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Wave packet - Integration by parts

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi ,

    I am reading a little on introductory QM , initial chapters on waves.
    They have given an integral for a wavepacket , assuming at t= 0.
    Which is: ψ(x,0) = [itex]\int A cosk'x dk'[/itex] (I don't know how to define limits to the integral in Latex upper = k+Δk , lower limit = k-Δk)
    which gives ψ(x,0) = S(x)coskx , where S (x) = 2AΔK sin(Δkx)/(Δkx).

    2. Relevant equations

    Integration by parts ?

    3. The attempt at a solution
    Now I am totally brain stuck , it surely can't be anything complex so I should be able to do it by parts.
    Let u =cosk'x and dv = dk'

    Or am I overlooking something ?
     
  2. jcsd
  3. Sep 8, 2012 #2

    vela

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    Yeah, it's just a straightforward integration. Remember x acts like a constant here because you're integrating with respect to k'. When you integrate cos k'x, you get sin k'x/x.
     
  4. Sep 8, 2012 #3

    vela

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    You can look at the code in this post to see how to write the integral with limits.
    [tex]\int_{k-\Delta k}^{k+\Delta k} A\cos k'x\,dk'[/tex] You also want a backslash before cos to get the function name to typeset correctly, and a little space (\,) before dk' makes it look better.
     
  5. Sep 8, 2012 #4
    It really is simple. x is constant during integration.
     
  6. Sep 16, 2012 #5
    Thanks for the responses everyone but I still don't get it.
    [tex]\int_{k-\Delta k}^{k+\Delta k} A\cos k'x\,dk'[/tex]
    As it has been mentioned that X and A are constant, so I end up with :

    Asink'x/x but when I plug in the limits I get:

    2Asin(Δkx)/x... how do they end up with the cos. identity ?
     
  7. Sep 16, 2012 #6

    vela

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    After you integrate, you have
    $$\left.\frac{A}{x}\sin k'x\right|_{k-\Delta k}^{k+\Delta k} = \frac{A}{x}[\sin ((k+\Delta k)x) - \sin((k-\Delta k)x)].$$ How did you manage to go from this last expression to what you got?
     
    Last edited: Sep 16, 2012
  8. Sep 16, 2012 #7
    Yes. I did get the above step after which I expanded the brackets , which resulted in the cancelling of -kx +kx terms , right ?
     
  9. Sep 16, 2012 #8

    vela

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    Are you claiming that sin(a+b)-sin(a-b) = 2 sin b by simply subracting the arguments of the sines?
     
    Last edited: Sep 16, 2012
  10. Sep 16, 2012 #9
    Oh no, I was greatly mistaken !
    I just realized sum formula is all I need....

    I feel so stupid :

    Just how the hell am I going to become a physicist if I keep doing such silly mistakes...
     
  11. Sep 16, 2012 #10
    Ok.. so I have just used the identified that I previously mentioned.. I still don't end up with the form in the OP.


    $$\left.\frac{A}{x}\sin k'x\right|_{k-\Delta k}^{k+\Delta k} = \frac{A}{x}[\sin ((k+\Delta k)x) - \sin((k-\Delta k)x)].$$

    Now sin (a+b) = sinAcosB+sinBcosA , sin (a-b) = sinAcosB-sinBcosA

    I used the above to only end up with

    [itex]2A sin(Δkx) coskx / x [/itex]

    They seem to have two extra Δk for some reason..
     
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