Find the momentum-space wave function, $\mathit{\Phi }\mathbf{(}\mathit{p}\mathbf{,}\mathbf{}\mathit{t}\mathbf{)}$,for a particle in the ground state of the harmonic oscillator. What is the probability (to 2significant digits) that a measurement of p on a particle in this state would yield a value outside the classical range (for the same energy)? Hint: Look in a math table under "Normal Distribution" or "Error Function" for the numerical part-or use Mathematica.

Any function (say $g\left(x\right)$) can be expressed in terms of the eigenfunctions $f_p\left(x\right)$of the momentum operator, for a function$g\left(x\right)$,

$g\left(x\right)={\int }_{-\infty }^{\infty }c\left(p\right)f_p\left(x\right)dp$

Any function (say $g\left(x\right)$) can be expressed in terms of the eigenfunctions $f_p\left(x\right)$of the momentum operator, for a function $g\left(x\right)$,

$\begin{array}{rcl}g\left(x\right)& =& {\int }_{-\infty }^{\infty }c\left(p\right)f_p\left(x\right)dp\\ & =& \frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{\infty }c\left(p\right)e^{ipx/\hslash }dp\end{array}$

Where $c\left(p\right)$is the inverse Fourier transform:

If $g$is a wave function $\Psi (x,t)$, we can view $c(p,t)$as the momentum-space wave function, often given the symbol $\Phi (p,t)$. That is:$\Phi (p,t)=\frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{\infty }\Psi (x,t)e^{-ipx/\hslash }dx$

The ground state wave function for the harmonic oscillator:${\psi }_0(x,t)={\left(\frac{m\omega }{\pi \hslash }\right)}^{1/4}{e}^{-m\omega x^2/2\hslash }{e}^{-i\omega t/2}$${\psi }_0(x,t)={\left(\frac{m\omega }{\pi \hslash }\right)}^{1/4}{e}^{-m\omega x^2/2\hslash }{e}^{-i\omega t/2}$

Let $\alpha =(m\omega /\pi \hslash {)}^{1/4}$and $\beta =m\omega /2\hslash$, thus we get

${\Phi }_0(p,t)=\frac{\alpha e^{-i\omega t/2}}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{\infty }{e}^{-ipx/\hslash }{e}^{-\beta x^2}dx$ ...... (1)

The value of the integral, first we complete the square of the quadratic equation as:

$\begin{array}{rcl}{\int }_{-\infty }^{\infty }{e}^{-ipx/h}{e}^{-\beta x^2}dx& =& {\int }_{-\infty }^{\infty }\exp \left\{-\beta \left(x^2+\frac{ip}{h\beta }x\right)\right\}dx\\ & =& {\int }_{-\infty }^{\infty }\exp \left\{-\beta \left(x^2+\frac{ip}{h\beta }x+{\left(\frac{ip}{2h\beta }\right)}^2\right)+\beta {\left(\frac{ip}{2h\beta }\right)}^2\right\}dx\\ & =& {\int }_{-\infty }^{\infty }\exp \left\{-\beta {\left(x+\frac{ip}{2h\beta }\right)}^2+\beta {\left(\frac{ip}{2h\beta }\right)}^2\right\}dx\\ & =& e^{\beta {(ip/2h\beta )}^2}{\int }_{-\infty }^{\infty }\exp \left\{-\beta {\left(x+\frac{ip}{2h\beta }\right)}^2\right\}dx\end{array}$

let $y=x+ip/2\hslash \beta$, so $dy=dx$, thus we get

${\int }_{-\infty }^{\infty }{e}^{ipx/\hslash }e{e}^{\beta x^2}dx=e^{\beta {(ip/2\hslash \beta )}^2}{\int }_{-\infty }^{\infty }{e}^{\beta v^2}dy$

Substitute ${\int }_{-\infty }^{\infty }{e}^{\beta v^2}dy=\sqrt{\frac{\pi }{\beta }}$in above equation

${\int }_{-\infty }^{\infty }{e}^{-ipx/h}{e}^{-\beta x^2}dx=\sqrt{\frac{\pi }{\beta }{e}^{\zeta {(im/2hj)}^2}}$

Substitute into equation (1),

${\Phi }_0(p,t)=\frac{\alpha e^{-i\omega t/2}}{\sqrt{2\pi h}}\sqrt{\frac{\pi }{\beta }}{e}^{\beta {(ip/2\hslash \beta )}^2}$

Now substitute with $\beta$and $\alpha$to get:

${\Phi }_0(p,t)=\frac{1}{{\left(\pi m\omega \hslash \right)}^{1/4}}{e}^{-p^2/2m\omega \hslash h}{e}^{-i\omega t/2}$ ...... (2)

This is the required momentum space wave function.

Integrate the modulus squared of the wave function outside the classical range but before we need to find this range, the ground state energy is $E_0=\hslash \omega /2$so the maximum ground state momentum is $p_0=\sqrt{2m{E}_0}= \sqrt{m\hslash \omega }$, therefore the classical range is from $-\sqrt{m\hslash \omega }$to $\sqrt{m\hslash \omega }$, so the probability outside this range is from $-\infty$to $\infty$, but since the probability is an even function then, we can just integrate from to $\infty$and multiply the integral by 2 , as:

$\text{Prob}=2{\int }_{\sqrt{m\hslash \omega }}^{\infty }{\left|{\Phi }_0(x,t)\right|}^{2}dp$

Substitute from equation 2, we get,

$\begin{array}{rcl}\text{Prob}& =& \frac{2}{\sqrt{\pi m\hslash \omega }}{\int }_{\sqrt{m\hslash \omega }}^{\infty }{e}^{-p^2/m\hslash \omega }dp\\ & =& 1-erf\left(1\right)\\ & =& 0.15\\ & =& 0.15\end{array}$

Thus, the probability of measurement of particle is 0.15.