Perturbation theory question (in Quarks & Leptons)

gop
Messages
55
Reaction score
0
Hi

I'm referring to the book Quarks and Leptons (Halzen, Martin). On pages 79-82 nonrelativistic perturbation theory is investigated (i.e. by using the Schroedinger equation, which is first order in time). On Page 85, however, the transition amplitude (T_fi) is used that has been derived on pages 79-82. However, in this chapter (chapter 4) the Klein-Gordon equation is investigated (which is relativistic as well as second order in time).
How can this be right?

thx
 
Physics news on Phys.org
why should it be wrong?
 
Well , the derivation of T_fi (on pages 79-82) uses the fact that the Schroedinger equation is first order in time. If a second order equation is substituted, a differential equation for the coefficients (a_f) results that is quite different from the simple differential equation that results from a first order equation.

In addition it is written in the book that these derivation is (only) a recapitulation of non-relativistic perturbation theory. However, then it is used in the next chapter for the (relativistic) Klein Gordon equation (without any mention as to why this is applicable).

thx
 
Quantum mechanics is always first order in time. In QFT you need to distinguish between the field and the wave-function, which actually turns into a wave-functional. The equations of motion for the field is turned into a Lagrangian, and then into a Hamiltonian, and it is this Hamiltonian that generates the first-order equations for the wave-functional. It is customary, however, to starting hiding the state when talking about QFT, and instead refer only to the operators. In that case, the field is a field of operators, and one can write a 2nd order differential equation for it; states are then referred to by the operators used to construct them from the vacuum state.
 
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