Transition Probability

[P3X-018]

At page 190 of Bransden & Joachain (see the page from here (http://books.google.com/books?id=i5IPWXDQlcIC&printsec=frontcover&dq=B.+H.+Bransden,+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 d\omega . Hence they can't be the same dimensionally. How can this be right?

[Cyosis]

I am not sure what the question is, but they are both dimensionless. Probability doesn't have a unit.

[P3X-018]

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 d\omega term which has the same power in both, hence (4.39) is short by a factor of 1/sec.

[phsopher]

In any case (4.38) is not the probability, it's the amplitude.

[P3X-018]

But it's supposed to be dimensionless like the probability, hence must have the same dimensions.

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

c_b = \int_0^{\infty} f(\omega) \, d\omega

while (4.39) has the form

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

which certainly can't be right. What's going on?

[Redbelly98]

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.

[P3X-018]

Loudon (http://books.google.com/books?id=AEkfajgqldoC&printsec=frontcover&dq=Rodney+Loudon&hl=da#PPA54,M1) avoids this problem by taking the sum over frequencies first after squaring the amplitude, but he still mentions, after eq. (2.3.11), that the integral should be over the excitation amplitudes.
How is the derivation then consistent?