# Find the energy eigen value given wave function

I'm given the ground state wave function $\psi(x)=A\operatorname{sech}(bx)$. Potential is not given but told that it goes to 0 at $\infty$. How to find the eigen value of energy in this state?

My approach so far: Using $\psi(x)$ in TISE, $$\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right]\psi(x) = E\psi(x)$$

EDIT after suggestions: $$\frac{\hbar^2}{2m}Ab^2\operatorname{sech}(bx)(2\operatorname{sech}^2(bx) -1) + V(x)A\operatorname{sech}(bx) = EA\operatorname{sech}(bx)$$ Evaluating at $\infty$ $E=-\frac{\hbar^2b^2}{2m}$

Oh, i have messed up by converting hyperbolic to exponential. Thanks. A little surprising that it has got the same ground state energy of a $\delta$ potential

• The RHS doesn't depend on x – Phoenix87 Feb 16 '15 at 8:43
• Your equation appears to be incorrect – hft Feb 16 '15 at 8:44

First, you need to fix your equation for E. You seem to have divided out by $\psi$ in the V and E terms, but not in the kinetic term... among other issues (the kinetic term should end up proportional to sech^2-tanh^2)... Just recheck the derivatives.