Tag Info

New answers tagged

0

Use a power series $ R(r) = \sum_{n=0}^\infty a_n r^n$. Doing so, you'll find an equation which connets the coefficients $a_n$, $a_{n+1}$ and $a_{n+2}$. By looking at the behaviour for big $n$, one can conclude that the series has to stop at some point $n=N$ (or else it will grow like $\exp()$ and prevent normalization). This $N$ is a new quantum number. For ...


1

Your main question was "Is there an example where the overlap between a bound state and the scattering states makes a measurable contribution to the energy in the perturbative regime?" Actually, I disagree with the statement of the other answer that the scattering states must be included in the perturbative calculations only if the result is to be highly ...


1

The answer to your question comes from math, not from physics. Consider the wave equation in 1+1 dimensions: $$\partial^{2}_{x}\phi(x,t) - \partial^{2}_{t}\phi = 0$$ One obvious way to solve this is to convert the PDE in two variables into two ODEs in one variable. This is done by assuming that the solution can be written as some sum $$\phi(x,t) = ...


1

The quantum numbers serve to enumerate the solutions to the time-independent Schrodinger equation (Energy eigenstates) which are also eigenstates of the angular momentum (squared) operator and have a definite component of angular momentum in some direction labeled $ z $. When solving the problem, it just turns out to be handy to label the solutions with the ...


0

The easiest way to see that $p^4$ is spherically symmetric is to view it in momentum space. If you apply $p^4$ to a momentum eigenstate $|p\rangle$ the result clearly only depends on the magnitude of the momentum vector of the state and not on its orientation, so if $R$ is a rotation operator we have \begin{equation} p^4 R|p\rangle = Rp^4|p\rangle ...



Top 50 recent answers are included