Can two different arrangements produce exactly the same diffraction intensities, even with perfect measurements? Yes, in a specified scattering model. The difficulty can be missing structural information rather than insufficient precision.
The short answer is that intensity records the squared magnitude of a combined wave. In the simple periodic model below, that quantity depends on a ledger of directed pair separations. Two arrangements can have the same separation ledger without being the same arrangement. They then give the same intensity at every integer diffraction order in the model.
We will build and check one complete example on twelve equally spaced positions around a repeating cell. Arrangement A occupies positions 0, 1, 4 and 6. Arrangement B occupies 0, 1, 3 and 7. Their intensity records agree exactly, but their adjacent gaps show that no translation or reflection turns one into the other. These are mathematical point-scatterer constructions, not proposed chemical structures or laboratory measurements.
Start with the X-ray Crystallography subject guide for the experimental setting, structure factors and validation. This article makes one part of the inverse problem inspectable: why an exact intensity match can leave more than one non-equivalent candidate, and what kind of added evidence could distinguish the stated pair.
eduKateSG · X-ray Crystallography
Find your next learning step
Build two complete separation ledgers and test exactly which measurements distinguish the remaining candidates.
ROUTE 1 · CHAPTERS 1–5
Build the forward model
Define a periodic cell and derive intensity from ordered pair differences.
ROUTE 2 · CHAPTERS 6–10
Prove and inspect the match
Compare ledgers, rule out trivial copies and calculate intensities and phases.
ROUTE 3 · CHAPTERS 11–15
Recover information and test limits
Invert the ledger, preserve assumptions and distinguish exact ambiguity from noise.
ROUTE 4 · CHAPTERS 16–20
Change evidence carefully
Test contrast, uniform scaling, computation, search scope and structural validation.
ROUTE 5 · CHAPTERS 21–25
Transfer and continue
Change the origin and cell, solve an independent example and follow specialist sources.
Chapter contents
Build the forward model · Chapters 1–5
1. Put the cell on a clear coordinate system
2. Define what the model calls an intensity
3. Expand the square to reveal pair separations
Prove and inspect the match · Chapters 6–10
6. Count Arrangement B independently
7. Prove the arrangements are not just shifted copies
8. Derive the full intensity sequence
Recover information and test limits · Chapters 11–15
11. Explain the word homometric without overextending it
12. Recover the ledger from a complete intensity record
13. Keep a contrast experiment's added assumptions visible
14. Distinguish an exact ambiguity from a noisy fit
15. Test whether a proposed extra datum distinguishes the pair
Change evidence carefully · Chapters 16–20
16. Change one scattering weight and recalculate
17. Check a common change that does not help
18. Use an occupancy vector for a reproducible calculation
19. Do not mistake a unique search result for a unique structure
Transfer and continue · Chapters 21–25
21. Work a changed-origin transfer
22. Solve a new pair-ledger check with fewer prompts
23. Repair, stabilise and extend your reasoning
CHAPTER 1 OF 25 · Build the forward model
1. Put the cell on a clear coordinate system
Imagine twelve marked positions around a circle, labelled zero through eleven. The circle represents periodic coordinates, not a bent physical molecule. After position eleven, the next position is zero in the next copy of the repeating cell.
A position label x corresponds to fractional coordinate x/12 along a one-dimensional cell. Thus position six is halfway through the cell. Four positions are occupied by identical unit point scatterers, and the remaining positions are empty. Every occupied position has the same real scattering weight one in this model.
The repeating cell, equal grid spacing, identical weights and ideal coherent summation are supplied assumptions. Real scattering factors, thermal motion, chemical constraints, dimensions and experimental corrections are outside this small construction unless added explicitly later.
Write A = {0, 1, 4, 6} and B = {0, 1, 3, 7}. Curly brackets indicate sets of occupied positions; the order in which those labels are written has no physical meaning. The spatial order around the cell does matter, and we will use it to test whether the two sets are genuinely different beyond a changed coordinate origin.
