4.5. THE HYDROGEN MOLECULAR ION55Now evaluate the expectation energy:(E)=(ax(ℓ−x)|H|ax(ℓ−x))=|a|2(Biggx(ℓ−x)vextendsinglevextendsinglevextendsinglevextendsinglevextendsingle−¯h22m∂2∂x2vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglex(ℓ−x))BiggYou can substitute in the value of|a|2from the normalization requirement above and applythe Hamiltonian on the function to its right:(E)=30ℓ5¯h2m(x(ℓ−x)|1)The inner product is by definition the integralintegraltextℓ0x(ℓ−x) dx, which was given to beℓ3/6. Sothe final expectation energy is(E)=¯h2102mℓ2versus¯h2π22mℓ2exact.The error in the approximation is only 1.3%!That is a surprisingly good result, since theparabolaax(ℓ−x) and the sinea′sin(πx/ℓ) are simply different functions. While they mayhave superficial resemblance, if you scale each to unit height by takinga= 4/ℓ2anda′= 1,then the derivatives atx= 0 andℓare 4/ℓrespectivelyπ/ℓ, off by as much as 27%.If you go the next logical step, approximating the ground state with two functions as
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