Physics, 05.10.2019 02:30 serenityburke

The wavefunction of a particle confined to the x axis is ψ = e-x for x > 0 and ψ = ex for x < 0. show that the wavefunction is normalized and calculate the probablilty of finding the particle betweenx-0.5 and x 0.5.

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Answer from: tommy6668

First part

The wavefunction is

$\psi = \left \{ {{e^{-x} \ \ x\ge0} \atop {e^{x} \ \ x$

As we know, the wavefunction is normalized if:

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx =1$

Lets prove it:

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = \int\limits^{0 } _{-\infty } {|\psi(x)|^2} \, dx + \int\limits^{\infty } _{0 } {|\psi(x)|^2} \, dx$

$\ \int\limits^{0 } _{-\infty } {(e^{x} )^2} \, dx + \int\limits^{\infty } _{0 } {(e^{-x}) ^2} \, dx$

Lets take the first integral

$\ \int\limits^{0 } _{-\infty } {(e^{x} )^2} \, dx$

we can use the substitution

$x = -y \\\\dx = - dy$

when x= 0 then y = 0, and when $x= - \infty$ , $y=\infty$

$\ \int\limits^{0 } _{\infty } {(e^{-y} )^2} \, (-dy)$

$\ - \int\limits^{0 } _{\infty } {(e^{-y} )^2} \, dy$

inverting the limits of integration to take care of the minus sign

$\ \int\limits^{ \infty} _{0 } {(e^{-y} )^2} \, (dy)$

but this is the second integral!!!

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = 2 * \int\limits^{\infty } _{0 } {{e^{-2x} } \, dx$

now, we can see that

$\frac{d}{dt} e^{-2x} = -2 e^{-2x}$

$\frac{1}{-2} \frac{d}{dt} e^{-2x} = e^{-2x}$

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = 2 * \int\limits^{\infty } _{0 } { \frac{1}{-2} * \frac{d}{dt} e^{-2x} } \, dx$

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = \frac{2}{-2} * \int\limits^{\infty } _{0 } { \frac{d}{dt} e^{-2x} } \, dx$

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = -1 * [ \right e^{-2x} \left ] ^{\infty } _{0 }$

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = -1 * [ ( \lim_{x\rightarrow \infty } e^{-2x} ) -e^{-2*0} ]$

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = -1 * ( 0 - 1)$

$\int\limits^{\infty } _{-\infty } {|\psi(x)|^2} \, dx = 1$

This is what we wanted to show!

Second part

We just need to obtain $p(|x| \le 0.5)$ :

$p(|x| \le 0.5) = \int\limits^{0.5 } _{-0.5 } {|\psi(x)|^2} \, dx$

We can use the same approach:

$\int\limits^{0.5 } _{-0.5 } {|\psi(x)|^2} \, dx = \int\limits^{0 } _{-0.5 } {|\psi(x)|^2} \, dx + \int\limits^{0.5 } _{0 } {|\psi(x)|^2} \, dx$

$\ \int\limits^{0 } _{-0.5 } {(e^{x} )^2} \, dx + \int\limits^{0.5 } _{0 } {e^{-x} ^2} \, dx$

$p(|x| \le 0.5) = 2 * \int\limits^{0.5} _{0 } {{e^{-2x} } }\, dx$

$p(|x| \le 0.5) = 2 * \int\limits^{0.5} _{0 } {\frac{1}{-2} \frac{d}{dt} e^{-2x} } \, dx$

$p(|x| \le 0.5) = -1 * [ \right e^{-2x} \left ] ^{0.5} _{0 }$

$p(|x| \le 0.5) = -1 * (e^{-1} - 1)$

$p(|x| \le 0.5) = 0.6321$

Answer from: Quest

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justin can't pay the full balance:

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Answer from: Quest

My answer is opptionc, just check with anyone else