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Your final expression is correct except that $V$ should be in front since it does not commute with $Θ_0$ in general. In potential scattering ${Θ_0}V$ is often a compact operator and for large positive imaginary part of $z=E+iℏε$ its norm becomes arbitrarily small so the series converges. In certain cases one can do things a little differently by ...


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It all depends on what application you are interested in. When discussing chiral perturbation theory it is implied that you are interested in low energy processes. If you are considering processes such as low energy (~100 MeV) pion scattering, pion decay, etc. then the typical energies involved are O(100 MeV) which is much smaller then $\Lambda_{QCD}$. This ...


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1) Your perturbation operator does not conserve the particle number of the phonons, so only even powers of it will contribute to equilibrium expectation values. Since you are interested only in the ground state, which doesn't have any phonons excited, this means that you have to create a phonon first. After that either another phonon can be created or the ...


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Non zero quark masses break chiral symmetry. This breaking is assumed to be small enough (certainly for the up and down but also for the strange quark) to do a meaningful Taylor like expansion in it. The relevant scale to which the masses need to be compared is about 1 GeV. It is no guarantee the series will converge but there are predictions that follow ...


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So, you want to diagonalize an operator on a subspace. Specifically you want diagonalize $W_f^n$, the restriction of $W_f$ to the eigen-space associated to $E_n^0$. [Since] $W_f$ doesn't depend of the proton spin, it's possible to divide by 2 the dimension of the problem ($\frac{g_n}{2} X \frac{g_n}{2}$ matrix instead of $g_n X g_n$ matrix). If an ...


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Only the full hamiltonian is observable, as in "corresponds to a physical quantity that can be observed." The "free" and "perturbed" parts are a convenient split when we do a calculation but are not separately observable. In fact it would be surprising if they were, since the split between free and perturbed was arbitrary, it is our choice how to make the ...


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A real linear combination of any two observables is an observable. Suppose $A,B$ are observables, and you have a Hamiltonian $H$. Let $C=A+B$, and note: $C^\dagger=A^\dagger+B^\dagger=A+B=C$ $[H,C]=[H,A+B]=[H,A]+[H,B]=0$ So theoretically speaking, yes, it would be. (The result simply extends to multiplication by a real constant)


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Yes. If $A$ and $B$ are observables, then so is $A+B$. In this case, since $H_0$ and $H$ are observables, so is $H-H_0 = W$.


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I found an answer myself and I would like to share it via this answer. The process of arriving to this Hamiltonian is described in details in the following book: G.L. Bir, G.E. Pikus "Symmetry and strain-induced effects in semiconductors" The process is described in chapter 15 below the topic "Perturbation theory for the degenerate case" The approach the ...


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Comments to the question (v3): Yes, given a Hamiltonian $H$, the semiclassical WKB method can give a qualitative but not a quantitative prediction for the ground state energy $E_0$. The WKB prediction (incl. the metaplectic correction/Maslov index) for the harmonic oscillator (HO) happens to be exact due to a hidden supersymmetry, cf. this Phys.SE post. ...


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Renormalization group is usually used because of some divergence your are avoiding in perturbation theory(such as the divergence of your greens function). In perturbation theory your still expanding in terms of the small parameter before you apply the renormalization group action. Therefore your expansion is still valid. Also just because your parameter ...



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