Question:
Suppose we have a particle in 1-dimension, with wavefunction $ Ae^{-\frac{|x|}{2d}} $. What is the probability to find the particle in the interval $ [0, d] $?
Solution:
The born's interpretation tell us the probability is:
$ \begin{eqnarray*} P &=& \int\limits_{0}^{d}{\psi(x, t)^2dx} \\ &=& \int\limits_{0}^{d}{(Ae^{-\frac{|x|}{2d}})^2dx} \\ &=& A^2\int\limits_{0}^{d}{(e^{-\frac{x}{d}})dx} \\ &=& A^2(-de^{-\frac{x}{d}})|_0^d \\ &=& A^2(-de^{-1}-(-d)) \\ &=& A^2d(1-e^{-1}) \\ \end{eqnarray*} $
If we wish, we can substitute the $ A = \frac{1}{\sqrt{2d}} $ in it.
Suppose we have a particle in 1-dimension, with wavefunction $ Ae^{-\frac{|x|}{2d}} $. What is the probability to find the particle in the interval $ [0, d] $?
Solution:
The born's interpretation tell us the probability is:
$ \begin{eqnarray*} P &=& \int\limits_{0}^{d}{\psi(x, t)^2dx} \\ &=& \int\limits_{0}^{d}{(Ae^{-\frac{|x|}{2d}})^2dx} \\ &=& A^2\int\limits_{0}^{d}{(e^{-\frac{x}{d}})dx} \\ &=& A^2(-de^{-\frac{x}{d}})|_0^d \\ &=& A^2(-de^{-1}-(-d)) \\ &=& A^2d(1-e^{-1}) \\ \end{eqnarray*} $
If we wish, we can substitute the $ A = \frac{1}{\sqrt{2d}} $ in it.
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