Question

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.

Answer 1

First part

The wavefunction is

`\psi = \left \{ {{e^(-x) \ \ x\ge0} \atop {e^(x) \ \ x<0}} \right.`

As we know, the wavefunction is normalized if:

`\int\limits^(\infty ) _(-\infty ) \, dx  =1`

Lets prove it:

`\int\limits^(\infty ) _(-\infty ) \, dx  = \int\limits^(0 ) _(-\infty ) \, dx  + \int\limits^(\infty ) _(0 ) \psi(x) \, 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!!!

so

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

now, we can see that

`(d)/(dt) e^(-2x) = -2 e^(-2x)`

so

`(1)/(-2) (d)/(dt) e^(-2x) = e^(-2x)`

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

`\int\limits^(\infty ) _(-\infty ) \, dx  = (2)/(-2)  *   \int\limits^(\infty ) _(0 ) {  (d)/(dt) e^(-2x) } \, dx`

`\int\limits^(\infty ) _(-\infty ) \, dx  = -1  *  [ \right e^(-2x) \left ] ^(\infty ) _(0 )`

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

`\int\limits^(\infty ) _(-\infty ) ^2 \, dx  = -1  * ( 0 - 1)`

`\int\limits^(\infty ) _(-\infty ) \psi(x) \, 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 ) \, dx`

We can use the same approach:

`\int\limits^(0.5 ) _(-0.5 ) \, dx  = \int\limits^(0 ) _(-0.5 ) \psi(x) \, dx  + \int\limits^(0.5 ) _(0 ) ^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 ) {(1)/(-2) (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`