Question

About how long would it take for a (full) can of beer at room temperature to topple over spontaneously, as a result of quantum tunneling? Hint: Treat it as a uniform cylinder of mass m, radius R, and height h. As the can tips, let x be the height of the center above its equilibrium position (h/2) .The potential energy is mgx, and it topples when x reaches the critical value ${\mathbf{X}}_{\mathbf{0}}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{h}\mathbf{/}\mathbf{2}\mathbf{)}}^{\mathbf{2}}}\mathbf{-}\mathit{h}\mathbf{/}\mathbf{2}$. Calculate the tunneling probability (Equation
8.22), for E = 0. Use Equation 8.28, with the thermal energy $\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{)}{\mathbf{mv}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{\left)}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}\mathbf{\right)}$to estimate the velocity. Put in reasonable numbers, and give your final answer in years.

$\mathbf{T}\mathbf{\approx }{\mathbf{e}}^{\mathbf{-}{\mathbf{2}}_{\mathbf{T}}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{with}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\gamma }\mathbf{\equiv }\frac{\mathbf{1}}{\sout{\mathbf{h}}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{a}}\left|p\left(x\right)\right|\mathbf{dx}\mathbf{.}\mathbf{.}\mathbf{.}$ (8.22).

tau=$\frac{\mathbf{2}{\mathbf{r}}_{\mathbf{1}}}{\mathbf{V}}{{\mathbf{e}}^{\mathbf{2}}}_{\mathbf{T}}$ (8.28).

Short answer

The tunneling probability to estimate the velocity is$\Gamma ={10}^{5\times {10}^{30}yr}$

Step-by-step solution

Tunneling probability

The tunneling probability is a ratio of barrier of squared amplitudes of the wave of the wave past the barrier to the incident wave. The tunnelling probability depends on the energy of the incident particles relative of the incident particle relative to the height of the barrier and on the width of the barrier.

Step 2: Calculating the tunneling probability to estimate the velocity

The tunneling probability:

$\mathrm{T}={\mathrm{e}}^{-2\mathrm{\gamma }}$

Where $\gamma$is the tunneling factor and it is given by:

$\mathrm{\gamma }=\frac{1}{\sout{\mathrm{h}}}{\int }_0^{{\mathrm{x}}_0}\sqrt{2\mathrm{m}(\mathrm{V}-\mathrm{E}}\mathrm{dx}.$

Let x be the center of the mass of the can, then its potential energy is:

$\mathrm{V}\left(\mathrm{x}\right)=\mathrm{mgx}$ (Half the diagonal).

$$\begin{gathered} \frac{\sqrt{2\mathrm{m}}}{\sout{\mathrm{h}}}\sqrt{\mathrm{mg}}{\int }_0^{{\mathrm{x}}_0}{\mathrm{X}}^{1/2}\mathrm{dx}={\overline{)\frac{\mathrm{m}}{\sout{\mathrm{h}}}\sqrt{2\mathrm{g}}\frac{2}{3}{\mathrm{x}}^{3/2}}}_0^{{\mathrm{x}}_0} \\ =\frac{2\mathrm{m}}{3\sout{\mathrm{h}}}\sqrt{2\mathrm{g}}{\mathrm{x}}_0^{3/2} \\ \mathrm{T}\approx {\mathrm{e}}^{-2\mathrm{\gamma }},\mathrm{with}\mathrm{\gamma }\equiv \frac{1}{\sout{\mathrm{h}}}{\int }_0^{\mathrm{a}}\left|\mathrm{p}\left(\mathrm{x}\right)\right|\mathrm{dx}... \end{gathered}$$

Estimate:

$$\begin{gathered} :\mathrm{h}=10\mathrm{cm},\mathrm{R}=3\mathrm{cm},\mathrm{m}=300\mathrm{gm};\mathrm{let}\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2.\mathrm{Then}{\mathrm{x}}_0=\sqrt{9+25}-5=0.83\mathrm{cm};\mathrm{and} \\ \mathrm{\gamma }=\frac{\left(2\right)\left(0.3\right)}{\left(3\right)\left(1.05\times {10}^{-34}\right)}\sqrt{\left(2\right)\left(9.8\right)}{\left(0.0083\right)}^{3/2}=6.4\times {10}^{30} \end{gathered}$$

Frequency of attempts: say f=v/2R . We want the product of the number of attempts (ft) and the probability of toppling at each attempt (T), to be 1:

$$\begin{gathered} \mathrm{t}\frac{\mathrm{V}}{2\mathrm{R}}{\mathrm{e}}^{-2\mathrm{\gamma }}=1 \\ \Rightarrow \mathrm{t}=\frac{2\mathrm{R}}{\mathrm{v}}{\mathrm{e}}^{2\mathrm{\gamma }}. \end{gathered}$$

Step 3: Estimating the thermal velocity

Estimating the thermal velocity: $\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}$(I’m done with the tunneling probability; from now on T is the temperature, 300 K) $\Rightarrow \mathrm{v}=\sqrt{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}/\mathrm{m}.}$

$$\begin{gathered} \mathrm{t}=2\mathrm{R}\sqrt{\frac{\mathrm{m}}{{\mathrm{k}}_{\mathrm{B}}{\mathrm{T}}^{{\mathrm{e}}^{2\mathrm{\gamma }}}}} \\ =2\left(0.03\right)\sqrt{\frac{0.3}{\left(1.4\times {10}^{-23}\right)\left(300\right)}} \\ =5\times {10}^8{\left({10}^{\mathrm{loge}}\right)}^{13\times {10}^{30}} \\ =\left(5\times {10}^8\right)\times {10}^{5.6\times {10}^{30}\mathrm{S}.} \\ =16\times {10}^{5.6\times {10}^{30}\mathrm{yr}} \end{gathered}$$

Don’t hold your breath.

[Actually, to tunnel all the way through the classically forbidden region, the center of mass must not only rise from 0 to $x_0$, but also drop back to 0. This doubles γ, and makes the final exponent]