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An Integral question - PHYSICS (if someone does physics can you check it for me pls)

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Please I am just going to list the question then show my solving, It is a physics question but I need help on integration, on parts of the question, thank you

    quesiton:

    A particle in the infinite square well, (i.e [tex] V(x) = \left{ \begin{array}{ccc} 0 & \textrm({ if 0 <= x <= a} \\ \infty & otherwise \end{array}\right [/tex]
    has the initial wave function

    [tex] \Psi(x,0) = Asin^3(\frac{\pi x}{a} \textrm{ for (0<=x<=a) } [/tex]

    Determine A, find [tex] \Psi(x,t) \textrm{ and calculate <x> as a function of time, what is the expectation value of the energy? (hint: sin^n \theta and cos^n \theta can be reducted by repeated application of trigonometric sum formulas, to linear combinatiosn of sin(m\theta) and cos(m\theta), with m = 0,1,2....n }
    [/tex]

    2. Relevant equations

    [tex] 1 = \int_{-\infty}^{\infty} |\Psi(x,0)|^2 dx [/tex]



    3. The attempt at a solution

    My attempt, well first I normalize it,

    so the solution to the question is,


    [tex] \int_0^a |A sin^3(\frac{\pi x}{a}|^2 = 1 [/tex]

    now this brings up something I do not know how to solve,
    since the sin would go to the ^6 power

    I was told by my course instructor that there are many ways to solve integrals of this type,
    and the question states Sin^n theta can be reduced by repeated application of the trigonometric sum formulas, to linear combinations of sin

    so I have to use some trigonometric identities to solve/simplify the integral basically
     
    Last edited: Apr 30, 2010
  2. jcsd
  3. Apr 29, 2010 #2
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    Ok, so what is the problem you are having? Look up the trigonometric sum formulas and find one that matches sine squared.
     
  4. Apr 30, 2010 #3
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    I need to know the integral of sin^3 x

    to reduce it to sin^2

    [tex] \int sin^3x dx = \int [(sin^2 x dx)(sin x) [/tex]

    umm

    using the trigonometric identity (the only one I know, [tex] sin^2 x + cos^2 x = 1 [/tex]

    I'm going to substitute in [tex] 1-cos^2 x = sin^2x [/tex]

    so uh, [tex] sin^3 x = (1-cos^2 x) sin x [/tex]

    can I just move the sin to the other side? so it cancels out and becomes

    [tex] \frac{sin^3 x}{sin x} = (1-cos^2 x) [/tex]
    [tex] sin^2 x = 1-cos^2x [/tex]

    oh wow, it works and just becomes the trig identity lol
    jeese well that's helpful
    yesssssss

    now i make it simplier?
     
  5. Apr 30, 2010 #4
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    The problem in your original post had a sine squared, now you're working on sine cubed...which is it?

    The hint you posted says use the trig sum identities. If you don't know them, look them up.
     
  6. Apr 30, 2010 #5
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    so my equation

    [tex] \int_0^a sin^3\frac{\pi x}{a} [/tex]

    [tex] \int_0^a sin^2(\frac{\pi x}{a} = \int_0^a (1-cos^2\frac{\pi x}{a} ) [/tex]

    now I need to find a trig identity for sin^2 x to make it simplier
     
  7. Apr 30, 2010 #6
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    Sorry it was a typo

    it was A sin^3 x
     
  8. Apr 30, 2010 #7
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    You're confusing yourself. LOOK UP THE TRIGONOMETRIC SUM IDENTITIES.
     
