Nanomechanics is barely mechanical quantummechanics

Math

Let us define "quantumness" as the ratio of the energy quantisation (the minimum allowed energy steps) to the average thermal energy in a single degree of freedom:
Quantumness:

$$Q = \frac{\Delta E}{E_T}$$
First we'll need the thermal energy:
Equipartitioning: $$E_T = \frac{1}{2}k_BT \quad$$
The size of the energy quanta depends on the system under consideration.

Reciprocative linear motion

The uncertainty relation:

$$\Delta x \Delta p \geq h \quad$$ Newton: $$\Delta E = \frac{\Delta p^2}{2m} \quad$$
Quantumness: $$Q = \frac{h^2}{k_B} \frac{1}{m \Delta x^2 T}$$

Reciprocative circular motion

Here alpha is the fraction of a full circle that is passed through in a rotative oszillation.
For a normal unidirectional rotation alpha must be set to 2pi.

The uncertainty relation:

$$\alpha \Delta L \geq h \quad$$ Newton: $$\Delta E = \frac{\Delta L^2}{2I} \quad$$
Quantumness: $$Q = \frac{h^2}{k_B} \frac{1}{I \alpha^2 T}$$

Values

Average thermal energy per degree of freedom:

$$E_{T=300K} = 414 \cdot 10^{-23} J$$

rotative (full 360°)

$$L_0 = \hbar = 1.054 \cdot 10^{-34} {kg m^2} / s$$

$$L_0 = I \omega_0 = 2 m r^2 \omega$$

N₂ nitrogen molecule:

$$\quad 2r = 0.11 nm \quad m_N = 2.3 \cdot 10^{-26} kg$$

$$\omega_0 = 2 \pi f = 7.5 \cdot 10^{11} s^{-1}$$

$$f_0 = 119GHz$$

$$E_0 = I \omega_0^2 /2 = L_0 \omega_0 /2$$

Size of energy quanta:

$$E_0 = 3.95 \cdot 10^{-23} J$$

$$Q \lt 1/100$$

Note that this is a single free floating molecule. In advanced nano-machinery there are axles made of thousands and thousands of atoms which are in turn stiffly integrated in an axle system made out of millions of atoms. This is making energy quantisation imperceptible even at liquid helium temperatures.

linear

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Discussion

There are three parameters that can be changed to get something to behave more quantum mechanically.
The three options are:

• (1) lowering temperature
• (2) lowering inertia
• (3) decreasing the degree of freedom