Respuesta :
First part
The wavefunction is
[tex]\psi = \left \{ {{e^{-x} \ \ x\ge0} \atop {e^{x} \ \ x<0}} \right.[/tex]
As we know, the wavefunction is normalized if:
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx =1[/tex]
Lets prove it:
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = \int\limits^{0 } _{-\infty } {|\psi(x)|^2} \, dx + \int\limits^{\infty } _{0 } {|\psi(x)|^2} \, dx [/tex]
[tex]\ \int\limits^{0 } _{-\infty } {(e^{x} )^2} \, dx + \int\limits^{\infty } _{0 } {(e^{-x}) ^2} \, dx [/tex]
Lets take the first integral
[tex]\ \int\limits^{0 } _{-\infty } {(e^{x} )^2} \, dx [/tex]
we can use the substitution
[tex]x = -y \\\\dx = - dy[/tex]
when x= 0 then y = 0, and when [tex]x= - \infty[/tex] , [tex]y=\infty[/tex]
[tex]\ \int\limits^{0 } _{\infty } {(e^{-y} )^2} \, (-dy) [/tex]
[tex]\ - \int\limits^{0 } _{\infty } {(e^{-y} )^2} \, dy [/tex]
inverting the limits of integration to take care of the minus sign
[tex]\ \int\limits^{ \infty} _{0 } {(e^{-y} )^2} \, (dy) [/tex]
but this is the second integral!!!
so
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = 2 * \int\limits^{\infty } _{0 } {{e^{-2x} } \, dx [/tex]
now, we can see that
[tex]\frac{d}{dt} e^{-2x} = -2 e^{-2x}[/tex]
so
[tex] \frac{1}{-2} \frac{d}{dt} e^{-2x} = e^{-2x}[/tex]
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = 2 * \int\limits^{\infty } _{0 } { \frac{1}{-2} * \frac{d}{dt} e^{-2x} } \, dx [/tex]
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = \frac{2}{-2} * \int\limits^{\infty } _{0 } { \frac{d}{dt} e^{-2x} } \, dx [/tex]
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = -1 * [ \right e^{-2x} \left ] ^{\infty } _{0 } [/tex]
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = -1 * [ ( \lim_{x\rightarrow \infty } e^{-2x} ) -e^{-2*0} ] [/tex]
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = -1 * ( 0 - 1) [/tex]
[tex]\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = 1 [/tex]
This is what we wanted to show!
Second part
We just need to obtain [tex]p(|x| \le 0.5)[/tex]:
[tex]p(|x| \le 0.5) = \int\limits^{0.5 } _{-0.5 } {|\psi(x)|^2} \, dx [/tex]
We can use the same approach:
[tex]\int\limits^{0.5 } _{-0.5 } {|\psi(x)|^2} \, dx = \int\limits^{0 } _{-0.5 } {|\psi(x)|^2} \, dx + \int\limits^{0.5 } _{0 } {|\psi(x)|^2} \, dx [/tex]
[tex]\ \int\limits^{0 } _{-0.5 } {(e^{x} )^2} \, dx + \int\limits^{0.5 } _{0 } {e^{-x} ^2} \, dx [/tex]
[tex] p(|x| \le 0.5) = 2 * \int\limits^{0.5} _{0 } {{e^{-2x} } }\, dx [/tex]
[tex] p(|x| \le 0.5) = 2 * \int\limits^{0.5} _{0 } {\frac{1}{-2} \frac{d}{dt} e^{-2x} } \, dx [/tex]
[tex] p(|x| \le 0.5) = -1 * [ \right e^{-2x} \left ] ^{0.5} _{0 } [/tex]
[tex] p(|x| \le 0.5) = -1 * (e^{-1} - 1) [/tex]
[tex] p(|x| \le 0.5) = 0.6321 [/tex]