CHAPTER 2 OF 25 · Build the forward model
2. Define what the model calls an intensity
At integer order h, define the combined amplitude F_h as the sum of exp(2πihx/12) over the four occupied positions. The imaginary unit i allows each term to represent a unit wave with a position-dependent phase. The intensity in our chosen normalisation is I_h = |F_h|².
If complex numbers are unfamiliar, think of four arrows of length one on a plane. The position and order determine each arrow’s angle. Add their horizontal components and their vertical components. Squaring the length of the resulting arrow gives intensity.
Adding intensities of the four arrows separately would give four every time and would discard interference. The model adds waves first, then squares the magnitude. That operation is responsible for the structural information that remains and the phase information that is lost.
The IUCr introduction to calculating structure factors develops the wave-summing background. We use an intentionally reduced version with equal unit weights so that the entire calculation can be checked by hand. Our intensity numbers are normalised model values, not instrument counts.
CHAPTER 3 OF 25 · Build the forward model
3. Expand the square to reveal pair separations
Multiply F_h by its complex conjugate. One factor contributes exp(2πihx/12), while the conjugate contributes exp(−2πihy/12). Each product is exp[2πih(x − y)/12]. Summing over every ordered pair x, y therefore produces the intensity.
The important quantity is the directed difference x − y modulo twelve. Modulo twelve means that differences separated by a whole cell are treated as equivalent. A difference of minus one is represented by eleven, and a difference of thirteen by one.
Let C_d count the ordered pairs with difference d modulo twelve. Then I_h = Σ C_d exp(2πihd/12), where d runs from zero through eleven. The separation-count ledger C is the discrete periodic autocorrelation of the occupied pattern.
This derivation explains why two patterns with the same C have the same intensity at every integer h. It does not yet say that different patterns with the same C exist. We will establish that by counting the pairs and separately proving that our arrangements cannot be matched by trivial coordinate changes.
Each of the four occupied positions can pair with itself. Every self-pair has difference zero. Thus C_0 = 4 for both arrangements. No other ordered pair has difference zero because no position is occupied twice.
There are four choices for x and four for y, giving sixteen ordered pairs in total. Four are self-pairs and twelve connect distinct positions. The sum of all C_d must therefore be sixteen. This is a useful checksum independent of the detailed ledger.
An ordered pair differs from an unordered pair. The pair from position one to position four has the opposite directed difference from the pair from four to one. Both contribute to the intensity expansion. Counting each unordered pair only once would lose half the cross-pair terms unless a corresponding cosine representation were used consistently.
For these real equal weights, the count at difference d equals the count at twelve minus d. That symmetry ensures the imaginary terms cancel in the intensity sum. It is another check, but symmetry alone is not enough to show that two complete ledgers are identical.
CHAPTER 5 OF 25 · Build the forward model
5. Work through Arrangement A's differences
Take x from {0, 1, 4, 6}. For x = 0, subtracting y = 0, 1, 4, 6 gives differences 0, 11, 8 and 6 modulo twelve. For x = 1, the differences are 1, 0, 9 and 7.
For x = 4, the differences are 4, 3, 0 and 10. For x = 6, they are 6, 5, 2 and 0. Collect the sixteen entries rather than treating each row as a different kind of measurement.
Zero appears four times, six appears twice, and each other difference appears once. The total is four plus two plus ten, which is sixteen. The opposite-difference pairs also match: one and eleven, two and ten, three and nine, four and eight, five and seven.
A good hand check is to choose one entry and trace its actual ordered pairs. Difference six comes from 0 − 6 and 6 − 0 because minus six and plus six coincide modulo twelve. Its double appearance is not a duplicated mistake. It is a consequence of the half-cell separation.
For B = {0, 1, 3, 7}, the x = 0 row gives differences 0, 11, 9 and 5. The x = 1 row gives 1, 0, 10 and 6. The x = 3 row gives 3, 2, 0 and 8. The x = 7 row gives 7, 6, 4 and 0.
