Should we think a free particle as a particle in an infinitely big box?

1. Sep 8, 2008

nathatanu0

I've found that the momentum expectation of a particle inside an one dimensional finite box of length 'L' is '0'... we can say the probability of going right=that of the left... so sum up to zero. Now I calculate the same(<p>) for a free particle in one dimension if I think that free particle is not exactly free but kept inside an infinitely long box(L-->infinity)... the result should be the same... does that happen.... does it make any sense to draw analogy between the two.... ?? please help.

atanu

2. Sep 8, 2008

Avodyne

This is a subtle question. For the finite box, I assume you required the wave function to vanish at the walls; in other words, you imposed the boundary conditions $\psi(0)=0$ and $\psi(L)=0$. But there is another kind of finite box: a circle instead of a line-segment. Then the wave function is required to be periodic: $\psi(x+L)=\psi(x)$. (Actually, you can have periodicity up to a phase instead,$\psi(x+L)=e^{i\phi}\psi(x)$, but I want to ignore that possibility.) In this case, you can have a free-particle wave function that has nonzero $\langle p\rangle$, such as $\psi(x)=L^{-1/2}\exp(2\pi i n x/L)$, where $n$ is an integer (positive, negative, or zero); then $\langle p\rangle=2\pi\hbar n/L$. We could now consider the limit of $L\to\infty$; if we also take $n\to\infty$, we end up with nonzero $\langle p\rangle$. But for the hard-wall box, we would always get $\langle p\rangle=0$. So the results depend on the boundary conditions, even for an infinite box.

3. Mar 4, 2009

sanjibghosh

it can be think as
for free particle $$\Delta$$X=infinity