Electron density in a crystal is *tri*-periodic. It can be calculated using the following
equation:

1/V

h
Σ

k
Σ

l
Σ

in which, the summations are over positive and negative indices, and

V — unit cell volume

For an introductory exercise, all of the important concepts can be addressed in *one*
dimension. To use a real molecule in the analogy, however, we need to generalize a bit and
make the problem *mono-periodic*. [*n.b.* 'mono-periodic' and 'one-dimensional'
are *not* synonymous, but are often used interchangeably. It is usually easy to tell
what is meant from the context.]

In a mono-periodic 'crystal' with repeat distance 'a', the above equation for electron density reduces to:

1/a

h
Σ

Using Euler's formula, *e ^{ iφ}* = cos(

1/a

h
Σ

The electron density equation is thus a sum of different frequency sinusoids, *i.e.*, a
*Fourier series*.

Consider a hypothetical mono-periodic crystal of carbon dioxide, a linear centrosymmetric
molecule. A ball-and-stick representation of CO_{2} looks like this:

A 1–D plot of its electron density looks *qualitatively* something like this:

The oxygen atoms have larger peaks than carbon because oxygen has eight electrons, while carbon only has six. The electron density drops down to a flat baseline on either side but in the bonds it is non-zero.

In such a made-up crystal, the molecules are arranged in a line. Five in a row might look like this:

For which, the electron density 'wave' would be:

Since it is periodic, we can model it using the above *Fourier* summation. Assuming a repeat
distance of 5.00Å and one molecule per 'unit cell', the electron density looks like this:

The area under the curve must equal the number of electrons in the unit cell. For a single
CO_{2} molecule, that's 22e^{–}.
A properly scaled plot of electron density is thus *quantitative*. The question is,
what superposition of waves will approximate this curve ?

Before answering that, here's a quick review of some properties of sines and cosines:

Sine and cosine waves differ in *phase* by a quarter of a period (*i.e.*, 90° or
*π*/2 radians). In crystallography it is more natural to use fractions of a period
than degrees or radians. The above plot shows that over one period, the cosine curve is
symmetric, while the sine is anti-symmetric, *i.e.*,

Since our example is centrosymmetric, the sine terms all cancel, so we can disregard them. For
cosines, there are only two possible *phases* that preserve centrosymmetry: zero (as above)
or shifted by half a period. The two possibilities are drawn below.

Since cosines range from –1 → +1, the area under either curve is zero. Such plots can be
scaled by multiplying the cosine by a constant multiplier, or *coefficient*:

The coefficients can be any real number. Negative coefficients invert the plot and therefore
have the effect of phase-shifting the wave by half a period. Thus, the two allowed phases
(for centrosymmetry; non-centrosymmetry is more complicated) arise from waves having either
positive or negative coefficients. In addition to amplitude and phase, cosine (and sine) waves
can have different *frequency*:

In the uppermost of the above two plots, *positive*-coefficient *odd*-frequency waves
(blue) have minima at the centre and *positive*-coefficient *even*-frequency waves
(orange) have maxima. In the lowermost plot for *negative* coefficients the situation is
reversed.

Since our example is centrosymmetric, the anti-symmetric nature of the sine function, in our case

dictates that all the sine functions cancel, leaving

1/a

h
Σ

The term in the above summation for *h* = 0 is a special case because the cosine evaluates to
exactly 1. Thus, the value for *F* (0) can be removed from the summation. This quantity,
*F* (0) is simply the number of electrons in the unit cell. For a real crystal in 3D,
this quantity is *F* (000). It is the amount needed to shift the electron density plot
*en masse* upwards so that the electron count over the whole unit cell is correct. For one
CO_{2} molecule, that's 22e^{–}.

Lastly, due to the cosine function being symmetric, *i.e.*,

The summation need only cover the positive *h* indices, (or negative, or a unique unsigned
combination thereof) *viz*.

