Recent content by spocchio

Consider the variational principle used to obtain that in the vacuum the Einstein tensor vanish. So we set the lagrangian density as L(g,\partial g)=R and asks for the condition 0 = \delta S =\delta\int{d^4 x \sqrt{-g}L} proceeding with the calculus I finally have to vary R such that...

Huh, what should be the different, as I just understood, with the geodesics equation I have a first order differential equation that should give me a "line" in the manifold given an initial point and an initial vector (velocity). As i understood light cones are given by imposing...

I had this doubt studing GR, but let's consider SR for semplicity, where g_{\mu\nu}=\eta_{\mu\nu}the geodesics are 0=ds^2=dt^2-dr^2 we obtain the constraint we obtain the constraint r=(+/-)t So it is a well known light cone, but in SR we have that a (test?)particle can always move in a...

Oh so you talk about the spin of the proton, that's another interesting subject about coupling. I expressed myself badly, i was asking about spin of ELECTRON (i supposed chemists don't work with nuclear spin)

As I know from the laws of chemistry,to form H_2 the H atoms must have opposite spin. What if they have the same spin? I mean, can an H atom rotate to swap his spin (i'm not expert of the dynamics of spin), or there is no way to couple them? In the case I have a gas of H with all of them...

As I know from the laws of chemistry,to form H_2 the H atoms must have opposite spin. What if they have the same spin? I mean, can an H atom rotate to swap his spin (i'm not expert of the dynamics of spin), or there is no way to couple them? In the case I have a gas of H with all of them...

in particular, i wonder if the trasposition super operator is basis independent or not. We can always write an operator W as \hat{W}=\sum_{i,j} c_{i,j} |i\rangle\langle j| and for the transposed we obtain \hat{W}^T=\sum_{i,j} c_{j,i} |i\rangle\langle j| we obtain a relation true for each...

nice question, I don't know the answer, what should change? can we say that is not the knowledge of H(q,p) that gives the desired entropy?. In conclusion (in statistical mechanics) only rho(q,p) that determinate uniquely the entropy, with the request that \frac{d \rho}{dt}=0.

Remember that a state in the superposition in a certain basis can be a well state in another basis. I think the problem comes when you consider a preferred base for your visualization, that is the q basis. for example you can say that the state |\psi\rangle = |q_1\rangle isn't in a...

let suppose you have of H: \hat{H}|E_i\rangle=E_{H,i}|E_i\rangle If you can diagonalize H and rho together there is no reason to think that the set of |E_i\rangle that can be diagonalized at the same time with H are ALL the |E_i\rangle that are diagonal for \hat{H}. in general the...

This is my apparently-trivial problem, that probably means i don't understood what entropy is. We all have faced the statement that entropy always increase. With the power of statistical mechanics we can calculate the variation of entropy from two states, considering the difference of entropy...

The solution are the same under your approimation: since \frac{d}{dt}p=-mg you have that dp \cong dt and so, in an approximation of O(t^2) to check the solution you have to discard O(p^2) \psi(p(t)) = p^2 - 2 g m p t - \frac{i t p^4}{2m} + O(t^2) = - 2 g m p t + O(t^2) + O(p^2) \psi(p(t)) =...

let's take another example,this time we'll consider a well known quantum problem and it's related Hamiltonian. Consider a quantum particle in free fall: \hat{H}=\frac{1}{2m}\hat{p}^2+mg\hat{q} let's see the scrodingher equation in\lvert p \rangle...

I do not think that all the Hamiltonians can accommodate your request that you can find a \frac{d}{dt}\vec{x} = \vec{X} for each \hat{H} . But I'll show you a class of Hamiltonians wherewith you're able to do it, unlucky these Hamiltonians don't have a quantum physical sense, but it's stll...

Hi, I've studied a bit nuclear physics, and I don't get how the mathematics under the creation and destruction of particles is handled. if shrodingher equation leads a state vector in the usual way Hpsi=ih d(psi)/dt let's suppose I've N particles and psi=psi(q1...qn) so, if a new particle...