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The wavefunction of a particle of mass m in a unidimensional potential field U(x)=kx^(2)//2 has in the ground state the form Psi(x)=Ae^(-alphax^(2)), where A is a normalization factor and alpha is positive constant. Making use of a Schrodinger equation, find the constant alpha and the energy E of the particle in this state. |
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Answer» Solution :The Schrodinger equation is `(d^(2)Psi)/(dx^(2))+(2m)/( ħ^(2))(E-(1)/(2)kx^(2))Psi=0` we are GIVEN `Psi=Ae^(-alphax^(2)//2)` Then `Psi''= -alphaxAe^(-alphax^(2)//2)` Substituting we find that following equation must hold `[(alpha^(2)x^(2)-prop)+(2m)/( ħ^(2))(E-(1)/(2)kx^(2))]Psi=0` since `Psi!=0`, the bracket must vanish identicall. This means that the cofficient of `x^(2)` as well the term independent of `x` must vanish. we get `prop^(2)=(mk)/( ħ^(2))` and `prop=(2ME)/( ħ^(2))` Putting `k//m= OMEGA^(2)`, this LEADS to `prop=(m omega)/( ħ^(2))`and `E=( ħ^(2)omega)/(2)` |
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