Problems on quantum field operators in QFT

Dyson
Messages
21
Reaction score
0
Hello!

I met some annoying problems on quantum field operators in QFT.They are as follows:

(1)The quantum field operator( scalar field operator, for example),is often noted as

φ(r,t). Can it be interpreted as like this: φ(r,t) is the coordinate represetation of a

more abstract,theoretical operator φ in Hilbert space?

(2)Is there exist the represetation(such as momentum, coordinate...) which is similar to

that in quantum mechanics in QFT?

Thanks for reply!
 
Physics news on Phys.org
Regarding 1) no, r is an "index", not an entity living in a Hilbert space. This becomes clear if one derives quantum field theory of a scalar field from infinitly many coupled harmonic oscillators.

Regarding 2) yes, these are the so-called creation an annihilation operators obtained by Fourier transform of the field operators, usually denoted as a^\dagger(p) and a(p)
 
tom.stoer said:
Regarding 1) no, r is an "index", not an entity living in a Hilbert space. This becomes clear if one derives quantum field theory of a scalar field from infinitly many coupled harmonic oscillators.

Regarding 2) yes, these are the so-called creation an annihilation operators obtained by Fourier transform of the field operators, usually denoted as a^\dagger(p) and a(p)

Thank you very much!
 
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