Recent content by Lagrange fanboy

Feymann said that it's the case because the region isn't simply connected, but I don't see how that leads to the gradient theorem breaking down, as it only requires the scalar potential to be differentiable along the path.

In Feymann's seminar on superconductivity, there was this equation (21.28) ##\oint_C \nabla \theta\cdot dl = \frac q \hbar \Phi##. But the gradient theorem demands that ##\oint_C \nabla \theta\cdot dl=0##

Wow now I'm really confused. I read on wikipedia that the postulates of quantum mechanics include: Quantum states are unit vectors in some Hilbert space Observables are Hermitian operators acting on said Hilbert space Probability of results are calculated from inner products between the...

Still, if ##Ae^{\frac{ipx}{\hbar}}## is the wave function for ##|p\rangle## wrt. position basis, then I'd expect ##\int_{-\infty}^\infty \langle p|x\rangle \langle x|p\rangle dx = 1##.

But the norm of the wavefunction won't be 1 then, shouldn't this cause a problem down the line when calculating probabilities?

So on page 256 of Quantum Mechanics - The Theoretical Minimum, it says that the wave function of a momentum eigenvector, with respect to the position eigenbasis is ##\psi_p(x)=Ae^{\frac{ipx}{\hbar}}##, and ##A## must be ##\frac{1}{\sqrt{2\pi}}## to keep it a unit vector. However why must...

The difference between the rate shuttlecock decelerates and the expected rate due to gravity will be due to drag. You might also find motion history image useful.

I was reading page 33 of https://staff.fnwi.uva.nl/p.j.c.spreij/onderwijs/TI/mtpTI.pdf when I saw this claim: Given measurable spaces ##(\Omega_1,\Sigma_1), (\Omega_2,\Sigma_2)## and the product space ##(\Omega_1\times \Omega_2, \Sigma)## where ##\Sigma## is the product sigma algebra, the...

Goldstein's Classical Mechanics makes the claim (pages 382 to 383) that given coordinates ##q,p##, Hamiltonian ##H##, and new coordinates ##Q(q,p),P(q,p)##, there exists a transformed Hamiltonian ##K## such that ##\dot Q_i = \frac{\partial K}{ \partial P_i}## and ##\dot P_i = -\frac{\partial...