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20

An orbit is stable because of conservation of angular momentum. Suppose we start with an object in an exactly circular orbit and slow it down slightly. That means it is moving at less than orbital velocity so it starts to fall inwards. However as its distance to the Sun decreases the tangential component of its velocity has to increase to conserve angular ...


13

This was something that confused me for awhile as well until I found this great set of notes: homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf Let me just briefly summarize what's in there. The free Klein-Gordon field satisfies the field equation $(\partial_{\mu} \partial^{\mu} +m^2) \phi(x) = 0$ the most general solution to this equation ...


13

I) The lowering of the ground state energy is a special case of the more general phenomenon of level repulsion (because the excited energy levels by definition must be larger than the ground state energy). II) Level repulsion is not just a quantum phenomenon. It also happens for purely classical systems, e.g. two coupled oscillators, as mentioned in the ...


11

The criterion you mention is roughly the threshold for the formation of the Coulomb gap in the Hubbard model or the local moment in the Anderson model. It is a common break-down region for many approaches starting from one of the limits (insulator/local moments versus conductor/mixed valence). For perturbation theory in $U$, see the PRB 36, 675 (1986) by ...


8

A nonperturbative theory means a theory where all the results can be calculated in principle to arbitrary accuracy on a computer. This is really just the same as a well-defined theory, a theory which is mathematically ok, and which makes sense, which isn't incomplete in some way at some high energy or when you do measurements at some high accuracy. An ...


8

The gravitational potential is what is known as a central force, which means that "how strong" the potential is only depends on how far away you are, and not on what angle you are relative to it. Having said that, gravitational systems are often treated in terms of an effective potential (full explanation provided on the Wikipedia page) which look like this ...


7

No. There is nothing wrong with perturbation theory, or with theories with known, restricted accuracy. The point of theory is to explain the results of observation from as simple an initial theoretical standpoint as possible. Therefore: Since experiment always has a finite uncertainty, one can only ask that theory match the experimental value within its ...


7

I've been thinking about divergent series on and off, so maybe I could chip in. Consider a sequence of numbers (in an arbitrary field, e.g. real numbers) $\{a_n\}$. You may ask about the sum of terms of this sequence, i. e. $\sum a_n$. If the limit $\lim_{N\rightarrow\infty} \sum^N |a_n|$ exists then the series is absolutely convergent and you may talk ...


7

The reasons were given here. Essentially, at tree level you recover classical results. Loop corrections are proportional to powers of $\hbar$ and these are quantum terms.


6

There is a very easy way to see this and it is through an $\hbar$ series. This claim can be traced back to Sydney Coleman and states that in the ultraviolet one is doing an expansion with $\hbar$ going to zero. A previous answer cited these lectures on classical fields but I would like to start from the generating functional of the scalar field theory and ...


6

Most of the difficulties of these calculations comes from the fact that people write the Lagrangian directly in terms of the metric perturbation $h_{\mu\nu}$. It is much simpler to write it in terms of the difference of connections tensor $F_{\mu\nu}{}^\beta$, i.e., $$(\nabla_\mu-\bar{\nabla}_\mu)A_\nu = \mathcal{F}_{\mu\nu}{}^\beta{}A_\beta,$$ where ...


6

In order to do perturbation the expansion parameter needs to be small. Otherwise the the system will be strongly coupled and you're in the non-perturbative regime. It's the same as for instance in QM: for perturbative calculations the pertubation must be small.


6

Good question. You are correct in that without any restoring force, an object balanced precariously in an equilibrium position will be unstable. In physics, we use the scalar quantity of "potential" to find the equilibrium positions. These will be the maxima and minima in the potential field. The negative gradient of the potential gives the force. You've ...


6

Qualitative discussion (almost math free) The real key to understand orbits is the conservation of angular momentum. A two body orbit is nice this way insofar as it is a planar system and we get an easy expression for the angular momentum (we'll assume a satellite much, much less massive than the primary and not bother with the canonical transformation ...


4

You're looking for quantum hadrodynamics. See, e.g. Serot and Walecka.


