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Particle in a Box

  1. Mar 1, 2006 #1


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    The wave function of an electron in a one dimensional hydrogen atom of width [tex]L = 2a_0[/tex] is [tex]\psi(x) = A_n sin (\frac{n \pi x}{L}) [/tex]

    If the particle is constrained to lie within the box, what is the value of [tex]A_1[/tex]?

    Is this a really easy problem? I might be banging my head for no reason here, I don't know. First of all, the question is ambiguous. What is this box they speak of? I am presuming it's a box of width L, but I don't know for sure. Why the value of just A1? I didn't think the max amplitude changed as n changed.

    I'm not looking for an answer here, I'd just like someone to guide me in the right direction.

  2. jcsd
  3. Mar 1, 2006 #2

    Meir Achuz

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    The integral of |\psi|^2 from 0 to L must equal one. Just let n=1 and integrate.
  4. Mar 2, 2006 #3
    In case this is where your confusion lies: n is just a quantum index. Specifically, it's an index of energy.

    As you will learn if you study QM, the energy of this system is quantized. This means you cannot have any arbitrary energy. Instead, only certain discreet values are allowed.

    This is where the n comes into play. As Meir Achuz already pointed out, when they ask you to find A1 they want you to find A when n=1.

    What they're asking you to do is essentially normalize the wave function.
  5. Mar 2, 2006 #4
    ek, another way to think about this is this:
    Because the particle is confined to an infinite potential well, it must be in that box between 0 and L. It cannot escape because the potential barrier at x=0 and x=L is infinite.

    So translating that into math, the probability of finding the particle between 0 and L is 1. So [tex]\int_0^L \psi^{*}\psi = \int_0^L |\psi|^2 = 1[/tex]
  6. Mar 2, 2006 #5


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    Thank you all. I got the answer. Much appreciated.
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