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7

It is "easy" to show that the cross-section of the 2 reactions: $$ (1)~~~\nu_{\mu} + d \to \mu^- + u~~~~~~(2)~~~\bar{\nu}_{\mu} + u \to \mu^+ + d$$ are different. The naive calculation gives a factor $\sigma_1/\sigma_2 = 3$ as shown below. The fact that the figure in the particle data group gives actually almost a factor 2 needs to take into account the ...


3

You need to distinguish between the virtual gravitons that appear in a quantum field theory calculation of a gravitational interaction and real gravitons that form a gravitational wave. The sort of experiment you're describing requires the emission of real graviton i.e. the emission of a gravitational wave. The trouble is that the coupling constant for the ...


3

The short answer is that the deformation parameters are related to the diagonal matrix elements of the multipole operators, while the reduced transition probabilities $B(El)$ are related to the off-diagonal matrix elements. As a specific example, the quadrupole moment (directly related to the deformation parameter $\beta$) of a nucleus in its ground state ...


3

The answer is that in general cross section of particles and antiparticles are different. For this case in specific there is one point that Dirac particle have four possible states: due two possible ranges for energy E<0, E>0, and two possible range for helicities h>0 and h<0. In the standard model of particles only the neutrino with h<0 interact ...


1

As you guessed, the probability density is not the same as transmission. In order to have non-zero transmission there has to be nonzero gradient: $$j=\frac{1}{2m}(\psi^\dagger \hat{p}\psi - \psi \hat{p}\psi^\dagger)$$ In the case of a constant probability, there is no gradient. You only have particle "buildup". The reason for this is that you are solving ...


1

A single photon with 3-momentum $p=\hbar k$ incident on the system is scattered to an outgoing state with momentum $\hbar k'$. If the interaction operator between the photon and the sample is $U$, and the initial and final states are labelled by $|k\rangle$ and $|k'\rangle$ respectively (we ignore photon polarization/spin), then the amplitude for this ...



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