Again zero appears four times, six twice and every other difference once. The entire directed periodic ledger agrees with A’s ledger, not merely its total or a few selected distances.
| Directed difference d | Count for A | Count for B |
|---|---|---|
| 0 | 4 | 4 |
| 1 | 1 | 1 |
| 2 | 1 | 1 |
| 3 | 1 | 1 |
| 4 | 1 | 1 |
| 5 | 1 | 1 |
| 6 | 2 | 2 |
| 7 | 1 | 1 |
| 8 | 1 | 1 |
| 9 | 1 | 1 |
| 10 | 1 | 1 |
| 11 | 1 | 1 |
The row-by-row calculation matters because it provides a reproducible route to the claim. “The distances look similar” would not establish equality. Neither would comparing only the shortest separation or an average separation. The intensity depends on the complete weighted ledger in the model.
CHAPTER 7 OF 25 · Prove and inspect the match
7. Prove the arrangements are not just shifted copies
Read the adjacent gaps around the periodic cell. For A, moving through positions 0, 1, 4, 6 and back to twelve gives gaps 1, 3, 2 and 6. For B, the gaps are 1, 2, 4 and 5.
Translating the entire pattern changes where the list starts but preserves its cyclic gap sequence. Reflecting reverses the sequence, also preserving the multiset of gap lengths. A’s multiset is {1, 2, 3, 6}; B’s is {1, 2, 4, 5}. Since those multisets differ, neither a translation nor a reflection can transform A into B.
This argument uses a necessary invariant. If two gap multisets agree, that alone would not always prove congruence; their cyclic order may still matter. Here their disagreement is sufficient to rule congruence out.
We have now established two separate facts: equal separation ledgers, and non-equivalent arrangements under the stated trivial transformations. The first gives equal intensities. The second shows that the ambiguity contains structural information beyond an arbitrary choice of origin or handed coordinate direction.
CHAPTER 8 OF 25 · Prove and inspect the match
8. Derive the full intensity sequence
Our common ledger has one count at every difference, plus three extra counts at zero and one extra at six. Therefore I_h equals the sum of all twelve roots exp(2πihd/12), plus three, plus exp(πih).
When h is not a multiple of twelve, the twelve-root sum is zero. The last term is (−1)^h. Thus odd orders have intensity two and even orders that are not multiples of twelve have intensity four. At multiples of twelve, all four scatterer waves align, giving amplitude four and intensity sixteen.
The complete repeating sequence for h = 0 through eleven is therefore 16, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2. The next twelve orders repeat these model values. Negative orders give the same intensities as their positive counterparts under our real-weight assumptions.
The zero order is a mathematical checksum in this construction. We do not claim that an experimental direct beam is collected as an ordinary structural reflection in the same way. The proof of ambiguity concerns the specified forward model and its integer-order intensity values.
CHAPTER 9 OF 25 · Prove and inspect the match
9. Check one order using wave components
At h = 1, A’s four terms correspond to angles zero, thirty, one hundred and twenty, and one hundred and eighty degrees. Their horizontal sum is (√3 − 1)/2 and their vertical sum (1 + √3)/2.
Squaring and adding those components gives intensity two. The resultant points at seventy-five degrees under our positive-exponent phase convention. This is a complete amplitude with a magnitude and a phase, even though intensity retains only the squared magnitude.
For B, the angles are zero, thirty, ninety and two hundred and ten degrees. The thirty-degree and two-hundred-and-ten-degree arrows cancel. The remaining sum is 1 + i, with intensity two and phase forty-five degrees.
The two resultant arrows therefore have the same length and different directions at this order in this fixed coordinate system. This is a concrete demonstration of the missing phase information. It does not mean one phase value alone would generally solve an unknown structure, especially when the origin has not been externally fixed.
CHAPTER 10 OF 25 · Prove and inspect the match
10. Keep phase convention and origin explicit
Changing the sign in the exponential conjugates these structure factors. The quoted phases then change sign, while the intensities remain unchanged. A phase comparison needs one consistent convention; an intensity comparison does not depend on that sign choice here.
