Recent content by Siberion

do you suggest using that relation on the last part, or did I make a wrong choice from the beginning?

Sorry, I think my reasoning was erroneous (it still might be) I think it is better to say that the initial state of the system can be described by the singlet state (because they point in opposite directions at t = 0, and the wave function needs to be antisymmetric): \Psi(0) = 1/ \sqrt{2} (...

Thank you very much for your reply. I am getting some problems with the algebra: As the particles are fixed, the hamiltonian is going to be equal to the potential energy: \hat{H} = \hat{V} Generally speaking, \hat{\sigma_1} = \sigma_{1x} \hat{i} + \sigma_{1y} \hat{j} + \sigma_{1z} \hat{k}...

Homework Statement Consider two spin 1/2 particles interacting through a dipole-dipole potential \hat{V} = A\frac{(\hat{\sigma_1} \cdot \hat{\sigma_2})r^2 - (\sigma_1 \cdot \vec{r})(\sigma_2 \cdot \vec{r})}{r^5} If both spins are fixed at a distance d between each other, and at t = 0 one of...

Such a beautiful and elegant explanation! This was indeed a very interesting problem. It had me thinking for a long time, dealing with logarithmic solutions and trying to understand what they really meant... By the due date of the homework (yesterday), my professor commented on how this...

Hmm, I found a similar example of electromagnetic interaction in order to find the Lorentz Force. Here \dot{p}\neq mv, and the force F actually corresponded to the derivative of the linear momentum, which differed form \dot{p}. At this point I'm kind of lost, I was thinking about a...

Hello, thanks for your replies. Yes this is exactly what I did. You are right, I commited a manipulation failure, but as far as I can tell it doesn't affect the equation that much. This is the procedure I followed: From eq. (1), we have p= m\dot{q} e^{q/a} . Deriving this equation...

Thanks for your reply. The Hamilton equations of motion read: \dot{q}=\frac{∂H}{∂p} \dot{p}=-\frac{∂H}{∂q} From which I obtained the following equations: \dot{q}=\frac{p}{m}e^{-\frac{q}{a}} (1) \dot{p}=\frac{p^2}{2am}e^{-\frac{q}{a}} (2) From (1), I obtained an expression for p...

Homework Statement Consider the following Hamiltonian H=\frac{p^2}{2m}e^{\frac{-q}{a}} a: constant m: mass of the particle q corresponds to the coordinate, and p its momentum. note: q' stands for the derivative of q. a) Prove that for p(t) > 0 this system seems to describe a particle...

Haha, my apologies, I got confused with the word commonly used in spanish "par". In english, it would indeed be an even function.

Homework Statement Considering the following integral, I = \int^\infty_{-\infty} \frac{x^2}{1+x^4} I can rewrite it as a complex contour integral as: \oint^{}_{C} \frac{z^2}{1+z^4} where the contour C is a semicircle on the half-upper plane with a radius which extends to infinity. I can...

Hmm, I think I solved the problem. We can take the Lagrange equations in the form that follows from the D'Alembert principle, and supose that the generalized force is derivable not from the potential V but from a more general function U, which is usually called generalized potential. Then...

Homework Statement Consider a particle of mass m and electric charge e subject to a uniform electromagnetic field (E(x,t),B(x,t)). We must remember that the force they exert is given by: F = cE(x,t) + ex' \times B(x,t) A principle of action that represents such particle subject to the...

Thanks a lot TSny, according to what I calculated, using E-L equations, its derivative indeed equals to zero. I got lost trying to rearrange the Hamiltonian, I should have checked the derivative of Q at first.

Homework Statement Consider a particle of mass m and electric charge e moving in a uniform magnetic field given by B = Bẑ. Then the Lagrangian is given by: L = \frac{m}{2}(x'^2 + y'^2 + z'^2) + \frac{Be}{2}(xy' - yx') Prove that Q={L} \cdot{B} + \frac{e}{2}((r \times B)(r \times B)) is a...