[This problem generalizes Example 1.2.] Imagine a particle of mass m and energy E in a potential well V(X), sliding frictionlessly back and forth between the classical turning points (a and b in Figure 1.10). Classically, the probability of finding the particle in the range dx (if, for example, you

Question

[This problem generalizes Example 1.2.] Imagine a particle of mass m and energy E in a potential well V(X), sliding frictionlessly back and forth between the classical turning points (a and b in Figure 1.10). Classically, the probability of finding the particle in the range dx (if, for example, you took a snapshot at a random time t) is equal to the fraction of the time T it takes to get from a to b that it spends in the interval dx:

[latex] ρ(x) d x=\frac{d t}{T}=\frac{(d t / d x) d x}{T}=\frac{1}{v(x) T} d x [/latex] (1.41)

where v(x) is the speed, and

[latex] T=\int_{0}^{T} d t=\int_{a}^{b} \frac{1}{v(x)} d x [/latex]. (1.42).

Thus

[latex] ρ(x) = \frac{1}{v(x)T} [/latex] (1.43).

This is perhaps the closest classical analog to [latex] \left|\Psi \right|^2 [/latex].

(a) Use conservation of energy to express v(x) in terms of E and V(x) .

(b) As an example, find ρ(x) for the simple harmonic oscillator, [latex] V(x) = kx^2/2 [/latex]. Plot ρ(x), and check that it is correctly normalized.

(c) For the classical harmonic oscillator in part (b), find [latex] \left\langle x\right\rangle ,\left\langle x^2\right\rangle [/latex] and [latex] σ_x [/latex].

Accepted Answer

(a)

[latex] \frac{1}{2} = m v^2 + V = E [/latex] → [latex] v(x) = \sqrt{\frac{2}{m} (E-V(x))} [/latex].

(b)

[latex] T=\int_{a}^{b} \frac{1}{\sqrt{\frac{2}{m}\left(E-\frac{1}{2} k x^{2}\right)}} d x=\sqrt{\frac{m}{k}} \int_{a}^{b} \frac{1}{\sqrt{(2 E / k)-x^{2}}} d x [/latex].

Turning points: v = 0 ⇒ [latex] E = V = \frac{1}{2}k b^2 [/latex] ⇒ [latex] b = \sqrt{2E/k}[/latex]; a = -b.

[latex] T=2 \sqrt{\frac{m}{k}} \int_{0}^{b} \frac{1}{\sqrt{b^{2}-x^{2}}} d x=\left.2 \sqrt{\frac{m}{k}} \sin ^{-1}\left(\frac{x}{b}\right)\right|_{0} ^{b}=2 \sqrt{\frac{m}{k}} \sin ^{-1}(1) [/latex].

[latex] = 2 \sqrt{\frac{m}{k} } \Bigl(\frac{\pi }{2}\Bigr) = \pi \sqrt{\frac{m}{k} } [/latex].

[latex] \rho(x)=\frac{1}{\pi \sqrt{\frac{m}{k}} \sqrt{\frac{2}{m}\left(E-\frac{1}{2} k x^{2} 1\right)}}= \frac{1}{\pi \sqrt{b^2 -x^2} } [/latex].

[latex] \int_{a}^{b} \rho(x) d x=\frac{2}{\pi} \int_{0}^{b} \frac{1}{\sqrt{b^{2}-x^{2}}} d x=\frac{2}{\pi}\left(\frac{\pi}{2}\right)=1 [/latex].

(c) [latex] \left\langle x\right\rangle = 0 [/latex].

[latex] \left\langle x^{2}\right\rangle=\frac{1}{\pi} \int_{-b}^{b} \frac{x^{2}}{\sqrt{b^{2}-x^{2}}} d x=\frac{2}{\pi} \int_{0}^{b} \frac{x^{2}}{\sqrt{b^{2}-x^{2}}} d x [/latex].

[latex] =\left.\frac{2}{\pi}\left[-\frac{x}{2} \sqrt{b^{2}-x^{2}}+\frac{b^{2}}{2} \sin ^{-1}\left(\frac{x}{b}\right)\right]\right|_{0} ^{b}=\frac{b^{2}}{\pi} \sin ^{-1}(1)=\frac{b^{2}}{\pi} \frac{\pi}{2}=\frac{b^{2}}{2}= \frac{E}{k} [/latex].

[latex] \sigma_{x}=\sqrt{\left\langle x^{2}\right\rangle-\langle x\rangle^{2}}=\sqrt{\left\langle x^{2}\right\rangle}=\frac{b}{\sqrt{2}}= \sqrt{\frac{E}{k}}[/latex].

Question 1.11: [This problem generalizes Example 1.2.] Imagine a particle o...

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