4

I thinnk the earliest papers where this was written down were by DeWitt. But for a reference easily available via the arXiv look at hep-th/9411092. Eq (2.17) has the expansion you want and eq. (2.18) even has it to fourth order in $h$.


4

I think "quasi-circular" is a misleading name for this problem. Perhaps "quasi-elliptical" would be better? I say this because this problem does, in fact, contain a closed circular orbit (the radius of which you have called $r_c$). For that orbit you can find the period using Kepler's second law, which gives the result you show. An interesting way to ...


4

As I mentioned in the comments, the assertion that $E_n^{(1)}\equiv0$ cannot hold in general since a scalar perturbation does not obey it. For the particular case you mention, a linear perturbation on a harmonic oscillator, however, it does hold. The simplest way to see this is that the perturbation can be included in the oscillator potential to give ...


4

The scattering states must be included in the perturbative calculations if the result is to be highly accurate. In particular, it is not justified to ignore the continuous spectrum at energies close to the dissociation threshold. The Hilbert space in the position representation is the space of square integrable functions on $R^3\setminus\{0\}$ with respect ...


4

First, a correction. The first formula is the probability, not probability amplitude. And it's computed at the leading order only, "linearized" in a sense, so of course it is only a good approximation for $P_{f\leftarrow i}\ll 1$. When the probability becomes comparable to one, subleading and higher-order corrections become important because one must also ...


4

My approach would be: first determine the time evolution of $\hat{x}(t)$ and $\hat{p}(t)$. For $\hat{x}$ you have $$ \frac{d}{dt}\hat{x}_H(t) = i[H_H,\hat{x}_H(t)] = \frac{i}{2m} [\hat{p}_H(t)^2,\hat{x}_H(t)] = \frac{\hat{p_H(t)}}{m} $$ and for $p$ you have (assuming $0\leq t \leq T$) $$ \frac{d}{dt}\hat{p}_H(t) = i[H_H(t),\hat{p}_H(t)] = -m\omega_0^2 ...


4

Is there somewhere in the paper that they say that the third derivative vanishes, or invoke its vanishing as an approximation? In general, you can't make tensorial objects by differentiating the metric. To get a tensor by differentiating a tensor, you have to take a covariant derivative. But the covariant derivative of the metric vanishes identically. ...


4

The idea is the following: Consider a series expansion in elementary charge squared, $e^2$: \begin{equation}S(e^2)=\alpha_0+\alpha_2 e^2+\alpha_4 e^4+\;\ldots\end{equation} Assuming finite convergence radius, $S(e^2)$ is analytic at $e=0$. This leads to the analyticity of $S(-e^2)$ (i.e. $e\rightarrow ie$). A theory with imaginary charges possesses an ...


4

I) This is discussed around eq. (23.7.1) on p. 462 in Ref. 1. The task is to perform the path integral $$\tag{1} \int_{BC} [d\phi]e^{\frac{i}{\hbar}S[\phi]} ~=~\sum_{\nu}\int\! du \int_{BC_0} [d\phi_q]e^{\frac{i}{\hbar}S[\phi_{cl}+\phi_{\nu,u}+\phi_q]} $$ over fields $\phi$ with some (possible inhomogeneous) boundary conditions $BC$. This is done by ...


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 ...


3

Yes, your second guess is more or less correct. In GR, perturbing the metric is the usual way of doing perturbation theory. One writes for the true metric $g_{\mu\nu}$ an expansion of the form $$ g_{\mu\nu} = g^{(0)}_{\mu\nu}+h_{{\mu\nu}}+O(h^2), $$ where $g^{(0)}_{\mu\nu}$ is known and usually called background metric. One then substitutes this into the ...


3

Suppose that the Hamiltonian is written as $$ H = H^{(0)} + V $$ It is assumed that both $H^{(0)}$, the unperturbed hamiltonian, and $H$, the perturbed hamiltonian, are self-adjoint. In particular, there is an orthonormal basis $\psi^{(0)}_m$ for the Hilbert space consisting of eigenvectors of $H^{(0)}$, and a an orthonormal basis $\psi_m$ consisting of ...



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