Pencil on tip
A pencil sits on its point in an unstable equilibrium. The pencil is 18 cm long and has a mass of 7 grams. From a quantum mechanical perspective this pencil has to fall. Classically it could remain in unstable equilibrium. Please estimate the time necessary for the pencil to land on the table, from its initial upright unstable position on its point.
Small oscillations about all positions exist in quantum mechanics. . The uncertainty principle states this. Because of this, an unstable equilibrium position in quantum mechanics can not be maintained. After a certain time, a pencil that sits on its tip is going to fall on the table, on all cases after a certain time.
The uncertainty principle reads for complementary variables. Complementary variables have their commutator nonzero. Examples of complementary variables are $x$ and $p$ , $E$ and $t$ or $\phi and $x$. The general equation for the uncertainty principle is
$\Delta(a)\Delta(b) \geq \hbar/2$
Continue reading on Heisenberg uncertainty principle.
The unstable equilibrium
The uncertainty principle for angles ($\phi$) and positions ($L$) (length of pencil) is written as
The uncertainty principle for linear momentum ($P$) and position ($L$) (length of pencil) is
$\Delta(P)*\Delta(L) >= \hbar/2$
The pencil is positioned as well as possible on its tip and needs to fall. The uncertainty in its position is $L=18 cm$.
$\Delta(\phi) = (\hbar/2) /L$ (1)
Also regarding its momentum one can derive
$\Delta(P) = (\hbar/2) /L$
$\Delta(V) = (\hbar/2) /(L*m)$ where V is speed.
Assuming that the initial speed is equal to the uncertainty $V0=\Delta(V) =(\hbar/2)/(L*m)$
the relation between initial speed and angle is:
($R =L$ assuming initially the pencil is positioned on its tip)
From the equation (1) we have:
$L*\Delta(\phi) = (\hbar/2)$
$(\hbar/2) /(L*m) = (\hbar/2) / \Delta(t)$
$\Delta(t) =L*m =0.007*0.18 =0.00126 seconds$
This expressed to one significant figure is
$\Delta(t)=0.001 sec =1 ms$