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Nov
4
answered Can we regard field operator $\Psi (x)$ as $a_{x}^{\dagger }$ ,$a_{x}$?
Nov
4
comment Are there QFTs in which a field cannot produce a real particle?
Howard Georgi has written some articles over the last few years about "unparticle physics".
Nov
4
comment Statistics of many body systems in pure states
@lionelbrits: I'm not sure why one cannot pose a similar question in the quantum mechanics of one mode (single "particle", but multiple levels). Maybe I don't understand the essence of your question then.
Nov
3
comment Statistics of many body systems in pure states
If I made a caricature of the problem (Single mode, 2 levels (fermionic), high temperature limit so approximately equal probabilities and "maximum" thermal entropy), you're asking about the entropy of $|\psi\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}$ vs $|\psi\rangle \langle\psi| = \frac{1}{2}|0\rangle \langle0| + \frac{1}{2} |1\rangle \langle 1| $
Nov
3
comment Statistics of many body systems in pure states
What you're thinking about is the difference between states defined as a coherent sum -vs- a decoherent sum :-? I wonder if the coupling of the modes to the cavity will "decohere" the system into a mixed state rather than a pure state (by "dissipating" off-diagonal elements) -- I think this might happen if the the modes are eigenmodes of the free theory but not the interacting theory.
Nov
2
comment Spontaneous symmetry breaking to subspace not giving massless bosons
BTW (general comment), you would get 2 massive bosons and one massless if the Higgs were in the adjoint (triplet) representation of $SU(2)$. It's easy to see that from similarity to $SO(3)$ -- any direction is left invariant under a particular axis of rotation. Since experiments show 2+1 massive bosons and 1 massless, we need an extra U(1), but to give mass to 3 bosons, the Higgs should be in the fundamantal (doublet) representation of $SU(2)$.
Nov
2
comment What is the meaning of a state in QFT?
BTW, we needn't necessarily talk of relativistic theories. QFT is perfectly applicable in non-relativistic contexts as well, eg: cond-mat.
Nov
2
revised What is the meaning of a state in QFT?
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Oct
31
answered What is the meaning of a state in QFT?
Oct
31
comment Is it THE Higgs?
To add to Anna's answer, it seems like it's the Higgs of the Standard Model, but it's quite plausible that there will be other "similar" particles in some viable extensions of the Standard Model.
Oct
30
revised Renormalization, integrating out high momenta Wilson way
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Oct
30
revised Renormalization, integrating out high momenta Wilson way
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Oct
30
revised Renormalization, integrating out high momenta Wilson way
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Oct
30
answered Renormalization, integrating out high momenta Wilson way
Oct
30
comment How does path integral formulation explain bound states?
The question I posed above is problematic to me, since the final state is not a "fluctuation" off the QFT vacuum, but a bound state. Maybe I'm missing something, but I don't know how to describe it in terms of the creation and annihilation operators of the proton and electron.
Oct
30
comment How does path integral formulation explain bound states?
Once I can compute that, I could use the optical theorem to calculate the one-loop correction from such bound states to the Moller-like scattering of the electron and proton. Going that way, I think there might be a path (forgive the pun) to incorporate the effects of bound states into the formalism.
Oct
30
comment How does path integral formulation explain bound states?
What if I posed the question thus: Consider a QFT of a proton field and an electron field. Initial state: Free electron and free proton --> Final state Hydrogen atom in the 1s state. How would one calculate this transition amplitude?
Oct
27
reviewed Approve suggested edit on Effect of doping on the width of depletion layer of PN-junction diode
Oct
23
comment Is there some special case where a fermion can mediate a force?
Oops, I didn't mean to put a vev $\langle \rangle$ above. @arivero: Yes, that's what I was wondering -- the composite force mediating degree-of-freedom is again bosonic, now!
Oct
22
comment Is there some special case where a fermion can mediate a force?
I wonder if the fermion loop is equivalent to a composite bosonic operator $\langle \bar{\psi} \psi \rangle$ mediating the force.