Calculate the probability of finding a particle in the classically forbidden region of a harmonic oscillator for the states n = 0, 1, 2, 3, 4. Are these results compatible with their classical counterparts?

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The classical turning points are defined by [latex]E_{n} =V(x_{n} )[/latex] or by [latex]\hbar \omega (n+\frac{1}{2} )=\frac{1}{2}m\omega ^{2} x^{2}_{n}[/latex]; that is, [latex]x_{n}=\pm \sqrt{\hbar /(m \omega )} \sqrt{2n+1}[/latex]. Thus, the probability of finding a particle in the classically forbidden region for a state [latex]\psi _{n}(x) [/latex] is

[latex]P_{n} =\int_{-\infty }^{-|x_{n}|}\left|\psi _{n}(x)\right| ^{2} dx+\int_{|x_{n}|}^{+\infty }\left|\psi _{n}(x)\right| ^{2}dx=2 \int_{|x_{n}|}^{+\infty }\left|\psi _{n}(x)\right| ^{2}dx[/latex], (4.297)

where [latex]\psi _{n}(x) [/latex] is given in (4.172),

[latex]\psi _{n}(x)=\frac{1}{\sqrt{\pi }2^{n}n!x_{0}} e^{-x^{2}/2 x^{2}_{0}} H_{n}\left(\frac{x}{x_{0} } \right) [/latex]. (4.172)

[latex]\psi _{n}(x)=1/\sqrt{\sqrt{\pi }2^{n}n!x_{0} } e^{-x^{2} /2x^{2}_{0}}H_{n}(x/x_{0})[/latex], where [latex]x_{0}[/latex] is given by [latex]x_{0}=\sqrt{\hbar /(m\omega )}[/latex]. Using the change of variable [latex]y=x/x_{0}[/latex], we can rewrite [latex] P_{n}[/latex] as

[latex]P_{n}=\frac{2}{\sqrt{\pi }2^{n}n! } \int_{\sqrt{2n+1} }^{+\infty }e^{-y^{2}}H^{2}_{n}(x) dy[/latex], (4.298)

where the Hermite polynomials [latex]H_{n}(y)[/latex] are listed in (4.120).

[latex]H_{0}(y)=1,[/latex] [latex]H_{1}(y)=2y,[/latex]

[latex]H_{2}(y)=4y^{2} -2,[/latex] [latex]H_{3}(y)=8y^{2}-12y,[/latex]

[latex]H_{4}(y)=16y^{4}-48y^{2}-12y+12,[/latex] [latex]H_{5}(y)=32y^{5}-160y^{3}+120y.[/latex]

The integral in (4.298) can be evaluated only numerically. Using the numerical values

[latex]\int_{1}^{\infty } e^{-y^{2}}dy=0.1394, [/latex] [latex] \int_{\sqrt{3} }^{\infty }y^{2}e^{-y^{2}}dy=0.0495, [/latex] (4.299)

[latex]\int_{\sqrt{5} }^{\infty }(4y^{2}-2)^{2} e^{-y^{2}}dy=0.6740, [/latex] [latex]\int_{\sqrt{7} }^{\infty }(8y^{3}-12y)^{2}e^{-y^{2}}dy=3.6363, [/latex] (4.300)

[latex]\int_{\sqrt{9} }^{\infty }(16y^{4}-48y^{2}+12)^{2}e^{-y^{2}}dy=26.86, [/latex] (4.301)

we obtain

[latex]P_{0}=0.1573, [/latex] [latex]P_{1}=0.1116, [/latex] [latex]P_{2}=0.095 069, [/latex] (4.302)

[latex]P_{3}=0.085 48, [/latex] [latex]P_{4}=0.078 93. [/latex] (4.303)

This shows that the probability decreases as n increases, so it would be very small for very large values of n. It is therefore unlikely to find the particle in the classically forbidden region when the particle is in a very highly excited state. This is what we expect, since the classical approximation is recovered in the limit of high values of n.

Question 4.P.12: Calculate the probability of finding a particle in the class...

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