meopemuk said:
However, it is important that by spending some extra efforts we can always improve particle localization, and there is no any natural limit for such an improvement. So, theoretically, the assumption of the possibility of perfect localization seems to be reasonable.
That doesn't follow. Lower bounds are not always reachable; you could make all the improvements you want, but you need to do something fundamentally different if you actually want to arrive at the conclusion that perfect localization makes sense!
If we agree that such perfectly localized states should be present in our formalism, then we should conclude that the number of mutually orthogonal vectors in the (Hilbert) space of states of our particle is uncountable.
If we agree that such perfectly localized
states should be present in a continuum formulation of QM, then the current mathematical tools are inadequate
But if we don't insist that they are actual
quantum states, then there is no technical problem with using deltas or other generalized functions.
f we sum up squares of all these (seemingly vanishing) amplitudes over all points in space (there is an infinite uncountable number of them) we should get exacly "1" - the probability of finding the particle somewhere. So, basically, we have here an uncertainty of the type "zero * infinity".
Or, we could treat the situation the way we normally would:
<br />
\int_{-\infty}^{+\infty}<br />
\left| \int_{-\infty}^{+\infty} \delta(x - y) f(x) \, dx \right|^2 \, dy = 1.<br />
Now, if you were going to invoke nonstandard analysis, and define things on a hyperfinite lattice (rather than on a continuum) whose points are separated by
1/H (for some transfinite
H), and range over the interval [-H, H], then, we have
<br />
\langle \varphi(x) \mid \psi(x) \rangle = <br />
\int_{-\infty}^{+\infty} \varphi(x)^* \psi(x) \, dx \approx<br />
\sum_{n = -H^2}^{H^2} \frac{1}{H} \varphi(n/H)^* \psi(n/H),
where the approximation symbol denotes the fact they are infinitessimally close. In particular, we have
\sum_{n = -H^2}^{H^2} \frac{1}{H} |\varphi(n/H)|^2 \approx 1.
Now, we could define a function
d(x) by:
<br />
d(x) =<br />
\begin{cases}<br />
H & x = 0 \\<br />
0 & x \neq 0<br />
\end{cases}<br />
If you have a standard point
a, then you can find a point
b of the lattice that is infinitessimally close to
a, and we have that
\langle d(x - b) \mid \varphi(x) \rangle \approx<br />
\sum_{n = -H^2}^{H^2} \frac{1}{H} d(n/H - b)^* \varphi(n/H)<br />
= \varphi(b) \approx \varphi(a)
You could take it's square root to get a unit vector, if you so desired. Then, it is true that
<br />
\begin{multiline}<br />
\begin{split}<br />
\sum_{n = -H^2}^{H^2} \left| \langle \sqrt{d(x - n/H)} \mid \varphi(x) \rangle \right|^2<br />
&=<br />
\sum_{n = -H^2}^{H^2} \left| \frac{1}{\sqrt{H}} \langle d(x - n/H) \mid \varphi(x) \rangle \right|^2<br />
\\&=<br />
\sum_{n = -H^2}^{H^2} \frac{1}{H} \left|\varphi(n/H) \right|^2<br />
\\&\approx 1.<br />
\end{split}<br />
\end{multiline}<br />
However, you are wrong on a different point: this sum must be taken over all the points of the lattice. In particular, each ordinary point
a has
lots of nonstandard points near it, and we must also sum over transfinite points, which are not near any standard point.
In other words, you cannot use nonstandard analysis to say that you have one basis vector per standard point.
