PROFESSOR: So what do we do?
We are going to sum over final states
the probability to go from i to final at time t0
to first order.
Since the sum of our final states is really a continuum,
this is represented by the integral
of the f i t0 1, multiplied by the number of states
at every little interval.
So this will go rho of Ef dEf.
So this is what we developed about the number of states.
So I'm replacing this--
I have to sum but I basically decide
to call this little dN, the little number of states
in here, and then I'm going to integrate this probability, so
the number of states over there, and therefore the dN
is replaced by rho times dEf.
So then this whole transition probability will be 4 integral,
I'm writing now the integral, VfI squared, sine squared,
omega f i t0 over 2 Ef minus Ei squared, rho of Ef, d of Ef.
And you would say at this moment,
OK, this is as far as you go, so that
must be Fermi's golden rule, because we
don't know rho of Ef, it's different in different cases,
so we have to do that integral and we'll get our answer.
But the great thing about this golden rule
is that you can go far and you can do the integral.
Now I don't even know, this VfI also depends on the energy,
how am I ever going to do the integral?
That seems outrageous.
Well, let's try to do it, and part of the idea
will be that we're going to be led to the concept
that we already emphasized here because
of this suppression, that only a narrow band of states
contribute, and in that narrow band,
if the narrow band is narrow enough, in that region
VfI high be approximately constant
in a narrow enough region, and rho
will be approximately constant.
So we'll take them out of the integral,
do the rest of the integral, and see later
whether the way we're doing the integral
shows that this idea is justified.
So I'll just-- you know, sometimes you
have to do these things, of making the next step,
so I'll do that.
I'll take these things out, assuming they're
constant enough, and then we'll get 4 VfI squared, rho of E,
what should I put here?
E sub i, is that right?
Because if it's all evaluated at the initial energy Ei,
if only a narrow band will contribute,
I'll put an h squared here so that this will become omega fi,
and now I will integrate over the sum range of energies
the function sine squared omega f t0 over 2,
over omega fi squared dEf.
So I just took the thing out of the integral
and we're going to hope for some luck here.
Whenever you have an integral like that it probably
is a good idea to plot what you're integrating
and think about it and see if you're going
to get whatever you wanted.
Look, I don't know how far I'm going to integrate,
I probably don't want to integrate
too far because then these functions that I took out
of the integral are not constants,
so let's see what this looks like,
the integrand, this function here.
Well, sine squared of x over x squared
goes to 1, you know when--
this we're plotting as a function of omega fi.
Why?
Time is not really what we're plotting into this thing,
we're plotting-- we're integrating our energy,
Ef, omega fi is Ef minus Ei, so omega is the variable you
should be plotting, and when omega goes to zero,
this whole interval goes like t0 squared over 4,
and then sine squared of x over x squared does this thing,
and the first step here is 2 pi over t0, 2 pi over t0
and so on.
And now you smile.
Why?
Because it's looking good, this thing.
First what's going to be this area?
Well, if I look at this lobe, roughly, I
would say height t0 squared with 1 over t0, answer proportional
to t0.
This whole integral is going to be proportional to t0.
The magic of the combination of the x squared growth,
t0 squared and the oscillation is
making into this integral being linear
in t0, which is the probability the transition [INAUDIBLE]
is going to grow linearly is going
to be a rate, as we expected.
So this is looking very good.
Then we can attempt to see that also most of the contribution
here happens within this range to the integral.
If you look at the integral of sine squared x over x squared,
90% of the integral comes from here.
By the time you have these ones you're up to 95%
of the integral.
Most of the integral comes within those lobes.
And look what I'm going to say, I'm going to say,
look, I'm going to try to wait long enough,
t0 is going to be long enough so that this narrow thing is going
to be narrower and narrower and therefore most of the integral
is going to come from omega fi equal to 0,
which means the f equals to Ei.
If I wait long enough with t0, this is very narrow,
and even all the other extra bumps are already
4 pi over t0 over here is just going
to do it without any problem, it's going to fit in.
So another way of thinking of this
is to say, look, you could have argued that this is going
to be linear in t0 if you just change variables here, absorb
the t0 into the energy, change variable,
and the t0 will go out of the integral in some way,
but that is only true if the limits go from minus infinity
to plus infinity.
