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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 wavefunction, 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 $Q_n^m(x)$.

Question:

If we have a slightly different potential

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

does the Schrödinger can also be solved analytically?

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