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Homework Help: Transition Probability

  1. May 24, 2009 #1
    At page 190 of Bransden & Joachain (see the page from http://books.google.com/books?id=i5...nsden,+Charles+Jean+Joachain&hl=da#PPA190,M1"), there are 2 expressions for the transition probability, (4.38) and it's absolute value squared in (4.39).
    Is it just me or are the 2 term dimensionally different? Obviously everything from (4.38) is squared in (4.39) except the [itex] d\omega [/itex]. Hence they can't be the same dimensionally. How can this be right?
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. May 24, 2009 #2


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    I am not sure what the question is, but they are both dimensionless. Probability doesn't have a unit.
  4. May 24, 2009 #3
    Cyosis: Yes they are supposed to be dimensionless, but I just looked at the difference between the 2 expressions.

    Note that every term from (4.38) appears squared in (4.39) except for the [itex]d\omega [/itex] term which has the same power in both, hence (4.39) is short by a factor of 1/sec.
  5. May 24, 2009 #4
    In any case (4.38) is not the probability, it's the amplitude.
  6. May 24, 2009 #5
    But it's supposed to be dimensionless like the probability, hence must have the same dimensions.

    Ok to make it more clear, then (4.38) has the form

    [tex] c_b = \int_0^{\infty} f(\omega) \, d\omega [/tex]

    while (4.39) has the form

    [tex] |c_b|^2 = \int_0^{\infty} |f(\omega)|^2 \, d\omega [/tex]

    which certainly can't be right. What's going on?
    Last edited: May 24, 2009
  7. May 24, 2009 #6


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    That's weird, it certainly does look dimensionally inconsistent, regardless of what cb represents.

    I'm assuming ω means what it usually does and has units of s-1.
  8. May 25, 2009 #7
    Last edited by a moderator: Apr 24, 2017
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