Question #109168
An electron has the following wave function:
¥ (x) = Ae-x (1-e -x)
Determine
i) the normalization constant A, and
ii) the expectation value of x
1
Expert's answer
2020-04-13T10:17:54-0400

The wave function is


ψ(x)=Aex(1ex).ψ(x)=Ae^{-x}(1-e^{ -x}).

Determine the normalization constant A:


1=+ψ(x)ψ(x)dx, 1=0+A2e2x(1ex)2 dx, 1=A2(112),A2=12. 1=\int^{+\infty}_{-\infty}\psi^*(x)\psi(x)dx,\\ \space\\ 1=\int^{+\infty}_{0}A^2e^{-2x}(1-e^{-x})2\space dx,\\ \space\\ 1=A^2\bigg(\frac{1}{12}\bigg),\\ A^2=12.\space\\

Determine the expectation value of x:


<A^>=+ψ(x)A^ψ(x)dx, <x>=120+xe2x(1ex)2 dx, <x>=1312.<\hat{A}>=\int^{+\infty}_{-\infty}\psi^*(x)\hat{A}\psi(x)dx,\\ \space\\ <x>=12\int^{+\infty}_{0}xe^{-2x}(1-e^{-x})2\space dx,\\ \space\\<x>=\frac{13}{12}.

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