Problem:
The ground state wavefunction for a particle in a shallow potential well in 1 dimension is of the form: $ A e^{-\frac{|x|}{2d}} $. Given that the particle must be found somewhere in the range $ x \in (-\infty, \infty) $, the born rule then places a constraint on the modulus of A. Assuming that A is real and positive, what is the value of A?
Solution:
The born rule requires the squared wavefunction is the probability density function of finding the particle, which means it is simply an integration problem.
$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{\psi^2(x)dx} &=& 1 \\ \int\limits_{-\infty}^{\infty}{(Ae^{-\frac{|x|}{2d}})^2dx} &=& 1 \\ A^2\int\limits_{-\infty}^{\infty}{e^{-\frac{|x|}{d}}dx} &=& 1 \\ 2A^2\int\limits_{0}^{\infty}{e^{-\frac{x}{d}}dx} &=& 1 \\ 2A^2(-de^{-\frac{x}{d}})|_{0}^{\infty} &=& 1 \\ 2A^2(0 - (-d)) &=& 1 \\ A^2 &=& \frac{1}{2d} \\ A &=& \frac{1}{\sqrt{2d}} \end{eqnarray*} $
That is the answer we wanted.
The ground state wavefunction for a particle in a shallow potential well in 1 dimension is of the form: $ A e^{-\frac{|x|}{2d}} $. Given that the particle must be found somewhere in the range $ x \in (-\infty, \infty) $, the born rule then places a constraint on the modulus of A. Assuming that A is real and positive, what is the value of A?
Solution:
The born rule requires the squared wavefunction is the probability density function of finding the particle, which means it is simply an integration problem.
$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{\psi^2(x)dx} &=& 1 \\ \int\limits_{-\infty}^{\infty}{(Ae^{-\frac{|x|}{2d}})^2dx} &=& 1 \\ A^2\int\limits_{-\infty}^{\infty}{e^{-\frac{|x|}{d}}dx} &=& 1 \\ 2A^2\int\limits_{0}^{\infty}{e^{-\frac{x}{d}}dx} &=& 1 \\ 2A^2(-de^{-\frac{x}{d}})|_{0}^{\infty} &=& 1 \\ 2A^2(0 - (-d)) &=& 1 \\ A^2 &=& \frac{1}{2d} \\ A &=& \frac{1}{\sqrt{2d}} \end{eqnarray*} $
That is the answer we wanted.
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