Let $H_A \otimes H_B$ be your Hilbert space, and $O$ be an operator acting on this composite space. Then $O$ can be written has $$O = \sum_{i,j} c_{ij} M_i \otimes N_j$$ where the $M_i$'s and $N_j$'s ...

Below is a summary of my very limited understanding of what the elliptic genus is. I'll first give you the mathematical definition, followed by an explanation of how it appears naturally in physics. ...

Indeed the structure is similar for a very good reason: Both of these are particular examples of a beautiful concept in mathematics. You have some underlying manifold $M$. In addition to the ...

The two quantities don't correspond because they are conserved quantities corresponding to different symmetries. One is a symmetry from shifting your field, the other from shifting space-time itself. ...

One can directly show that it's conserved along geodesics as Dvij has done, but it's also illuminating to note that this is just a special case of Noether's theorem. In particular whenever a metric ...

You aren't "replacing a mathematical proof". What the statements you are referring to mean is that in tensor notation, the proof is immediate, so that nothing needs to be written down. This is because ...

By definition of the cross product and gradient operators, and summing over repeated indices we have, $$\nabla . (\nabla \phi \times \nabla \psi) = \partial_i (\epsilon_{ijk} \partial_j \phi \,\... View answer 4 votes Open up the chapter on perturbation theory in any quantum mechanics textbook and pick any Hamiltonian you see. Chances are an exact solution to that quantum system isn't known. Ask your Professor to ... View answer Accepted answer 4 votes Actually the equations of motion one ends up with are not manifestly the same: If I let$$L_1 = \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}}$$one finds that the Euler-Lagrange equation ... View answer 4 votes Adding to Lubos' answer, let me address this part more specificially: I am very confused about this - actually - I'm guessing that transformations that can be written A⊗B are a subset of all ... View answer Accepted answer 4 votes Sure, this kind of thing is well studied. The "space of metrics" of a manifold is usually called its moduli space. More precisely, if we have a manifold M, Let \text{Met}(M) be the set of all ... View answer 4 votes It's useful to know how matrix multiplication is defined: For n \times n matrices, A and B, denote the entry in the ith row and jth column by A^i_j and B^i_j respectively. Then for C ... View answer Accepted answer 3 votes First note the following: If \psi(x,0) is any normalizable wave-function in your Hilbert space, then \psi(x,t) = e^{-iHt} \psi(x,0) (I've set \hbar = 1) satisfies the time-dependent Schrodinger ... View answer Accepted answer 3 votes Elimination by the use of limit analysis is probably the best way to do it on the actual test, but here's the full solution: Let \theta' by the angle that light enters the cable. By Snell's law ... View answer 3 votes Let us check the claim that the earth is rotating about it's own axis. We may choose this axis to be the z-axis. The earth can be approximated by a sphere. Consider a pendulum living somewhere on ... View answer 2 votes A simple (but rather non-rigorous) way to think about this: Your Hilbert space \mathcal{H} is formed by square-integrable functions on the real line, that is elements of \mathcal{H} are given by ... View answer Accepted answer 2 votes If I understood correctly what I've been taught so far, in QFT one must find some way to quantize the fields obeying the field equation in question. This is correct. In this particular case you start ... View answer 2 votes Let$$|n\rangle = \frac{(a^{\dagger})^n}{\sqrt{n!}} |0\rangle$$be the n'th normalized eigenstate and recall the commutation relation for bosonic creation/annihilation operators, [a, a^{\dagger}] = 1... View answer Accepted answer 2 votes You can either vary the action directly, or apply the classical field theory Euler-Lagrange equations. The latter for a Lagrangian \mathcal{L}(\phi^{\alpha}, \partial_{\mu}\phi^{\alpha}) read$$\...

As requested in the comments by the user, I will write down the differential equation satisfied by the trajectory of a photon that travels along a radial path (no $\theta$ or $\phi$) dependence. By ...

Your sign is wrong when computing $$\frac{\partial{(B^2)}}{\partial{(\partial_{y} A_x)}}.$$ The only term in $B^2$ that contains $\partial_{y} A_x$ is $(B_z)^2 = (\partial_{x}A_y - \partial_{y} ... View answer 1 votes BBS omit to indicate which space they are tracing over in each step. Let me instead sketch why tracing over only the states created by the$\alpha_{-n}$'s in the last step is the correct thing to do. ... View answer Accepted answer 1 votes The proof of this identity is entirely analogous to how one would go about proving the Euclidean formula$\vec{a}.\vec{b} = |a||b| \text{cos}(\theta)$. What one has to do is as follows: Find a ... View answer Accepted answer 1 votes Note that the chain rule in this case, since$\mathcal{L} = \mathcal{L}(\varphi, \partial_{\mu} \varphi)$reads$$\partial_{\nu}\mathcal{L} = \frac{\partial \mathcal{L}}{\partial \varphi} \partial_{\... View answer Accepted answer 1 votes This simply follows from a resolution of the identity, written (formally) as$\int dy |y\rangle\langle y| = 1$:$\langle x|HU(t)|x\rangle = \int dy\langle x|H|y\rangle\langle y|U(t)|x\rangle$, which ... View answer 1 votes What's important when considering the behavior of tensors aren't specifically rotations, but any arbitrary coordinate transformation (a change of basis of your vector space). In general if you have a ... View answer 1 votes I didn't go through all of your work but to answer the question about how to solve that equation, change$\tan(\phi)$to$\frac{\sin(\phi)}{\cos(\phi)}$and take$\cos(\phi)$as a common denominator. ... View answer 0 votes This is something that really does depend on the specific problem you are studying. In general one can only say that we have some "space"/manifold of control parameters$\mathcal{M}$. Three different ... View answer 0 votes Actually this must be so, since in the$a \rightarrow \infty$limit, you better recover the quantum mechanics of a free particle moving on the real line. If you define$k_n = n\pi/a\$, one can see that ...