Answer to Question #152933 in Quantum Mechanics for adnan

Question #152933

A wave function is given to be 𝜓(𝑥) = 𝑁( 𝑥 /𝑥0 )^ 𝑛𝑒 −𝑥/𝑥0, 𝑁, 𝑛, 𝑥0 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 From Schrodinger wave equation find potential V(x) and the energy E for which 𝜓(𝑥) is an Eigen function.


1
Expert's answer
2020-12-28T10:27:18-0500

Let's write Schrodinger equation applied to ψ(x)\psi(x) :

22mψ+Vψ=Eψ-\frac{\hbar^2}{2m} \psi'' +V\psi = E\psi

First calculate ψ\psi'' :

ψ=Nex/x0(n(n1)x02(xx0)n22nx02(xx0)n1+1x02(xx0)n)\psi''=N e^{-x/x_0} (\frac{n(n-1)}{x_0^2} (\frac{x}{x_0})^{n-2}-2\frac{n}{x_0^2}(\frac{x}{x_0})^{n-1}+\frac{1}{x_0^2}(\frac{x}{x_0})^n)

ψ=Nex/x0(xx0)n(1x022nxx0+n(n1)x2)=ψ(1x022nxx0+n(n1)x2)\psi''=Ne^{-x/x_0} (\frac{x}{x_0})^n (\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2}) = \psi (\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})

Now we substitue this expression into the Schrodinger equation :

(22m(1x022nxx0+n(n1)x2)+V)ψ=Eψ(\frac{-\hbar^2}{2m}(\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})+V)\psi = E\psi

As ψ0\psi\neq 0 everywhere except for x=0x=0, we can simplify by ψ\psi and find

22m(1x022nxx0+n(n1)x2)+V=E\frac{-\hbar^2}{2m}(\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})+V = E

V=E+22m(1x022nxx0+n(n1)x2)V=E+\frac{\hbar^2}{2m}(\frac{1}{x_0^2}-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})

As V0V\to 0 when x+|x|\to+\infty , we find that

E=22mx02E=-\frac{\hbar^2}{2mx_0^2}

V=22m(2nxx0+n(n1)x2)V=\frac{\hbar^2}{2m}(-\frac{2n}{xx_0}+\frac{n(n-1)}{x^2})


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