  9. Apr 30, 2010 #8
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    [tex]
    \int_0^a sin^2(\frac{\pi x}{a}) = \int_0^a (1-cos^2(\frac{\pi x}{a} ))
    [/tex]

    Normalizing the wave function,
    [tex] \int_0^a \left| (1-cos^2(\frac{\pi x}{a})) \right|^2 =1 [/tex]

    i'm lost i've done something wrong

    i'm going to restart and do it properly
     
  10. Apr 30, 2010 #9
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    Found a trigonometric identity that I like

    [tex] sin 3\theta =3sin\theta -4sin^3 \theta [/tex]

    letting theta be my [tex] \frac{\pi x}{a} [/tex]

    [tex] sin 3(\frac{\pi x}{a}) = 3sin(\frac{pi x}{a})[/tex]

    re arranging it so sin^3 theta is the subject

    adding 4 sin^3 theta to each side

    [tex] sin 3\theta + 4sin^3 \theta = 3sin\theta [/tex]

    subtracting sin 3 theta from each side

    [tex] 4sin^3 \theta = 3sin\theta - sin 3\theta [/tex]

    diving the both sides by 4

    [tex] sin^3 \theta = \frac{3}{4} sin\theta - \frac{1/4}sin 3\theta [/tex]

    plugging back in my theta value I get

    [tex] sin^3 (\frac{\pi x}{a}) = \frac{3}{4}sin(\frac{\pi x}{a}) - \frac{1}{4} sin 3 (\frac{\pi x}{a}) [/tex]

    ok that seems better
     
    Last edited: Apr 30, 2010
  11. Apr 30, 2010 #10
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    now to normalize this wave function, I square it, and the absolute value =1

    [tex] \int_0^a \left| sin^3 (\frac{\pi x}{a}) \right|^2 dx= \left| \int_0^a \frac{3}{4}sin(\frac{\pi x}{a}) - \frac{1}{\4} sin 3 (\frac{\pi x}{a}) \right|^2 dx = 1[/tex]

    squaring each of the components
    Can someone check this part please, not sure how squaring in this context works

    [tex] \left( \frac{3}{4}sin(\frac{\pi x}{a}) - \frac{1}{\4} sin 3 (\frac{\pi x}{a}) \right)^2 [/tex]
    =
    [tex] \left( \frac{3}{4}sin^2(\frac{\pi x}{a})\right) - \left( \frac{1}{4} sin^2 (\frac{3\pi x}{a}) \right) [/tex]
    (can I just square the sin part? and leave the coefficients i.e 3/4 and 1/4 constant? without squaring them? )
     
  12. Apr 30, 2010 #11

    Cyosis

    User Avatar
    Homework Helper

    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    Hint:

    [tex](a+b)^2=a^2+b^2+2ab[/tex]
     
  13. May 2, 2010 #12
    Re: An Integral question - PHYSICS (if someone does physics can you check it for me p

    [tex] \left( \frac{3}{4}sin{\frac{\pi x}{a}} - \frac{1}{4}sin{3\frac{\pi x}{a} \right)^2 [/tex]

    =

    [tex] \left( \frac{3}{4}sin{\frac{\pi x}{a}} - \frac{1}{4}sin{3\frac{\pi x}{a} \right)\left( \frac{3}{4}sin{\frac{\pi x}{a}} - \frac{1}{4}sin{3\frac{\pi x}{a} \right) [/tex]

    =

    [tex] \left( \frac{3}{4}\frac{3}{4} sin^2{\frac{\pi x}{a} \right) + \left( \frac{1}{4}\frac{1}{4}sin^2 (3*3(\frac{\pi x}{a} ) ) \right) -2\left(\frac{1}{4}\frac{3}{4}sin^2 (3\frac{\pi x}{a} \right) [/tex]

    =

    [tex] \left( \frac{9}{14}sin^2 {\frac{\pi x}{a} \right) + \left( \frac{1}{16}sin^2 9\frac{\pi x}{a} \right) -2\left(\frac{3}{16} sin^2 3\frac{\pi x}{a} \right) [/tex]

    so my equation is

    [tex] 1 = A^2 \int_0^a \left[ \left( \frac{9}{14}sin^2 {\frac{\pi x}{a} \right) + \left( \frac{1}{16}sin^2 9\frac{\pi x}{a} \right) -2\left(\frac{3}{16} sin^2 3\frac{\pi x}{a} \right) \right]dx [/tex]

    what do I do from here?
     
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