Translating every position by a displacement s multiplies F_h by exp(2πihs/12). That rotates the resultant without changing its magnitude. An origin shift is therefore a trivial source of phase change. It should not be mistaken for a new physical structure.
Our A and B already passed the adjacent-gap test, so their ambiguity is not explained solely by translation or reflection. Nevertheless, translating B by one grid position changes its first-order phase from forty-five to seventy-five degrees, matching A’s first-order phase. This shows why an isolated phase value without an origin convention is not a general structural identifier.
If an added measurement supplies phase relative to a known external reference, state that reference. Then ask which candidates it actually excludes. A piece of information is useful because it partitions the allowed candidate set, not merely because it is called “phase information.”
CHAPTER 11 OF 25 · Recover information and test limits
11. Explain the word homometric without overextending it
Patterns sharing the relevant autocorrelation are called homometric. Our example uses a finite cyclic coordinate set representing a periodic cell. The directed differences wrap around the cell boundary, and that periodic convention is part of the definition of the comparison.
The research paper Homometric Point Sets and Inverse Problems, by Uwe Grimm and Michael Baake, discusses autocorrelation and nonuniqueness in diffraction inverse problems. Our twelve-site calculation offers a small, fully worked route into that mathematical idea.
Do not replace the periodic ledger by a list of ordinary unwrapped distances between four isolated points on a line. A separation of eleven in the periodic cell also represents a one-step separation in the opposite direction across the boundary. The isolated finite-line problem has a different forward model.
Likewise, matching one kind of pair summary does not guarantee matching every experiment. The word must remain attached to the specific correlation and scattering assumptions. Our conclusion is exact for this model, while broader structural claims require a broader set of evidence and constraints.
An explicit unwrapped-distance check shows how the boundary convention matters. As four isolated points on a line, A has unordered positive separations one, two, three, four, five and six. B has one, two, three, four, six and seven. These lists differ. In the twelve-site periodic model, however, a pair at separation five contributes directed differences five and seven; a pair at separation seven contributes seven and five. Their periodic contributions agree.
This is why a finite isolated-cluster calculation cannot be substituted for the repeating-cell calculation without rebuilding the forward model. Both coordinate pictures use the same printed labels, but they encode different relationships across the boundary. The distinction is mathematical, not an issue solved by rounding the labels more accurately.
When making your own example, write “periodic modulo twelve” beside the pair table and “integer order” beside the intensity formula. Those two statements travel together. If the sample is changed to a nonperiodic finite collection and continuously varying scattering coordinate, the agreement at the specified integer orders does not establish agreement everywhere in that changed model. Test the new predictions directly rather than extending the previous conclusion by analogy.
CHAPTER 12 OF 25 · Recover information and test limits
12. Recover the ledger from a complete intensity record
The forward relation takes twelve separation counts to twelve model intensities. It can also be inverted within this finite cyclic model. For each d, C_d equals one-twelfth of the sum of I_h exp(−2πihd/12) over h from zero through eleven.
The inversion follows from a simple cancellation. Substitute the forward expression for each I_h, then group the terms belonging to each difference e. The sum over h of exp[2πih(e − d)/12] is twelve when e equals d modulo twelve and zero otherwise. After division by twelve, only C_d remains.
This tells us exactly what a complete ideal intensity sequence recovers here: the periodic pair ledger. It does not automatically recover a unique occupancy set from that ledger. Our A and B are two different solutions to that second reconstruction step. The information loss sits between the ledger and the individual arrangement, even when the ledger itself is perfectly determined.
As a numerical checksum, sum the twelve intensities in our record. There is one value sixteen, six odd-order values two and five nonzero even-order values four. Their sum is forty-eight. Dividing by twelve gives four, which is C_0, the number of occupied positions for a binary occupancy vector.
For another check, the same inverse at difference six weights each intensity by (−1)^h. The even-order subtotal is thirty-six and the odd-order subtotal twelve, giving (36 − 12)/12 = two. That recovers the double half-cell count independently from the subtraction ledger.
