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Decoherence and cross section

  1. Oct 21, 2009 #1

    naima

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    Hello

    I am reading this book: "[URL [Broken] and the appearance of a classical word&f=false"]decoherence[/URL]
    I hope someone could help me.
    Please go to Appendix A1
    Joos introduces the notation f(q,q'). What is the dimension of f from his notation?
    Next page he writes that [tex]\int f^2 d\Omega d\Omega'[/tex] is a cross section.
    do you understand why?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 21, 2009 #2
    Can't see anything like that at A1...
     
  4. Oct 21, 2009 #3

    naima

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    F(q,q') is defined in (A1.12)
    Have you access to the book or just with my link (google book)?
     
  5. Oct 22, 2009 #4

    Avodyne

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    The scattering amplitude f has dimensions of inverse length. This is basic scattering theory; see any book on QM for the relation between the scattering amplitude and the cross section.
     
  6. Oct 22, 2009 #5

    naima

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    Good.

    Lets us come to what follows.
    [tex] \rho (x,x')[/tex] is the density matrix and we have
    [tex] \Delta \rho (x,x') =- \rho (x,x') F(x-x') \Delta t [/tex]
    F has the dimension of 1/T
    Joos wants to prove that the density matrix diagonalizes as time grows.
    [tex] F(x-x') = \int dq n(q) v(q) \int \frac{d\Omega d\Omega'}{2}(1 -exp[i(q-q')(x-x')])|f(q,q')|^2[/tex]
    Then he writes: for large distances x-x', F approaches
    [tex] 2 \pi \int dq n(q) v(q) \sigma(q)[/tex]
    with [tex] \sigma(q) [/tex] the total cross section

    So my question is: if f(q,q') has the dimension 1/L how can we get a cross section (dimension of a surface)?

    Thanks
     
  7. Oct 22, 2009 #6

    Avodyne

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    Argh! Sorry, I meant length, not inverse length. You can see this from A1.12, where T is dimensionless, and [itex]\langle q'|q\rangle=\delta^3(q'-q)[/itex] has dimensions of length cubed (q' and q are three-vectors, each with dimensions of inverse length), and [itex]\delta(q'-q)[/itex] has dimensions of length (q' and q are now magnitudes of three-vectors).
     
  8. Oct 23, 2009 #7

    naima

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    Thank you Avodyne.

    Your answer is very clear!
    <p|p'> is a product of 3 dirac

    I did the same mistake. I cannot easily understand how an amplitude <p |T| p'> has the dimension of a volume (and a density of probability a L^6 dimension)
     
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