What if we were interested in the distribution of momenta (p=m v), for the classical harmonic oscillator (Problem 1.11(b)), (a) Find the classical probability distribution ρ(p) (note that p ranges from -√2mE to +√2mE). (b) Calculate (p) , (p^2) , and σp. (c) What’s the classical uncertainty product

Question

What if we were interested in the distribution of momenta (p=m v), for the classical harmonic oscillator (Problem 1.11(b)),

(a) Find the classical probability distribution ρ(p) (note that p ranges from [latex] -\sqrt{2 m E} [/latex] to + [latex] \sqrt{2 m E}).

(b) Calculate [latex] \left\langle p\right\rangle , \left\langle ^2\right\rangle [/latex] , and [latex] σ_p[/latex].

(c) What’s the classical uncertainty product [latex] σ_xσ_p[/latex], for this system? Notice that this product can be as small as you like, classically, simply by sending E → 0 , But in quantum mechanics, as we shall see in Chapter 2, the energy of a simple harmonic oscillator cannot be less than [latex] \hbar \omega /2[/latex], where [latex] ω = \sqrt{k/m}[/latex] is the classical frequency. In that case what can you say about the product [latex] σ_xσ_p [/latex]?

Accepted Answer

(a)[latex] ρ(p) d p=\frac{d t}{T}=\frac{|d t / d p| d p}{T} [/latex].

where dt is now the time it spends with momentum in the range dp (dt is intrinsically positive, but dp/dt = F = -kx runs negative-hence the absolute value). Now

[latex] \frac{p^{2}}{2 m}+\frac{1}{2} k x^{2}=E \Rightarrow x=\pm \sqrt{\frac{2}{k}\left(E-\frac{p^{2}}{2 m}\right)} [/latex],

so

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

where [latex] c ≡ \sqrt{2 m E} [/latex]. This is the same as ρ(x) (Problem 1.11(b)), with c in place of b (and, of course, p in place of x).

(b) From Problem 1.11(c), [latex] \langle p\rangle=0,\left\langle p^{2}\right\rangle=\frac{c^{2}}{2}, \sigma_{p}=\frac{c}{\sqrt{2}}=\sqrt{m E} [/latex].

(c) [latex] σ_xσ_p = \sqrt{\frac{E}{k} } \sqrt{m E} = \sqrt{\frac{m}{k} } E = \frac{E}{ω}[/latex]. If [latex] E ≥ \frac{1}{2}\hbar \omega [/latex] then [latex] σ_xσ_p ≥ \frac{1}{2}\hbar [/latex] which is precisely the Heisenberg uncertainty principle!

Question 1.12: What if we were interested in the distribution of momenta (p...

Related Answered Questions