This derivation is optional algebra, but its conclusion is useful for all readers. A reconstruction can be exact at an intermediate level and still nonunique at the final level. Identify which representation has actually been recovered before saying the structure has been determined.
CHAPTER 13 OF 25 · Recover information and test limits
13. Keep a contrast experiment's added assumptions visible
The unequal-weight comparison used an independently identified position zero. That identification is extra information. If the experiment cannot distinguish which occupied site received the changed weight, one cannot simply select the convenient zero site in both candidates after seeing the result.
Write the added assumption before the recalculation: “This known site, in this fixed reference, has weight two.” Then compute each candidate’s prediction. If the site identity is itself uncertain, include that uncertainty as additional candidate cases and compare their predictions too.
This prevents a common inverse-problem mistake: using the desired structure to assign the new constraint, then treating the constraint as independent evidence for that structure. Independence is a property of how the information was obtained, not a label attached to a second calculation.
The mathematical result remains valuable. We have a fully specified changed model that separates the stated pair. Whether a real experiment can implement its assumptions is a separate question requiring physical evidence. Keeping that separation clear makes the proposed discriminator reviewable rather than merely persuasive.
CHAPTER 14 OF 25 · Recover information and test limits
14. Distinguish an exact ambiguity from a noisy fit
Two candidates can look similar because the measurements are imprecise, incomplete or noisy. Better precision may separate such near-matches. Our pair is stronger: its model intensities are exactly equal for every integer order, before noise is added.
Repeating the same intensity measurement with smaller uncertainty does not separate equal predictions. Increasing the number of integer orders also does not help within this unchanged ideal model because the common ledger proves agreement at all of them.
This does not make experimental quality unimportant. Good data are needed to test whether either candidate fits and whether the model is appropriate. The distinction is between uncertainty about a measured value and nonuniqueness of the map from structure to that value.
A useful diagnostic question is “What would the two candidates predict differently?” If the answer is “nothing in the proposed measurement under our assumptions,” precision alone is not the missing ingredient. Seek a changed constraint or a genuinely distinguishing observable rather than promising uniqueness from more repetitions of the same information.
CHAPTER 15 OF 25 · Recover information and test limits
15. Test whether a proposed extra datum distinguishes the pair
Suppose an independent observation determines that position four is occupied in the fixed coordinate system. A satisfies that statement and B does not. That datum distinguishes these two candidate sets immediately.
Now suppose the added statement is “there are four occupied positions.” Both already satisfy it. The datum may be useful against other candidates, but it does not distinguish this pair. The same is true of a newly measured integer-order intensity in the unchanged model.
An extra condition should therefore be tested by substituting both candidates. Write the predicted outcome beside each one and check whether they differ. Avoid assuming that an observation is decisive simply because it comes from a different instrument or uses a more technical name.
Even excluding B does not prove A is the only possible structure in a larger candidate universe. Another arrangement may satisfy all the retained constraints. Distinguishing a known pair and establishing global uniqueness are separate achievements. State which one the evidence supports.
CHAPTER 16 OF 25 · Change evidence carefully
16. Change one scattering weight and recalculate
Our equality relied on identical unit weights. As a changed model, suppose the occupied position zero is independently identified and given weight two, while every other occupied position keeps weight one. Use the same fixed reference in both candidates.
At h = 1, the new structure factor is the old F_1 plus one, because only the contribution at zero increases. For A, both components become (1 + √3)/2. The new intensity is 2 + √3, about 3.732051. For B, the sum becomes 2 + i and intensity five.
The modified measurement model now distinguishes the pair. We have shown this by calculation, rather than assuming that unequal weights always resolve every ambiguity. A different weighting choice would need its own check, and other candidate sets could still agree.
This is a mathematical contrast experiment, not a recommendation to modify an actual specimen or a claim about a specific chemical labelling method. Real changes to scattering contrast require their own physical model, sample suitability and expert experimental interpretation. The transferable idea is to test whether the proposed change alters the candidates’ predictions differently.
