quantum mechanics - Can we theoretically balance a perfectly symmetrical pencil on its one-atom tip?

I was asked by an undergrad student about this question. I think if we were to take away air molecules around the pencil and cool it to absolute zero, that pencil would theoretically balance.

Am I correct?

Veritasium/Minutephysics video on Youtube.

Answer

No. To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time.

Momentum and position form a conjugate pair. $\Delta x \Delta p \geq \hbar$.

Angular momentum and angular position form one too. $\Delta L \Delta \Theta \geq \hbar$

This doesn't guarantee that angular momentum and angular position will be non-zero. It is an uncertainty - The actual values can be anything, including 0.

But it does prevent you from arranging them both so the pencil stays upright. Furthermore, if you ask what the probability of finding both values very close to 0, you find that it is very small. In the limit, infinitely improbable.

If it turns out that $L = \Theta = \sqrt{\hbar}$, and you plug in reasonable values for the mass and length of the pencil, you will find it falls over in a few seconds.

Belated update

I was waiting until the weekend to add an update. By the time it got here, Floris had left very little to add. And he did a better job than I would have. Good answers.

A number of users felt that an ideal pencil sharpened to an atomic tip was not realistic. The pencil should have a flat on the bottom.

My own thought is that the pencil should be mounted on one of those massless, frictionless pulleys that seem to be so common in high school physics classrooms.

Never the less, a pencil with a flat can be treated semi classically. Because of the uncertainty principal, the pencil has an initial momentum, and therefore an initial energy. This will cause the pencil to tip. Which in turn will cause the pencil to rotate about an edge of the flat. The center of mass will rise until it is directly over the edge of the flat. If the initial "uncertainty" energy is larger than the energy needed to raise the center of mass, the pencil will tip over.

A quantum mechanical treatment would treat the region where the center of mass is over the interior of the flat as a potential well. There is a probability of tunneling out.

Both of these scenarios are treated in full detail (with diagrams in case my description is unclear) here. I found this link by following Floris' "interesting post that calculates the same thing." That post had some comments at the bottom. The very last comment contains the link.