In the ground state of the harmonic oscillator, what is the probability (correct to three significant digits) of finding the particle outside the classically allowed region?

A probability amplitude is a complex quantity used to describe system behaviour in quantum mechanics.

Use the formula,$P={\int }_{x_1}^{x_2}{y_0}^{*}{y}_{0}dx$

Here, $y_0$represent the wave function of the particle and $P$is the probability

Born's interpretation is that ${\left|{\Psi }_0(x,t)\right|}^2$represents the probability distribution for the particle's position in the ground state at time t. The likelihood that a particle will be found in a location that is traditionally permissible is ${\int }_{-a}^a\mid {\Psi }_0(x,t{)}^{2}dx$.

The probability that it’s not in this region is,

$1-{\int }_{-a}^a{\left|{\Psi }_0(x,t)\right|}^{2}dx=1-{\int }_{-a}^a{\Psi }_0(x,t){\Psi }_0^{*}(x,t)dx$

localid="1660906664910" $=1-{\int }_{-a}^a\left[-{\psi }_0\left(x\right)e^{-E_{0}t/\hslash }\right]\left[{\psi }_0\left(x\right)e^{i{E}_{0}t/\hslash }\right]dx$

$=1-{\int }_{-a}^a{\left[{\psi }_0\left(x\right)\right]}^{2}dx$

localid="1660906677303" $=1-{\int }_{-a}^a{\left[{\left(\frac{m\omega }{\pi {\displaystyle \hslash }}\right)}^{1/4}\exp \left(-\frac{m\omega }{2{\displaystyle \hslash }}{x}^2\right)\right]}^{2}dx$

Further evaluating,

localid="1660906688274" $=1-{\int }_{-a}^a\left[{\left(\frac{m\omega }{\pi {\displaystyle \hslash }}\right)}^{1/2}\exp \left(-\frac{m\omega }{{\displaystyle \hslash }}{x}^2\right)\right]dx$

localid="1660906698298" $=1-\sqrt{\frac{m\omega }{{\displaystyle \hslash }}{\int }_{-a}^a}\exp \left(-\frac{m\omega }{{\displaystyle \hslash }}{x}^2\right)dx$

Use

localid="1660906707184" $\xi =\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}x$

localid="1660906713573" $d\xi =\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}dx$

localid="1660906720474" $dx=\sqrt{\frac{{\displaystyle \hslash }}{m\omega }}d\xi$

Thus,

localid="1660906740360" $1-{\int }_{-a}^a{\left|{\psi }_0\left(x,t\right)\right|}^{2}dx=1-\sqrt{\frac{m\omega }{\pi {\displaystyle \hslash }}}{\int }_{-\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}}^{\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}}{e}^{-{\xi }^2}\left(\sqrt{\frac{{\displaystyle \hslash }}{m\omega }}d\xi \right)$

localid="1660906752800" $=1-\frac{1}{\sqrt{\pi }}{\int }_{-\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}}^{\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}}{e}^{-{\xi }^2}d\xi$

localid="1660906763118" $=1-\frac{2}{\sqrt{\pi }}{\int }_0^{\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}}{e}^{-{\xi }^2}d\xi$

Now the error function is defined as

localid="1660906772839" $1-{\int }_{-a}^a{\left|{\Psi }_0(x,t)\right|}^{2}dx=1-erf\left(\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}a\right)$

Thus the probability is,

localid="1660906782708" $1-{\int }_{-a}^a{\left|{\Psi }_0(x,t)\right|}^{2}dx=1-erf\left(\sqrt{\frac{m\omega }{{\displaystyle \hslash }}}a\right)$

localid="1660906791838" ${\psi }_0={\left(\frac{m\omega }{\pi {\displaystyle \hslash }}\right)}^{\frac{1}{4}}{e}^{-{\xi }^2/2}$

The energy of the ground state is known, so

localid="1660906804773" $\frac{{\displaystyle \hslash }\omega }{2}=\frac{1}{2}m{\omega }^2{a}^2$

Evaluate for a and get,

localid="1660906820214" $a=\sqrt{\frac{{\displaystyle \hslash }}{m\omega }}$

Thus the probability is,

$1-{\left.{\int }_{-i}^a{\Psi }_0(x,t)\right|}^{2}dx=1-erf1$

$\approx 0.157$

Therefore, the probability that the particle lies outside the classically allowed region in the ground state is $0.157$.