Fermi's Golden Rule & Transition Probability

plasmon
Messages
36
Reaction score
1
As i studied Fermi golden rule. It involves transition probability of an initial state transiting to a final state in case of constant potential. As i understand the product of time of transition and the frequency of transition should be very much larger than one in order that the dirac delta function appears.

Does this mean that the time of particle collision(for e.g two particles come from past and collide with each other) should be infinite in order that the energy of final and initial states are same.
 
Physics news on Phys.org
It means that after waiting a reasonably large time after the transition the final state has the same energy as the initial state.
Due to the perturbation term in your Hamiltonian your final state is a mixture of many energy eigenstates with energy peaked around the value of the initial state. In time, the coefficients belonging to the other states tend to zero.

So the earlier after the transition you measure the final state's energy, the more uncertainty you will have in prediciting your result.
 
Take a look into the derivation of transition probability in the case of the constant potential. The derivation shows that the time of the measurement is actually the time during which the interaction potential was switched on. It does not include the time after the potential was switched off.

The inequality that results in Dirac delta function is.

(Interaction Time)(Transition Frequency)>>1

Transition frequency= Difference in energy of final and initial state divided by Planck's constant.

There are two time scales involved here.

(i) Interaction time (Strength of interaction).

(ii) Transition time.

Now Since the interaction time cannot be infinite. What is the justification of applying Fermi Golden rule on elastic particle collisions in colliders, where we assume that the initial and final energy of particle is the same.

I have an idea the inequality actually means that

Interaction time>>>>Transition time (Interaction Time not is not infinite)

So in the end, only those states survives having energy same as the initial states.
 
Last edited:
Could you provide a link to said derivation? You keep talking about a constant potential, but what you describe is in fact a time-dependend potential.
 
Here, is one of the many links.

http://moleng.physics.upatras.gr/personnel/Koukaras/download/FermiGR.pdf
 
Last edited by a moderator:
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top