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To: 128a1-ax@socrates.berkeley.edu
cc: murayama@physics.berkeley.edu
Subject: Re: density of final states
Reply-to: murayama@lbl.gov
In-reply-to: Your message of "Fri, 12 Dec 1997 15:06:56 PST."
Date: Fri, 12 Dec 1997 16:42:57 -0800
From: Hitoshi Murayama
> Professor,
>
> This is what I originally wanted to ask you in office hours:
>
> In the nonrelativistic case, does d(Phi) =
>
> / 3 3 \
> | _____ d p_i d x_i | 3 3 / ___ \
> | | | ------------- | * (2*pi) * delta | p_tot - > p_j | ?
> | | | 3 | \ --- /
> \ i (2*pi*h_bar) / j
>
> Unfortunately, it came out like "Does d(Phi) still have a delta
> function?", and then you made some comments that blew me away, like
> " has a delta function in it." and "
>
> ___ _____
> > = | | (Integral of dp_i dx_i)
> --- | | --------
> f i
> "
>
> I still don't fully understand them, but anyway, I also wanted to ask what
> it is exactly that you sum when you sum over all final state
> configurations. Are you talking about summing |M|^2 to get sigma? And in
> the discrete case, is there an analogous expression using the Kronecker
> delta?
What you say is right. We sum over all possible final states with the
weight |M|^2 to obtain the total cross section sigma. I'm sorry to
have you "blown you away," by saying that the quantum amplitude
= i M (2pi)^4 delta^4(p_f - p_i),
already contains the delta function. Since any transition must
conserve momentum and energy, the delta function appears as a part of
the matrix element. You are also right to point out that the
Kronecker delta appears when you are dealing with discrete states.
> Also, when I talked about d(Phi) as a probability density,
> Cohen-Tannoudji talked in general about a transition from a discrete
> energy state to a continuum of energies, and he did talk about density of
> final states as if it were the probability density of finding a state
> there. Or maybe I got that idea from Professor McEuen last semester.
> I'll have to read it more carefully.
Correct.
> I think I tried to come up with an example about a photon kicking
> an atom out of the (discrete) bound states into the (continuous) free
> states, and you said that it can always get kicked out there, so the only
> probability density is |M|^2, and d(Phi) doesn't weigh into that, I think.
> But isn't the density of final states constant in that case? What if the
> final states are still bound states, but they're so close together that
> they look continuous?
Well, what matters is the product |M|^2 d(Phi). If photon comes in to
kick out an electron from a bound state, you sum over all possible
final continuum states of the electron to calculate the transition
rate. The sum involves the momentum integral over the final state
(continuum) electron: d(Phi). But you also need the matrix element M
which contains information about the initial electron wave function in
the bound state, strenght of the photon interaction, etc. If you
think about a different process knocking out electron from the bound
state, such as the scattering by a neutrino beam, |M| would be
different, even though you still need the same sum over the possible
electron final states. The phase space integral d(Phi) only knows
what states are out there with given momentum and energy; basically
determined by kinematics only. On the other hand, |M|^2 in general
contains information on the nature of particles, nature of
interactions etc.
If final states are discrete but pretty much continuous, you can
pretend that the sum over the final states is an integral.
Hitoshi Murayama
Assistant Professor of Physics
University of California
Berkeley, CA 94720