Schrodinger Equation for a central 2D potential

EliotHijano
Messages
16
Reaction score
0
Hello,
I would like to ask something about central potentials. When I am working in 3D, I haven´t got any problem solving the schrodinger equation since I use the following change of variables:
-\frac{\hbar^{2}}{2m}\nabla^{2}\Psi+V(r)\Psi=E\Psi
\Psi=\frac{\chi}{r}
With this change of variables, I can solve and understand the equation, but when I am working in 2D( usig r and \varphi) I can't solve the equation because the laplacian is different. The concrete potential distribution I want to study is the following:
V(\vec{r})=0 \stackrel{if}{\rightarrow} r<a
V(\vec{r})=\infty \stackrel{if}{\rightarrow} r>a
And I would like to solve the equation for the first and the second energy levels of the system. I would appreciate some tips, thank you.
 
Physics news on Phys.org
It's not as easy as 3D. The bound states are Bessel functions:
\psi=J_m(kr)\cos(m\theta), with the energy determined by
J_m(ka)=0.
 
Thank you very much clem, that will be really useful for me.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Thread 'Lesser Green's function'
The lesser Green's function is defined as: $$G^{<}(t,t')=i\langle C_{\nu}^{\dagger}(t')C_{\nu}(t)\rangle=i\bra{n}C_{\nu}^{\dagger}(t')C_{\nu}(t)\ket{n}$$ where ##\ket{n}## is the many particle ground state. $$G^{<}(t,t')=i\bra{n}e^{iHt'}C_{\nu}^{\dagger}(0)e^{-iHt'}e^{iHt}C_{\nu}(0)e^{-iHt}\ket{n}$$ First consider the case t <t' Define, $$\ket{\alpha}=e^{iH(t'-t)}C_{\nu}(0)e^{-iHt}\ket{n}$$ $$\ket{\beta}=C_{\nu}(0)e^{-iHt'}\ket{n}$$ $$G^{<}(t,t')=i\bra{\beta}\ket{\alpha}$$ ##\ket{\alpha}##...
Back
Top