The most common measure of the fluctuations of a set of numbers away from the average is the standard deviation, defined as follows.

(a) For each atom in the five-atom toy model of Figure 6.5, compute the deviation of the energy from the average energy, that is, $E_i-\overline{E},\text{for}i=1\text{to}5$. Call these deviations $\Delta E_i$.

(b) Compute the average of the squares of the five deviations, that is, $\overline{{\left(\Delta E_i\right)}^2}$. Then compute the square root of this quantity, which is the root-mean- square (rms) deviation, or standard deviation. Call this number ${\sigma }_E$. Does ${\sigma }_E$give a reasonable measure of how far the individual values tend to stray from the average?

(c) Prove in general that

${\sigma }_E^2=\overline{E^2}-(\overline{E}{)}^2$

that is, the standard deviation squared is the average of the squares minus the square of the average. This formula usually gives the easier way of computing a standard deviation.

(d) Check the preceding formula for the five-atom toy model of Figure 6.5.

The most common measure of the fluctuations of a set of numbers away from the average is the standard deviation.

(a) Consider the following toy energy model: there are two particles with E1 = 0 eV, two particles with E2 = 4 eV, and one particle with E3 = 7 eV, and their average energy is provided by:

$\overline{E}=\frac{2\left(0\mathrm{eV}\right)+2\left(4\mathrm{eV}\right)+1\left(7\mathrm{eV}\right)}{5}$

Where 5 is the total number of particles, so:

$\overline{E}=3\mathrm{eV}$

The deviation is given by $E_i-\overline{E}$, so:

$E_1-\overline{E}=-3\mathrm{eV}{E}_2-\overline{E}=1\mathrm{eV}{E}_3-\overline{E}=4\mathrm{eV}$

We have two particles with the first deviation and two particles with the second deviation.

(b)The square of these deviations is

$9{\mathrm{eV}}^{2}9{\mathrm{eV}}^{2}1{\mathrm{eV}}^{2}1{\mathrm{eV}}^{2}16{\mathrm{eV}}^2$

Average of these are:

${\sigma }_E^2=\frac{9{\mathrm{eV}}^2+9{\mathrm{eV}}^2+1{\mathrm{eV}}^2+1{\mathrm{eV}}^2+16{\mathrm{eV}}^2}{5}=7.2{\mathrm{eV}}^2$

The standard deviation is the square root of this value:

$$\begin{gathered} {\sigma }_E=\sqrt{7.2{\mathrm{eV}}^2}=2.683\mathrm{eV} \\ {\sigma }_E=2.683\mathrm{eV} \end{gathered}$$

(c) Standard deviation is:

${\sigma }_E^2=\overline{{\left(\Delta E_i\right)}^2}$

The average equals the sum over the number of the values, so:

${\sigma }_E^2=\frac{1}{N}\sum _i{\left(\Delta E_i\right)}^2\left(1\right)$

The deviation is:

$\Delta E_i=E_i-\overline{E}$

Substitute into (1)

$$\begin{gathered} {\sigma }_E^2=\frac{1}{N}\sum _i{\left(E_i-\overline{E}\right)}^2 \\ {\sigma }_E^2=\frac{1}{N}\sum \left(E_i^2+{\overline{E}}^2-2E_i\overline{E}\right) \\ {\sigma }_E^2=\frac{1}{N}\left(\sum _i{E}_i^2+{\overline{E}}^2-2\overline{E}\sum _i{E}_i\right) \\ {\sigma }_E^2=\overline{E^2}+{\overline{E}}^2-2{\overline{E}}^2 \\ {\sigma }_E^2=\overline{E^2}-{\overline{E}}^2 \end{gathered}$$

(d) We have 5 particles with energies of 0 eV, 0 eV, 4 eV, 4 eV, and 7 eV; their squares are 0 eV², 0 eV², 16 eV², 16 eV,² and 49 eV²; the average of these values is:

$\overline{E^2}=\frac{(0+0+16+16+49){\mathrm{eV}}^2}{5}=16.2{\mathrm{eV}}^2$

The average of the energies is:

$\overline{E}=\frac{(0+0+4+4+7)\mathrm{eV}}{5}=3\mathrm{eV}$

The standard deviation is:

${\sigma }_E=\sqrt{16.2{\mathrm{eV}}^2-(3\mathrm{eV}{)}^2}$

${\sigma }_E=2.683\mathrm{eV}$