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

nathatanu0
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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
 
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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.
 
it can be think as
for free particle \DeltaX=infinity
therefore, infinitely big box is not so big for free particle
 
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