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In general relativity, a diffeomorphism, i.e. gauge transformation, is infinitesimally represented by a shift on the manifold by a vector field, wherein $$X_a \to X_a + \xi_a$$ By definition, the metric tensor changes by a Lie derivative, $$g_{ab}\to g_{ab}+2\mathcal{L}_\xi g_{ab} = g_{ab} + \nabla_{(a}\xi_{b)}$$ Hence it induces a non-physical linear ...

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It sounds like you're trying to find the shifts in the energy levels caused by $H_V$. To find the energy shift that goes as the first order in your small parameter you compute $$E_n^{(1)} = \langle \psi_n^{(0)}|H_V|\psi_n^{(0)} \rangle \quad (1)$$ as you noted. In this expression $|\psi_n^{(0)}\rangle$ means "the $n^{\textrm{th}}$ eigenstate of the ...

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Spin1/2 particle Ususally, in this kind of Hamiltonian, people uses $s=s_z$, where $$s=s_z=\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right].$$ Then, your unperturbed hamiltonian $H_0$ is: $$H_0=-\mu s\cdot B_0 = -\mu \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right]B_{0,z}.$$ Then the eigen vectors of energy are: ...

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Since the 4-momentum must be conserved, there is no angular dependence in a 2-body decay in the CM frame. But If you are asking yourself what is the phase-space of such process, you should take a look at equation 4.72 of the book I mentioned in the comments: $$\Gamma(M\to 1 + 2)=\dfrac{|\vec{p_1}|}{32 \pi^2 M^2} \int \mathrm{d}\Omega ... 0 Here is a link that explains how to do it. You need to expand the Lagrangian around the steady solution. That should give you an easier set of differential equations for the small perturbation. Hope this helps. 1 then we can't state this basis spans the new Hilbert space There is no "new" Hilbert space. If the original basis of space L_2(x\in \mathbb{R}) consisting of eigenfunctions of \hat{H} is complete, any admissible function \psi can be expanded, eigenfunctions of \hat{H}+\hat{H}{}' including. 1 The point of introducing the coupling constant \lambda is that the perturbation series in \lambda might not have radius of convergence \geq 1, i.e. the power series might not be convergent at \lambda=1, and hence that it might not make sense to substitute \lambda=1. In fact, that's typically the case. Nevertheless, a divergent series still make ... 4 Firstly, I refer you to Prof. Binney's textbook (see below) which covers perturbation theory in quantum mechanics in explicit detail. When doing perturbation theory, we perturb the Hamiltonian H^{(0)} of a system which has been solved analytically, i.e. the eigenstates and eigenvalues are known. Specifically,$$H^{(0)}\to H^{(0)} + \lambda H'$$where ... 1 As far as I understand, the logic behind this is the following. We write down the Hamiltonian for the perturbed system as the Hamiltonian for the unperturbed one plus some perturbation $$H = H^{(0)} + H' \, .$$ Assuming that the perturbation is applied gradually we then introduce H(\lambda) operator ... 0 In case H' is small in some sense wrt. H_0 one usually writes$$H(\lambda)=H_0+{\lambda}H'. If the eigenvalues of $H_0$ are known one then obtains a perturbation series expressing the eigenvalues and eigenvectors of $H$ in terms of those of $H_0$. $\lambda$ is primarily introduced to keep track of terms. Complications occur in case an eigenvalue of ...

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