0

In case of the solution of Schrödinger equation in particle in a box we find that the probability of finding the particle at the middle of the box is always zero (for $n$ even) and the probability distribution has a node there.

Say initially the wave function is a superposition of all even-$n$ states. So we have a finite probability to find the particle on each side of the box but not at the middle therefore if I measure the particles position now and find it at the right side then after long time if I take another measurement and find the particle on the left side, I can't say that it has passed to the left side in the meantime because at the middle the particle can never be found.

So, my question is

  1. How can I find the particle on both sides but not at the middle, as if it is found to be present at both sides at two instances, it must pass through the centre?
  2. If this is not the case then what is the description of a particle in quantum mechanics? I mean which properties should a physical quantity have to be called a particle?
Qmechanic
  • 201,751
user157588
  • 854
  • The solution for the lowest energy state $\psi_1(x)\sim \sin(\pi x/L)$ for a well extending between $0$ and $L$. The probability density at $x=L/2$ is actually maximum, not $0$. More generally, $\psi_{n}(x)\sim \sin(n\pi x/L)$ will be max at $x=L/2$ every time $n$ is odd. The probability density is thus maximal at $x=L/2$ whenever $n$ is odd, although the density can also reach the maximal value at other points. In any case, it is certainly not $0$ at $x=L/2$ when $n$ is odd. – ZeroTheHero Oct 01 '17 at 16:17
  • I've edited my question,completely overlooked that – user157588 Oct 01 '17 at 16:21
  • You don't want to say "even" states either since the solutions are not even or odd w/r to the transformation $x\to -x$ (unless you choose to make your potential from $-L$ to $L$, which is not standard.) – ZeroTheHero Oct 01 '17 at 16:27
  • By even i mean n=2,4,6.... – user157588 Oct 02 '17 at 14:19
  • You might want to clarify as such in your question. – ZeroTheHero Oct 02 '17 at 14:53
  • I have removed the third sub-question as it is rather far removed from the first two. Feel free to ask it separately, though. – Emilio Pisanty Oct 02 '17 at 15:16
  • See also the multiple other duplicates of the existing questions, accessible via the Linked sidebar on the right. – Emilio Pisanty Oct 02 '17 at 15:25

0 Answers0