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Compare two rays scattered into parallel outgoing directions from neighboring lattice planes. In the diagram, $A$ lies on the upper plane and $B$ on the lower plane, directly beneath $A$ with $AB=d$. Draw through $A$ the incoming and outgoing equal-wave phase fronts, perpendicular to their respective rays. They meet the lower ray at $C$ and $D$. Each right triangle gives $CB=BD=d\sin \theta$, so the lower ray travels the extra distanceThe resulting phase difference is $\Delta \phi =2\pi \Delta \ell /\lambda$. Constructive interference requires $\Delta \phi =2\pi n$, which proves Bragg's law in this ray construction. Comparing equal-wave phase fronts, rather than arbitrary ray endpoints, is what makes the distance comparison meaningful.

The same geometry connects to the reciprocal lattice. With wavenumber $k=2\pi /\lambda$, the elastic Bragg scattering condition gives a transfer magnitude $|q|=2k\sin \theta$. A reciprocal vector normal to these lattice planes has $|q|=2\pi n/d$. Equating them again yields Bragg's law. The ray and reciprocal lattice descriptions therefore express the same wave phase matching.

Bragg's law selects directions of constructive interference in coherent elastic scattering from equally spaced lattice planes:Here $d>0$ is the perpendicular interplanar spacing, $\lambda >0$ the incident wavelength, $n$ the positive integer diffraction order, and $\theta$ the Bragg angle, measured between the incoming beam and the lattice planes. The scattering deflection, between incoming and outgoing propagation directions, is $2\theta$. Thus a plotted detector angle $2\theta$ must be halved before insertion into Bragg's law. The $\theta$ in the parent elastic Bragg scattering condition denotes the full deflection; the present $\theta$ denotes half of it. ^[ref]

The lattice planes organize wave interference throughout a crystal structure; they are not literal metal mirrors or separate reflecting surfaces. The picture assumes elastic scattering, hence an unchanged wavelength, and sufficiently stable relative wave phases across the contributing lattice planes. X-ray wavelengths comparable with atomic interplanar spacings make these directions experimentally accessible. The sine cannot exceed one: an order exists geometrically only if $n\lambda /(2d)\le 1$. For $\lambda >2d$, even the first diffraction order is inaccessible. A geometrically possible direction need not have nonzero wave intensity; the crystal lattice structure factor and crystal basis structure factor also matter. ^[ref]

For incident and outgoing unit propagation vectors ${\mathbf{s}}_{\mathrm{in}}$ and ${\mathbf{s}}_{\mathrm{out}}$, wavelength $\lambda >0$, and primitive lattice translation vectors $a_1,a_2,a_3$, the Laue equations areAll three must hold for the same wavelength and outgoing direction. The integers label a reciprocal lattice vector; they need not be coprime. For wavevectors ${\mathbf{k}}_{\mathrm{in},\mathrm{out}}=(2\pi /\lambda ){\mathbf{s}}_{\mathrm{in},\mathrm{out}}$, equivalentlyThis uses angular wavenumbers; the crystallographic convention without $2\pi$ instead uses $\mathbf{s}/\lambda$.^[ref]

For a short proof, translate a scatterer through $a_i$. Its incident wave phase changes by ${\mathbf{k}}_{\mathrm{in}}\cdot a_i$ and propagation to the same distant observation direction contributes $-{\mathbf{k}}_{\mathrm{out}}\cdot a_i$. The relative wave phase factor is therefore $\exp (-i\mathbf{G}\cdot a_i)$. It equals one for all lattice translations exactly when $\mathbf{G}$ belongs to the reciprocal lattice. A finite crystal lattice structure factor broadens the ideal maxima; the crystal basis structure factor can cancel them. Geometric phase matching is necessary for a sharp reflection, but does not guarantee a visible spot.

For elastic scattering, the additional constraint isIt is an Ewald sphere centered at $-{\mathbf{k}}_{\mathrm{in}}$ in the space of transfer vectors $\mathbf{G}$. The parent elastic Bragg scattering condition uses $q={\mathbf{k}}_{\mathrm{in}}-{\mathbf{k}}_{\mathrm{out}}=-\mathbf{G}$, so its displayed condition has the opposite sign. Both give Bragg's law for the same lattice planes.

In 1949, Clifford Shull and J. Samuel Smart compared powdered manganese(II) oxide at room temperature and $80\mathrm{K}$. Powder preparation supplies many crystal orientations. Their Fig. 1 records scattered-neutron intensity versus counter angle. Nuclear reflections remained, while the cold pattern gained strong reflections forbidden by the chemical unit cell. They indexed these with a magnetic unit cell having twice the chemical cell edge. The warm pattern already contained diffuse magnetic scattering: loss of sharp magnetic peaks does not mean loss of individual magnetic moments.^[ref]

The source was Oak Ridge's graphite-moderated reactor. In Early Development of Neutron Scattering (1994), Clifford Shull describes how the early two-axis instruments used a monochromating crystal to select the incident-neutron wavelength; a shielded neutron counter rotated around the specimen to record angular intensity. This specifies the source–monochromator–sample–counter path, without identifying which instrument revision produced the MnO trace.^[ref]

The control was the same compound above and below magnetic ordering, with retained nuclear peaks identifying the underlying chemical periodicity. The original paper labels an aluminium impurity peak and defers a complete magnetic-structure account. It supplies neither a chemical synthesis recipe nor detailed cold-stage construction. The inference is periodic magnetic order, rather than a claim that every possible structural change was excluded at arbitrary resolution.^[ref]

The connection to the measured angle is the elastic Bragg scattering condition. If the incident and outgoing wave vectors have equal magnitude $2\pi /\lambda$ and make an angle $2\theta$, their difference has magnitudeA longer real-space period supplies additional, smaller reciprocal lattice spacings, which can put magnetic reflections at new counter angles. In the one-dimensional toy model above, doubling the period changes the allowed spacing from $2\pi /a$ to $\pi /a$. The actual MnO indexing is three-dimensional; the toy row does not assert its reflection positions or relative peak strengths.

