could be possible for the delta distributions to play the same role as normalizable $\psi$'s. Or we could introduce new object, $\sqrt{\delta}$, to play the role of normalizable position eigenfunctions. But none of these seem to lead to much new understanding. One way to reconcile with this situation is to drop the idea that measurement of positions reduces the $\psi$ functions to eigenfunction of the operator $\hat x$. – Ján Lalinský Dec 18 '13 at 16:16
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Thank you very much for elucidating these issues! Do you happen to have any literature suggestions regarding this topic? I'd be particularly interested in seeing up to what level of mathematical rigor it is possible to introduce "normalizable position eigenstates" and what the implications of such an approach would be. – André Dec 18 '13 at 23:04
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In ordinary quantum theory, I do not know of of any good reference. In the quantum field theory, people talk about discretization much more - instead of continuous space, the theory is formulated for discrete lattice and the spacing is sent to zero only towards the end of the calculation. One important implication of this seems to be that once the space is discretized, it is hard to make the theory conform to the principle of relativity (lattice is not symmetric with respect to rotations, change of frame velocity brings length contractions of lattice etc.) – Ján Lalinský Dec 19 '13 at 01:36
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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