Question

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?

Short answer

Answer

The probability of finding the particle is$0.157$

Step-by-step solution

- Define probability in quantum mechanics

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

- Compute 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 $0.157$.