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63 questions
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Why does the Schrödinger equation work so well for the hydrogen atom despite the relativistic boundary at the nucleus?

I have been taught that the boundary conditions are just as important as the differential equation itself when solving real, physical problems. When the Schrödinger equation is applied to the ...
49 views

Symmetric potential well different solutions

I have solved $H|\psi\rangle=E_{n}|\psi\rangle$ with $V(x)=0$ from $-a<x<a$ and $\infty$ otherwise. If I propose a solution of the form $\psi(x)=A_{n}e^{ikx}+B_{n}e^{-ikx}$ I arrive to the ...
120 views

Why is the $k$-space in multiples of $2\pi/L$?

So when you find the solution to the Schrödinger equation you get that the wave function can have $k=n\pi/L$, $n=1, 2,3 \dots$ The problem I have is that when calculating the density of states of a ...
56 views

Green's function for infinite square well

The Green's function can be given in terms of left and right solutions. $G(x,x';k) = \frac{1}{W}\left(\Psi_{L}(x_{<})\Psi_{R}(x_{>})\right)$ But I don't understand how to determine these left ...
89 views

Near Earth's surface the Schrödinger equation of a freely falling particle takes the form, $$\frac {-\hbar^2}{2m} \frac {d^2 \psi (y)}{dy^2} + mgy\psi (y) = E \psi (y).$$ Putting $k=\frac {\sqrt {... 1answer 78 views Wavefunction of a shifted radial harmonic oscillator [duplicate] Suppose we know the solution of Schrodingers equation for a radial potential$V(r)$. Then the energy eigenstates are$\psi(r,\theta,\phi) = \frac 1ru(r)Y_\ell^m(\theta,\phi)$where the radial ... 1answer 28 views How does the image of$\sqrt{2/L} \ \sin\left({k_n x}\right)$satisfy the boundary conditions for the infinite square well? I understand mathematically how$\sqrt{2/L} \ \sin\left({k_n x}\right)$satisfies the boundary conditions for the infinite square well in terms of the fact that$\psi(0) = \psi(a) = 0$, and excuse the ... 1answer 109 views Boundary Condition for Dirac comb potential in solving independant Schrodinger Equation The Periodic potential is And, the general solution is: Then, boundary condition at$x=a$is: Where does$2\Omega u(a)$comes from? I know that boundary condition is just 1)$U(x<a)(a)=U(x>...
I have infinite annular potential well (scheme in the picture). Schrodinger equtation in the anullus (for $R_1 <r<R_2$ is $V=0$) with polar coordinates is \begin{equation} - \frac{ \hbar }{...