Louis Néel supplied the ordering theory; the 1949 diffraction work was a later experimental test. His 1970 Nobel Prize in Physics recognized antiferromagnetism and ferrimagnetism. Clifford Shull's neutron-diffraction award followed in 1994.^[ref]

In elastic scattering, the incoming and outgoing wave vectors have equal magnitude $k_0=2\pi /\lambda$. Their mutual angle is $2{\theta }_{hkl}$. The scattering vector $\mathbf{q}={\mathbf{k}}_f-{\mathbf{k}}_i$ therefore has magnitudeThe elastic Bragg scattering condition is $\mathbf{q}={\mathbf{G}}_{hkl}$, givingEquivalently, $G_{hkl}=2\pi /d_{hkl}$ turns this relation into Bragg's law, $2d_{hkl}\sin {\theta }_{hkl}=\lambda$, with the reflection order incorporated into the Miller indices.

In the conventional cubic unit cell, the four face-centered cubic lattice positions contributeThe two-atom atomic basis of a crystal gives the zinc blende structure factorMixed-parity Miller indices have $S_{\mathrm{FCC}}=0$. With all indices even or all odd, $S_{\mathrm{FCC}}=4$. For real atomic scattering factors,Hence all same-parity reflections are generally permitted in GaAs, although the last class is weak when the two atomic scattering factors nearly coincide. These are the structural selection rules; a particular experiment must also satisfy the elastic Bragg scattering condition.

Translate the scattering solution for the atom at the origin by $\mathbf{a}$. Its incident wave becomes $e^{i\mathbf{k}\cdot (\mathbf{r}-\mathbf{a})}$; multiplying the whole solution by $e^{i\mathbf{k}\cdot \mathbf{a}}$ restores the prescribed incident wave. At large $r$,Thus the outgoing amplitude acquires the phase $e^{i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot \mathbf{a}}$, with ${\mathbf{k}}^{\prime }=k\hat{\mathbf{r}}$.

In the single-scattering approximation, amplitudes from lattice sites add coherently. If $\mathbf{Q}=\mathbf{k}-{\mathbf{k}}^{\prime }$,For a rectangular block of a Bravais lattice with $N_j$ sites along primitive vectors ${\mathbf{a}}_j$ and $N=N_1{N}_2{N}_3$, the geometric series givesEach ratio has its limiting value $N_j$ when its argument is a multiple of $2\pi$. Thus the peaks have height $N$ and widths of order $1/N_j$ in those phase variables, at $\mathbf{Q}\cdot {\mathbf{a}}_j\in 2\pi \mathbb{Z}$, exactly the reciprocal lattice vectors.

With the incident-wave normalization used here, the Born approximation isFor the stated attractive three-dimensional delta potential,independent of scattering angle. This is the formal first Born amplitude, not a claim that an unregularized three-dimensional delta interaction defines an exact scattering Hamiltonian.

The 60-degree scattering angle is twice the Bragg angle, so $\theta =30^{\circ }$. First-order elastic Bragg scattering condition gives $2d\sin \theta =\lambda$, henceHigher orders would give $d=n\lambda$; the smallest, usual first-order spacing is the likely answer.

The reciprocal lattice is ${\Lambda }^{\ast }=\{G:G\cdot \ell \in 2\pi \mathbb{Z}\text{ for every }\ell \in \Lambda \}$. With oriented primitive-cell volume $v_c=a_1\cdot (a_2\times a_3)$, its basis is $b_1=2\pi (a_2\times a_3)/v_c$ and the cyclic variants; thus $b_i\cdot a_j=2\pi {\delta }_{ij}$.

The total potential is $V(x)={\sum }_{\ell }U(x-\ell )$. Using the Fourier transform convention $\tilde{U}(q)=\int e^{-iq\cdot x}U(x)d^{3}x$, translation gives $\tilde{V}(q)=\tilde{U}(q)\Delta (q)$, withThe given Born approximation therefore supplies the amplitude $\boxed{-m\Delta (q)\tilde{U}(q)/(2\pi {\hslash }^2)}$. A finite geometric series gives, when the denominators are nonzero,At reciprocal-lattice vectors each phase is one, so $|\Delta |=N_1{N}_2{N}_3$, the number of atoms. For large finite crystals these narrow peaks represent coherent Bragg scattering, with peak intensity proportional to the squared number of atoms; their finite width reflects the finite crystal size.

For the specified face-centered cubic lattice, $v_c=a^3/4$ andThe reciprocal lattice have coordinates $(2\pi /a)(h,k,l)$ with $h,k,l$ all even or all odd. The two shortest nonzero length families are $G_1=2\pi \sqrt{3}/a$ for $(111)$ and $G_2=4\pi /a$ for $(200)$. Elastic scattering satisfies $|q|=2|k|\sin (\theta /2)$. The assumed incident magnitude exceeds $G_2/2$, so both families are kinematically accessible, andThis is the angular prediction for suitable crystal orientations, or for a powder containing those orientations. A fixed single-crystal orientation must additionally satisfy the vector Elastic Bragg scattering condition $2k\cdot G+|G{|}^2=0$; the magnitude bound alone cannot guarantee both peaks for every incident direction.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.