So I cannot really integrate from minus infinity to plus
infinity in the final energies, but I don't need
to because most of the integral comes from this big lobe here,
and if t0 is sufficiently large, it is really within no energy
with respect to the energy, Ei.
So our next step is to simply declare
that a good approximation to this integral
is to integrate the whole thing from minus infinity
to infinity, so let me say this.
Suppose here in this range omega fi is in between 2 pi over t0
and minus 2 pi over t0.
What does this tell us that omega fi is in this region?
Well, this is Ef minus Ei over h bar so this actually tells you
Ef is in between Ei plus 2 pi h bar over t0,
and Ei minus 2 pi h bar over t0.
All right, so this is the energy range
and as t0 becomes larger and larger, the window for Ef
is smaller and smaller, and we have energy conserving.
So let's look at our integral again, the integral is I,
that's for the integral, this whole thing,
will be equal to integral dEf sine squared
of omega fi t0 over 2 over omega fi squared.
So what do we do?
We call this a variable, u, equal omega fi t0 over 2,
so that du is dEf t0 over 2 h bar,
because omega fi is Ef minus Ei, and Ef is
your variable of integration.
So you must substitute the dEf here
and the rest of the integron.
So what do we get from the dEf and the other part?
You get at the end h bar t0 over 2 integral from--
well, let's leave it, sine squared u over u squared du.
So look at this, the omega fi squared,
by the time you get here omega fi goes like 1 over time,
so when it's down here we'll give you a time squared,
but the dE gives you 1 over time so at the end of the day
we get the desired linear dependence on t0 here,
only if the integral doesn't have t0 in here,
and it will not have it if you extend it from minus infinity
to infinity.
And there's no error, really, in extending it
from minus infinity to infinity because you basically
know that n lobes are going to fit here
and are going to be accurate, because there is little energy
change if t0 is large enough.
If t0 is large enough, even a 20 pi h bar and a 20 by h bar
here, that still will do it.
So we integrate like that, we extend it,
and we get this whole integral has value pi,
so we get h bar t0 pi over 2, that's our integral, I.
So our transition probability, what is it?
We have it there, over there, we'll
have the sum over final states, i to f of t0, first order
is equal to the integral times this quantity,
so that quantity is h t0 pi over 2,
so it's 4, what do we have, VfI squared,
rho of Ei over h squared, then h bar t0 pi over 2.
So your final answer for this thing
is 2 pi over h bar VfI squared, rho of Ei t0.
So let's box, this is a very nice result,
it's almost Fermi's golden rule by now.
Let's put a time t here, t0 is a label,
not to confuse our time integrals or things like that,
so we could put the time, t, here,
is 2 pi over h bar VfI squared rho of Ei t.
From here we have a transition rate,
so a transition rate is probability of transition
per unit time, so a transition rate
would be defined as the probability of transition
after a time t, divided by the time t that has elapsed,
and happily, this has worked out so that our transition rate,
w is 2 pi over h bar VfI squared rho of Ei,
and this is Fermi's golden rule, a formula
for the transition rate to the continuum of final states.
You see, when I see [INAUDIBLE] it almost
seemed you still have to integrate,
there is a rho of E and let's integrate [INAUDIBLE]
but the interval has been done and it says transmission
amplitude squared evaluated at the state initial
and final with the same energy and final state,
and the rho evaluated at the energy of the initial state.
You don't have to do more with that.
So we have this formula, let's look at a couple more things.
Do units work out?
Yes, this is transition per unit, this is 1 over time,
this is energy squared, this is 1 over energy,
and this is an h bar, this will give you 1 over time,
so this thing goes well.
How about our assumptions?
This was calculated using some time t, we used to call it t0.
How large does it have to be?
Well, the larger it is the more accurate the integral is,
but you don't want to take it too large,
either, because the larger it is,
the transition probability eventually
goes wrong at first order of perturbation theory.
So this argument is valid if there is a time, t0,
that is large enough so that within this error bars, rho
and the transition matrix elements are constant so
that your integral is valid.
But this t0 being large enough should be small enough
that the transition probability doesn't become anywhere near 1.
That will happen in general or if VfI is sufficiently small,
so when VfI is sufficiently small, this will always hold,
and in physical applications this happens and it's OK.
So that's our presentation and derivation
of Fermi's golden rule, and we will turn now
to one application and we will discuss.