A Field operator eigenstates & Fock states(Hatfield's Sch rep)

ShayanJ
Science Advisor
Insights Author
Messages
2,801
Reaction score
606
In chapter 10 of his book "Quantum Field Theory of Point Particles and Strings", Hatfield treats what he calls the Schrodinger representation of QFT. He starts with a free scalar field and introduces field operators ## \hat \varphi(\vec x) ## and its eigenstates ## \hat \varphi(\vec x)|\phi\rangle=\phi(\vec x)|\phi \rangle ##. Then he says that the "coordinate" representation of the field state ## |\Psi\rangle ## is given by the wave-functional ## \Psi[\phi]=\langle \phi |\Psi\rangle##. I have two questions about this:

1) Is the ground state of the field, one of the eigenstates of ## \hat \varphi(\vec x) ##, i.e. is ## \hat \varphi(\vec x)|0\rangle=0 ## correct?( At the bottom of page 224 of his book "A modern introduction to quantum field theory", Maggiore calls ## \phi(\vec x)=0 ## the vacuum. It seems to me its related to my question. Is he correct or just being sloppy? )

2) What is the relationship between the Fock states that are created by ## \hat a_k ## and ## \hat a_k^\dagger ## and their superpositions that are created by ## \hat \varphi(\vec x) ## and ## \hat \varphi^\dagger(\vec x) ## and the eigenstates of ## \hat \varphi(\vec x) ##? Can we say that the eigenstates of ## \hat \varphi(\vec x) ## are some kind of a coherent state because ## \hat \varphi(\vec x) ## is an anihilation operator for Fock states?

Thanks
 
Physics news on Phys.org
I refer you to the book by Robin Ticciati "QFT for Mathematicians" chapter 2, hopefully after reading this chapter you'll understand your misconceptions. (I don't have time to copy the answer here, I am studyding for Stat Mech... :-)).
 
  • Like
Likes ShayanJ
Thanks. Ticciati's formula for the ground state was surprising for me. But there was no answer to my second question. Or at least I couldn't find it!
 
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