Semiclassical mechanics~

The basic outline of the solution is as follows :

• Calculate the mean and mean of squares of positions and from that calculate the standard deviation of position.
• Apply the same recipe for momentum to calculate the standard deviation of momentum.
• Apply Heisenberg's Uncertainty principle.

Throughout the problem, the downward direction will be considered positive.

Step 1 : Standard deviation of position :

Let $x$ be the position of the body at any time $t$ and let $T$ be the total time taken for the body to fall through a height $H$ . Then from the equations of free fall : $x=\frac12gt^2$ and $T=\sqrt{\frac{2H}{g}}$ .

Then, mean of position $\langle x\rangle = \frac { \int _{ 0 }^{ T }{ \frac { 1 }{ 2 } gt^{ 2 }dt } }{ \int _{ T }^{ 0 } dt } =\frac {\frac{ g }{ 6 } (t^{ 3 }) \Big|_0^T}{T}=\frac H3$ using the value of $T$ as above.

Similarly, the mean of squares of position $\langle x^2 \rangle = \frac { \int _{ 0 }^{ T }{ (\frac { 1 }{ 2 } gt^{ 2 })^2dt } }{ \int _{ T }^{ 0 } dt } =\frac {\frac{ g ^2}{ 20 } (t^{ 5 }) \Big|_0^T}{T}=\frac {H^2}5$ .

Thus standard deviation of position $\sigma _x = \sqrt{\langle x^2 \rangle - \langle x\rangle ^2} = \sqrt{\frac{H^2}{5}-\frac{H^2}{9}}=\sqrt\frac{4H^2}{45}=\frac{2H}{3\sqrt 5}$

Step 2 : Standard deviation of momentum :

Let $v$ be the velocity of the body at any time $t$ . Then, $p=mv=mgt$ .

Therefore, mean of momentum $\langle p\rangle = \frac { \int {mgt \; dt } }{ \int _{ T }^{ 0 } dt } =\frac {\frac 12 mg \; (t^{ 2 }) \Big|_0^T}{T}=mg \sqrt{\frac{H}{2g}}$

Similarly, the mean of squares of momentum $\langle p^2 \rangle = \frac { \int {m^2g^2t^2 \; dt } }{ \int _{ T }^{ 0 } dt } =\frac {\frac 13 m^2g^2 \; (t^{ 3 }) \Big|_0^T}{T}=\frac23m^2gH$

Thus standard deviation of momentum $\sigma _p = \sqrt{\langle p^2 \rangle - \langle p\rangle ^2} = \sqrt{\frac23m^2gH-\frac12m^2gH}=\sqrt{\frac16m^2gH}=m\sqrt{\frac{gH}{6}}$

Step 3 : Heisenberg's Uncertainty Principle :

We have, from definition : $\sigma_x \sigma_p \ge \frac{\hbar}{2}$

Putting everything in : $\frac { 2H }{ 3\sqrt { 5 } } \cdot m\sqrt { \frac { gH }{ 6 } } \ge \frac { \hbar }{ 2 }$

Squaring both sides and doing the necessary algebra yeilds : $H^3m^2g\ge\frac{135\hbar^2}{8} = 1.8767\times 10^{-67}$ which is of the form $a\times 10^b$ .

Thus, $a+|b|=1.8767+|-67|=\boxed{68.8767}$ .