CHAPTER 17 OF 25 · Change evidence carefully
17. Check a common change that does not help
Multiplying every scatterer’s weight by the same constant k multiplies every F_h by k and every intensity by k² for real k. Both candidates are scaled in exactly the same way. Their equality remains.
Similarly, adding the same known background intensity to both predicted records preserves equality. It may change the observed numbers but does not supply a structural discriminator between these two candidates. Better background estimation is valuable for data quality while leaving this exact ambiguity intact.
This distinction helps when planning a follow-up calculation. A changed number is not necessarily changed information about the candidates. Ask whether the difference between their predictions changes from zero to something observable under the revised model.
If a proposed procedure merely rescales or offsets both records identically, it cannot separate this pair by itself. Keep the symbolic equality in view. It is often easier to see that a transformation preserves equality before doing a long numerical calculation at every order.
CHAPTER 18 OF 25 · Change evidence carefully
18. Use an occupancy vector for a reproducible calculation
Represent A by a twelve-entry vector containing one at positions zero, one, four and six and zero elsewhere. Represent B in the same way with ones at zero, one, three and seven. The array index must match the coordinate label.
To compute C_d, loop over every occupied x and every occupied y, form (x − y) modulo twelve and increment the corresponding count. Check that the counts sum to sixteen and that C_0 is four. Compare the two entire vectors for exact integer equality.
For a second route, calculate each complex F_h directly and square its magnitude. Numerical floating-point results may be 1.9999999999999998 instead of two. Compare with a declared tolerance and retain the exact integer-ledger proof as the stronger explanation of equality.
A program should preserve both candidate arrays, the coordinate convention, weights and tested order range. Printing only “match” removes the evidence needed to audit the result. The calculation is short enough that a learner can also hand-check one row and one order before trusting the loop.
CHAPTER 19 OF 25 · Change evidence carefully
19. Do not mistake a unique search result for a unique structure
Suppose a program searches a restricted list and finds only A. That establishes uniqueness within that list under the search’s constraints and implementation, assuming it was exhaustive and correct. It does not establish uniqueness among all structures allowed by a broader physical problem.
The twelve-site model itself is restrictive: four identical scatterers must occupy distinct sites on a fixed grid. Real coordinates need not lie on that grid. Conversely, real chemistry can impose constraints absent from the point model. Both enlarging and narrowing a candidate universe can change the uniqueness question.
State the search space before reporting a unique result. Include which equivalences were removed, such as translations and reflections, and which priors were imposed. A chemically sensible candidate is not validated merely by being the first one returned.
Our counterexample does not show that all crystallographic problems are insoluble. It shows that an unrestricted promise of uniqueness from intensity alone fails for at least this explicit model. A rigorous counterexample challenges a universal claim without becoming a universal pessimistic claim of its own.
CHAPTER 20 OF 25 · Change evidence carefully
20. Connect the toy calculation to structural validation
The broader crystallographic workflow combines experimental data, model building, refinement, chemical knowledge and validation. Our construction isolates one information loss so that it can be inspected, rather than representing that entire workflow.
The existing article on diffraction and electron-density evidence explains the experimental sequence and the general phase problem. Use it when you need the wider distinction between a processed intensity dataset, a map and an atomic model.
An excellent intensity fit is evidence about compatibility with the measured data under a model. It is not, by itself, proof that every coordinate, chemical assignment or inferred function is unique. Independent constraints and validation questions remain meaningful even when residuals are small.
When reading a structural claim, ask which observations support the model, which assumptions supplied missing information and which alternatives were considered. Our exact pair gives a concrete reason for those questions: identical forward predictions can conceal different arrangements, so agreement must be interpreted in the context of the candidate space.
Translate A by two grid sites, wrapping modulo twelve. The new occupied set is {2, 3, 6, 8}. Its directed difference ledger is unchanged because subtracting two translated coordinates cancels the common shift.
Every intensity therefore remains unchanged. Its structure factor at order h is multiplied by exp(2πih × 2/12). At h = 1, the phase increases by sixty degrees under our convention, while the magnitude stays √2.
