Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 8, 2008 #1
    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.

  2. jcsd
  3. Sep 8, 2008 #2


    User Avatar
    Science Advisor

    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 [itex]\psi(0)=0[/itex] and [itex]\psi(L)=0[/itex]. But there is another kind of finite box: a circle instead of a line-segment. Then the wave function is required to be periodic: [itex]\psi(x+L)=\psi(x)[/itex]. (Actually, you can have periodicity up to a phase instead,[itex]\psi(x+L)=e^{i\phi}\psi(x)[/itex], but I want to ignore that possibility.) In this case, you can have a free-particle wave function that has nonzero [itex]\langle p\rangle[/itex], such as [itex]\psi(x)=L^{-1/2}\exp(2\pi i n x/L)[/itex], where [itex]n[/itex] is an integer (positive, negative, or zero); then [itex]\langle p\rangle=2\pi\hbar n/L[/itex]. We could now consider the limit of [itex]L\to\infty[/itex]; if we also take [itex]n\to\infty[/itex], we end up with nonzero [itex]\langle p\rangle[/itex]. But for the hard-wall box, we would always get [itex]\langle p\rangle=0[/itex]. So the results depend on the boundary conditions, even for an infinite box.
  4. Mar 4, 2009 #3
    it can be think as
    for free particle [tex]\Delta[/tex]X=infinity
    therefore, infinitely big box is not so big for free particle
  5. Mar 5, 2009 #4
    There is no reason to suppose that the (discrete set of) energy eigenstates of the square well will tend to the (continuum of) energy eigenstates for a free particle, but it is possible to write a general state as a sum of energy eigenstates and have this sum tend to an integral, and take apart the sines into e^ i k x and e ^ - i k x to see that the particle in an infinitely large box is like a free particle which has reflected off of infinity and so is in a superposition of +k and -k momenta. Such is nonrelativistic QM!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Should we think a free particle as a particle in an infinitely big box?
  1. Particle in Box (Replies: 3)

  2. Particle in a box (Replies: 3)