3.5.6.2 So­lu­tion pipee-b

Ques­tion:

Just for fun, plug macro­scopic val­ues, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 kg and $\ell_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 m, into the ex­pres­sion for the ground state en­ergy and see how big it is. Note that $\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.054 57 10$\POW9,{-34}$ J s.

An­swer:

$$E_1 = \frac{\hbar^2\pi^2}{2m\ell_x^2} = \frac{(\mbox{1.054 ... ...\;1\mbox{ kg}\;1\mbox{ m}^2} = \mbox{5.488 10$\POW9,{-68}$ J}$$

or 3.4 10$\POW9,{-49}$ eV. That en­ergy is much less than you could ever hope to ob­serve phys­i­cally. A sin­gle pho­ton of light would dwarf it by 50 or­ders of mag­ni­tude.