Finding Potential Energy for Sech Wave Function

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The discussion focuses on finding the potential energy V(x) for the wave function ψ(x) = sech(ax) and demonstrating that the ground-state energy E1 equals V(0)/2. The user differentiated the Schrödinger equation twice and substituted the result back into the equation, yielding a second derivative of the wave function. They derived the expression for V(x) and calculated V(0)/2 as (3ħ²a²)/(8m). The user seeks confirmation on the correctness of their approach and calculations.
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Homework Statement


For the (unnormalized) wave function ψ(x) = sech(ax), find the potential energy V (x), and show that the ground-state energy E1 is V(0)/2. The energies are in units of (hbar)^2a^2/2m.



Homework Equations



- \frac{\hbar}{2m}\frac{d^2 \psi}{dx^2}+V(x) \psi)=E \psi

The Attempt at a Solution



I differentiated the SE twice and sub it back into SE. Does that seem right to you?

\frac{d^2 \psi}{dx^2} = 0.5a^2 (cosh(2ax)-3)sech^3(ax)

\frac{-\hbar^2}{4m} a^2 (cosh(2ax)-3)sech^2(ax)+V(x)=E=\frac{\hbar^2 a^2}{2m}
 
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Also, I forgot to say that I got this :

V(0)/2=\frac{3 \hbar^2 a^2}{8m}
 
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