91Ó°ÊÓ

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:

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

Where 5 is the total number of particles, so:

$\stackrel{¯}{E}=3\mathrm{eV}$

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

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

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:

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

The standard deviation is the square root of this value:

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

(c) Standard deviation is:

${\mathrm{σ}}_E^2=\stackrel{¯}{{\left(\mathrm{Δ}{E}_i\right)}^2}$

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

${\mathrm{σ}}_E^2=\frac{1}{N}\underset{i}{∑}{\left(\mathrm{Δ}{E}_i\right)}^2\left(1\right)$

The deviation is:

$\mathrm{Δ}{E}_i=E_i-\stackrel{¯}{E}$

Substitute into (1)

$$\begin{gathered} {\mathrm{σ}}_E^2=\frac{1}{N}\underset{i}{∑}{\left(E_i-\stackrel{¯}{E}\right)}^2 \\ {\mathrm{σ}}_E^2=\frac{1}{N}∑\left(E_i^2+{\stackrel{¯}{E}}^2-2E_i\stackrel{¯}{E}\right) \\ {\mathrm{σ}}_E^2=\frac{1}{N}\left(\underset{i}{∑}{E}_i^2+{\stackrel{¯}{E}}^2-2\stackrel{¯}{E}\underset{i}{∑}{E}_i\right) \\ {\mathrm{σ}}_E^2=\stackrel{¯}{E^2}+{\stackrel{¯}{E}}^2-2{\stackrel{¯}{E}}^2 \\ {\mathrm{σ}}_E^2=\stackrel{¯}{E^2}-{\stackrel{¯}{E}}^2 \end{gathered}$$

(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:

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

The average of the energies is:

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

The standard deviation is:

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

${\mathrm{σ}}_E=2.683\mathrm{eV}$