1/a

[
h
Σ

for

To model electron density, consider first the lowest-frequency term only. A *negative*
coefficient, appropriately scaled and shifted upwards by *F* (0) to ensure an integral
over the unit cell of 22e^{–}, gives a *very* low-resolution electron density plot.
It peaks in the middle of the cell, which happens to be where we placed the molecule:

A better fit requires more terms, each with its own frequency, amplitude, and phase (sign).

The next step is to identify the dominant frequencies and their amplitudes. These come from the
Fourier transform of the actual electron density. The amplitudes, or *Fourier coefficients*,
correspond to peak heights in the Fourier transform, as shown below:

The connection to X-ray diffraction is that the Fourier coefficients are proportional to the
square root of diffracted intensities. These are the *structure factors*, *F* (*hkl* ),
in the first equation above. Here, the coefficients for frequencies 3 and 6 are tiny, so we'll
ignore them (for now). That leaves four waves, each having a positive or negative coefficient,
*i.e.*, two possibilities apiece. Unfortunately, the Fourier transform gives no phase (sign)
information. For four waves there are 2^{4} = 16 different combinations, *viz*.

+ + – + – – + – + + – – – – + +

+ – + + – + – – + – + – – + – +

+ – – + – + + – + – – – – + + +

Since there are only 16 combinations, we can easily plot them all and compare:

The shaded regions are *negative* and therefore *cannot* represent electron density.
Thus, all but two of the sign combinations can be rejected. Note also that flipping all of the signs
merely inverts the plot (prior to shifting it upwards to ensure an electron count of 22e^{–}).

The two plots that are *positive* everywhere have "– – + –" and "+ – + +" sign combinations.
These two plots are equivalent, but shifted relative to each other along *x* by half a unit
cell repeat:

The signs of odd coefficients are flipped, while the even coefficient signs remain the same.
The slight baseline ripple in the two plots above stem from ignoring the small coefficients
of order 3 and 6 (and higher!). Setting the first coefficient as either negative or positive
has the effect of *fixing the origin*. A real tri-periodic structure usually requires
setting the signs of three (well-chosen) reflections.

Atom coordinates can be extracted directly from such plots:

Thus, geometric parameters can be calculated. Here, the difference between carbon and either of the oxygen coordinates is 0.226. Since the unit cell length was 5.00 Å, the C—O interatomic distances (bond lengths) are:

For comparison, the *actual* bond length in CO_{2} is 1.16 Å. Not a bad result
for a made-up analogy! Especially since in routine X-ray crystal structures, the accuracy of
light atom positions is limited to about ±0.02 Å due to the way atomic scattering
factors are approximated (the *spherical atomic scattering factor approximation* ).

Also apparent is that symmetry is of *profound* importance. This mono-periodic CO_{2}
structure is symmetric by reflection through the points *x* = 0, 0.5, 1, *etc.*
Atom O1' is *equivalent* to O1 by reflection through *x* = 0.5, *viz.*

Due to symmetry, only the region from *x* = 0 → 0.5 needs to be specified, thereby
*halving* the computational burden. The smallest fragment needed to generate the whole
structure is thus *not* the unit cell, but in this case, just *half* of it.

The unique part is the *asymmetric unit* : the fundamental building block of crystal
structures. Different space groups have their own asymmetric units (details are in the International
Tables, vol. A). Atoms on a point-symmetry element (*x* = 0, 0.5, 1 in the above) occupy a
so-called *special position*. Atoms not on a special position occupy a *general position*.

For a *real* structure with thousands of diffraction maxima, the number of phase combinations
is *vast*. A typical structure might result in ~5000 reflections, giving ~2^{5000}
sign combinations. A bit of quick log base conversion maths tells us that's ~10^{1505}.
For comparison, the number of atoms in the known universe is ~10^{80}. Clearly, we can't
just guess the right phases! *That* is why the phase problem in crystallography is a *problem*.