Now compare this translated set with B. It remains non-equivalent to B because the cyclic gap multiset of A is unchanged. The original nontrivial ambiguity survives a mere coordinate translation.
This exercise separates two sources of nonuniqueness: a translation can change coordinates without changing the arrangement’s internal geometry, whereas A and B have different internal cyclic gaps. A learner should be able to explain both using differences, rather than relying only on a plotted picture or a remembered phase angle.
CHAPTER 22 OF 25 · Transfer and continue
22. Solve a new pair-ledger check with fewer prompts
Consider a smaller eight-site periodic cell with two identical unit scatterers. Candidate P occupies {0, 2}; candidate Q occupies {0, 3}. Before calculating any trigonometric values, list their directed differences.
For P, C_0 = 2, C_2 = 1 and C_6 = 1, with other entries zero. For Q, C_0 = 2, C_3 = 1 and C_5 = 1. The ledgers differ, so there is no ledger-based proof that all intensities match.
At h = 1, P has F = 1 + i and intensity two. Q has F = 1 + exp(3πi/4), giving intensity 2 − √2, about 0.585786. This one order distinguishes the candidates. Both have intensity four at zero order, showing why agreement at one order is insufficient.
The changed cell size tests whether you can reconstruct the phase denominator and modulo rule. Reusing twelve in an eight-site problem would apply the wrong forward model. Independent transfer begins by reading the new coordinate system, not by carrying the previous arithmetic unchanged.
CHAPTER 23 OF 25 · Transfer and continue
23. Repair, stabilise and extend your reasoning
If you lose cross-pairs, draw a four-by-four subtraction table and mark self-pairs. If your counts do not sum to sixteen, repair the ledger before evaluating exponentials. If you obtain complex intensities, check conjugation and opposite-difference symmetry.
To stabilise the distinction between equivalence and ambiguity, practise translations, reflections and non-congruent pairs. Use cyclic gaps to rule out trivial matches and the full directed ledger to test intensity equality. These are different checks serving different parts of the proof.
For extension, propose a new weight assignment and calculate whether it distinguishes A and B. Then propose a uniform scaling and explain symbolically why it cannot. The contrast tests whether you can recognise information gain rather than merely produce changed numerical values.
A strong independent explanation contains three linked statements: the intensity depends on the pair ledger; these two ledgers agree exactly; and these gap sequences cannot describe translated or reflected copies. Follow that explanation with a limit about real materials or the candidate universe. The proof and its scope belong together.
Does the example mean diffraction contains no structural information?
No. Different ledgers can produce different intensity sequences, as the eight-site exercise shows. The example establishes a particular nonuniqueness, not an absence of information.
Would more decimal places separate A and B?
Not under the unchanged ideal model. Their predictions agree exactly at every integer order. Better measurement can address uncertainty but cannot separate identical predictions by precision alone.
Are the two patterns mirror images?
No. Translation and reflection preserve the multiset of adjacent cyclic gaps, and our two multisets differ. That test rules out both possibilities for this pair.
Would one extra condition prove the complete structure?
It might exclude one candidate without excluding every other possible alternative. Test the condition against the allowed candidate set and distinguish pairwise discrimination from a global uniqueness claim.
CHAPTER 25 OF 25 · Transfer and continue
25. Continue with the wider evidence route
For experimental context, return to X-ray Crystallography and the existing guide to diffraction and electron-density evidence. The first provides a subject route; the second follows the broader path from measurement to structural interpretation.
For the wave calculation, use the IUCr teaching pamphlet on structure factors. For the research concept of homometry, read Grimm and Baake’s Homometric Point Sets and Inverse Problems, submitted in 2008. These sources support the specialist setting. The step-by-step twelve-site ledger, component checks and changed-context exercises here are teaching calculations, not reported experimental crystal structures.
The Gold Standard of Education connects this work with independent transfer. The How X Works hub offers other subjects where different hidden states can share an observable summary. Preserve the useful habit: write the forward model, compare complete predictions and identify exactly which new evidence would distinguish the remaining alternatives.
Contents · Previous chapter · Return to X-ray Crystallography