Why does the amplitude <p |T| p'> have the dimension of a volume?

[naima]

Hello

I am reading this book: "[URL 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 $$\int f^2 d\Omega d\Omega'$$ is a cross section.
do you understand why?

 

[mikeph]

Can't see anything like that at A1...

 

[naima]

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

 

[Avodyne]

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.

 

[naima]

The scattering amplitude f has dimensions of inverse length.

Good.

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

 

[Avodyne]

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

 

[naima]

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)