On a test, I'd probably use a geometrical method - essentially drawing and trying to see how each unit axis would change. This particular derivative is probably the trickiest one even using that method though.
However, you wanted an algebraic method. In one direction it looks like a trick to...
Have you tried going the other way? It is easier (at least more natural) to prove that statement that way, inserting the formulas for \hat{r} and \hat{\theta}.
Note that people usually call a the annihilation operator, and a^\dagger the creation operator. The reason it is worth to avoid confusion here, is that the annihilation operator a (which satisfies a|0\rangle=0, where |0 is the ground state) actually has an eigenstate, while the creation operator...
\beta >0, otherwise the free energy could be minimized by |\psi |\rightarrow \infty . This may not be obviously unphysical, but the Ginzburg-Landau model does after all aim to model the phase transition around |\psi |=0 .
\alpha <0, then. The parameter \alpha is indeed related to the...
It's not pretty, I agree, but then again the Levi-Civita symbol is what you use to hide the ugly things. It's possible to group the terms somewhat, but you can't remove degrees of freedom unless you now that one or more of the vector components happen to be zero... Just be happy that you don't...
Well, since the two free indices i and j are expressedly independent you can't set i=j (it's more interesting to calculate the commutator for two specific terms than the sum anyway). And the other four indices are already contracted, so you can't contract them with either i or j. You will just...
I'm pretty sure your textbook contains the definition of the Riemann tensor, and of the other tensors. Look it up. (You will only need to know the metric to calculate it.) Then, the Ricci tensor is related to the Riemann one through
R_{ij} = R^k_{\, ikj} ,
and, finally, the Ricci scalar is...
The \frac{\langle {p} \rangle t}{m} term you have corresponds to the linear term in the classical equation
\Delta x = v_0 t + \frac{a t^2}{2},
right? You obviously can't have a time dependence in \langle p \rangle here, since you already assumed that there isn't one, during the integration...
Well, the indices do run from 1 to 2n, and there is no deep physics hidden in calling the indices i and j, specifically. Could be any letter or symbol, really. However, the order is important.
Changing the order of the indices for a matrix actually is the same as taking the transpose. Think...
I think I was a bit sloppy last night, and yeah, you're quite right. When L is independent, then you have a conserved quantity. Goldstein calls this the energy function h, and when you go to the Hamiltonian formulation it's just the Hamiltonian H. Like you say, however, this is not necessarily...
It looks OK, yes. You have spherical symmetry, and the potential term behaves properly.
A constant of motion is really just some quantity that is conserved under the motion. Using Noether's theorem we know that it corresponds to a symmetry of the Lagrangian, and here you have rightly pointed...