The one dimensional time-independent Schrödinger equation reads:

\begin{equation} -\frac{h^2}{2m}\frac{d^2\psi}{dx^2}+U(x)~\psi=E~\psi \end{equation}

where $\psi(x)$ is the wave function, U(x) is the potential, E is the eigen-energy. $\hbar=1$ is the Plank constant, $m=1$ is the mass of the particle.

Under the potential of

\begin{equation} U(x)=(1+Tanh(x))(-1+Tanh(x)) \end{equation}

the Schrödinger can be solved analytically,

\begin{equation} \psi(x)=C_1P_1^{i\sqrt{2E}}(Tanh(x))+C_2Q_1^{i\sqrt{2E}}(Tanh(x)) \end{equation}

where P and Q are the associated Legendre polynomial $P_n^m(x)$ and associated Legendre function of the second kind $Q_n^m(x)$.


If we have a slightly different potential

\begin{equation} U(x)=(1+Tanh(x+1))(-1+Tanh(x-1)) \end{equation}

can the Schrödinger also be solved analytically?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.