Can you explain the working to this question? Probability - infinite potential well

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    Question 3, I copied the way to do it from the board last year but don't understand it. Can you explain why there's a subtitle 2 beside the psi? And explain why the working is worked out like this?

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    (Original post by Airess3)
    Question 3, I copied the way to do it from the board last year but don't understand it. Can you explain why there's a subtitle 2 beside the psi? And explain why the working is worked out like this?

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    Why  |\Psi|^2 ?

    It has to do with Born-interpretation of wavefunction.

     |\Psi|^2 is interpreted as the probability density that the quantum particle is at the position x.

    Then the probability of the quantum particle is found in the interval of  a \leq x \leq b is

     \int_a^b |\Psi(x)|^2 dx

    It seems that the wavefunction has something missing. See the link below.
    http://hyperphysics.phy-astr.gsu.edu...ntum/pbox.html

    You may want to ask your classmate for a copy of the solution. It seems that your integration is wrong.
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    (Original post by Eimmanuel)
    Why  |\Psi|^2 ?

    It has to do with Born-interpretation of wavefunction.

     |\Psi|^2 is interpreted as the probability density that the quantum particle is at the position x.

    Then the probability of the quantum particle is found in the interval of  a \leq x \leq b is

     \int_a^b |\Psi(x)|^2 dx

    It seems that the wavefunction has something missing. See the link below.
    http://hyperphysics.phy-astr.gsu.edu...ntum/pbox.html

    You may want to ask your classmate for a copy of the solution. It seems that your integration is wrong.
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    I asked the professor for some help but now I'm stuck on how he got from the 2nd last step to the last step (this is just part of the working):

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    How did he integrate that? Is there a specific formula that I can use? and how did he turn the 2/a into 1/a?*

    And now I've tried to do the rest of the working but the a's don't cancel out?

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    It's just normal integration. Note that there's a 2/a outside one bracket and a 1/2 immediately to its right. So multiplying them (during the integration step) gives 1/a. That's not affected by integration because x is the integration variable (not a).

    Regarding the a not cancelling: check your very last step.

    General tip, either use decimals or (better) only fractions. Something like '2*pi*(0.25a)/a' is unnecessarily complicated notation. (2*pi*a)/(4*a) is so much easier to clean up in a single step without a mistake. It's also faster to write in an exam.
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    (Original post by mik1a)
    It's just normal integration. Note that there's a 2/a outside one bracket and a 1/2 immediately to its right. So multiplying them (during the integration step) gives 1/a. That's not affected by integration because x is the integration variable (not a).

    Regarding the a not cancelling: check your very last step.

    General tip, either use decimals or (better) only fractions. Something like '2*pi*(0.25a)/a' is unnecessarily complicated notation. (2*pi*a)/(4*a) is so much easier to clean up in a single step without a mistake. It's also faster to write in an exam.
    Thanks, I understand it now. I cleaned up the last line and got 0.25-1/2pisin(0.5pi) and got 0.24.... and not the 0.091... it's supposed to be?
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    (Original post by Airess3)
    Thanks, I understand it now. I cleaned up the last line and got 0.25-1/2pisin(0.5pi) and got 0.24.... and not the 0.091... it's supposed to be?
    Make sure your calculator is in radian mode.
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    (Original post by Eimmanuel)
    Make sure your calculator is in radian mode.
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    Thanks! Got it now. *
 
 
 
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