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    Hi, I really can't get my head around how to work out particle in a box questions, could anyone help me with the following please

    Determine the probability of finding a particle in a 1-D box of size L in a region of size 0.01L at the locations x = 0, 0.25L, 0.5L, 0.75L and L when it is in its ground state. As percentages

    I've tried using the following equations:

    LaTeX Code: \\Psi (x)=LaTeX Code: \\sqrt{\\frac{2}{L}} sin(LaTeX Code: \\frac{nxPI}{L} )
    P=LaTeX Code: \\Psi (x,t)^2

    I'm not even sure if i've got the right equations there, i've tried loads of different ways of doing this and cannot get the right answers, which are 0%,1%,2%,1%,0%

    Cheers
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    Thats almost more of a philosophical question than mathematical. I would be tempted to say the probability is 0, as I don't think its conceptually possible to have a particle in 1D space, as particles are 3D in nature. But hey, I'm not at uni studying physics, so don't listen to me.
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    (Original post by yodude888)
    Thats almost more of a philosophical question than mathematical. I would be tempted to say the probability is 0, as I don't think its conceptually possible to have a particle in 1D space, as particles are 3D in nature. But hey, I'm not at uni studying physics, so don't listen to me.
    No there are mathematical answers to it, I just don't understand them lol. Cheers for the reply though
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    It's been a while since I did one like this, but isn't the formula for probability at a point x given by
    P(x)=(2/L).Sin^2(n.Pi.x/L)
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    Determine the probability of finding a particle in a 1-D box of size L in a region of size 0.01L at the locations x = 0, 0.25L, 0.5L, 0.75L and L when it is in its ground state
    Start with the wavefunction of the particle

     \phi(x) = \sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})

    The probability density is real space is found from this by

     p(x) = |\phi(x)\phi*(x)| = (\phi(x))^2 in this case as the expression is entirely real. (* deontes complex conjugate)

    Therefore

     p(x) = \frac{2}{L}sin^2(\frac{n\pi x}{L})

    This is the probability of finding the particle at position x. You want to know the probability of it being in some region, so you need to integrate over this region. In each case the region is 0.01L, as this is stated in the question. So for the first part the answer is

     p(0 \rightarrow 0.01L) = \int^{0.01L}_{0} \frac{2}{L}sin^2(\frac{n\pi x}{L}) dx

    with n = 1 as you're in the ground state.

    For the other parts, when integrating choose limits that are 0.005L above and below the point which you have been told to consider, e.g. integral limits for 0.25L should be 0.255L and 0.245L. This has the effect of creating a box in space that is centered on the point you have been told to consider and is 0.01L long.

    This is not done for x = 0 or x = L as the probability of the particle being outside the box (i.e. at x<0 or x>L) is 0 (I am here assuming it is an infinite potential outside the box; if not, then ignore this bit) and hence there is no point integrating there.

    Hope that makes some sense!
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    Forgot to say - to get as a percentage, divide each answer by the integral over the entire box i.e.

    \int^{L}_{0} \frac{2}{L}sin^2(\frac{n\pi x}{L}) dx
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    (Original post by spread_logic_not_hate)
    Start with the wavefunction of the particle

     \phi(x) = \sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})

    The probability density is real space is found from this by

     p(x) = |\phi(x)\phi*(x)| = (\phi(x))^2 in this case as the expression is entirely real. (* deontes complex conjugate)

    Therefore

     p(x) = \frac{2}{L}sin^2(\frac{n\pi x}{L})

    This is the probability of finding the particle at position x. You want to know the probability of it being in some region, so you need to integrate over this region. In each case the region is 0.01L, as this is stated in the question. So for the first part the answer is

     p(0 \rightarrow 0.01L) = \int^{0.01L}_{0} \frac{2}{L}sin^2(\frac{n\pi x}{L}) dx

    with n = 1 as you're in the ground state.

    For the other parts, when integrating choose limits that are 0.005L above and below the point which you have been told to consider, e.g. integral limits for 0.25L should be 0.255L and 0.245L. This has the effect of creating a box in space that is centered on the point you have been told to consider and is 0.01L long.

    This is not done for x = 0 or x = L as the probability of the particle being outside the box (i.e. at x<0 or x>L) is 0 (I am here assuming it is an infinite potential outside the box; if not, then ignore this bit) and hence there is no point integrating there.

    Hope that makes some sense!
    Thanks for the detailed reply, it's making more sense than it was before.
    Is below the right way to go, and if so what do I do with the 'L's

      \frac{2}{3}sin^3(\frac{\pi 0.255}{L}) dx  -  \frac{2}{3}sin^3(\frac{\pi 0.245}{L}) dx
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    Thanks for the detailed reply, it's making more sense than it was before.
    Is below the right way to go, and if so what do I do with the 'L's
    The L's will disappear when you put the integral limits in. I'll work through one for clarity:

     \int^{0.01L}_{0} \frac{2}{L}sin^2(\frac{n\pi x}{L}) dx

    Use  sin^2y = \frac{1-cos2y}{2} to get

     \frac{1}{L} \int^{0.01L}_{0} [1-cos(\frac{2n\pi x}{L})] dx

    which gives

     \frac{1}{L} [x - \frac{L}{2n\pi}sin(\frac{2n\pi x}{L})]^{0.01L}_0

    Now put in the limits to get

     \frac{1}{L} [0.01L - \frac{L}{2n\pi}sin(\frac{2n \pi \times 0.01L}{L})]

    We can now factorise out one of the L's, and cancel it with the one at the front. The other L (in the argument of the sin part) will cancel with that introduced in the limits, giving

     0.01 - \frac{1}{2n\pi}sin(2n \pi \times 0.01)

    Now put in n = 1 and work this out - thats almost the answer. To get the percentage, do the same integral but with the limits L and 0 (i.e. over the entire box) and divide this answer by that one.
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    That's beautiful. Maths has always been my weak point. Thanks again for all the help
 
 
 
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