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        <jats:article-title>Symmetries in transmission electron microscopy imaging of crystals with strain</jats:article-title>
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      <jats:ref id="">
        <jats:mixed-citation><jats:article-title>Symmetries in transmission electron microscopy imaging of crystals with strain</jats:article-title>,
            <jats:name><jats:surname>Maltsi</jats:surname>,
        <jats:given-names>Anieza</jats:given-names></jats:name><jats:name><jats:surname>Koprucki</jats:surname>,
        <jats:given-names>Thomas</jats:given-names></jats:name><jats:name><jats:surname>Mielke</jats:surname>,
        <jats:given-names>Alexander</jats:given-names></jats:name><jats:source/>,
            <jats:volume/>,
            <jats:page/> 
            (<jats:year/>).
            <jats:pub-id pub-id-type="doi"/></jats:mixed-citation>
      </jats:ref>
      <jats:abstract>
        <jats:p>Transmission electron microscopy (TEM) images of strained crystals often exhibit symmetries, the source of which is not always clear. To understand these symmetries, we distinguish between symmetries that occur from the imaging process itself and symmetries of the inclusion that might affect the image. For the imaging process, we prove mathematically that the intensities are invariant under specific transformations. A combination of these invariances with specific properties of the strain profile can then explain symmetries observed in TEM images. We demonstrate our approach to the study of symmetries in TEM images using selected examples in the field of semiconductor nanostructures such as quantum wells and quantum dots.</jats:p>
      </jats:abstract>
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    <jats:sec>Introduction<jats:p>In transmission electron microscopy (TEM), the main goal is to extract information on the specimen from the generated TEM images. This is particularly used for detecting shapes, sizes and composition of defects or inclusions like quantum wells and quantum dots (DT) in a larger specimen consisting of a regular crystalline material. However, there is no direct way to infer the inclusion properties from the TEM image. Hence, a commonly taken approach is to simulate the TEM imaging process with inclusions being described by parameterized data. Then, the comparison with experimental pictures can be used to fit the chosen parameters and deduce the desired data of the experimental inclusions.</jats:p><jats:p>A main feature in this process is symmetries for two reasons; first, the inclusions may have certain symmetries and second the TEM images may display symmetries that are related but not identical. The latter arises from the fact that the experimental set-up may have its own intrinsic symmetry properties. In the present work, we want to analyse these symmetries and explain why sometimes TEM images look more symmetric than the inclusion under investigation, or as the Curie's principle is stated in <jats:xref ref-type="bibr" rid="b0">[1]</jats:xref> (a3): the effect is more symmetric than its cause.</jats:p><jats:p>The interest in TEM image symmetries dates back to the 1960s-1970s, cf. <jats:xref ref-type="bibr" rid="b1">[2,</jats:xref><jats:xref ref-type="bibr" rid="b2">3]</jats:xref>, with the work focused mainly on the Reciprocity Theorem. It states that the amplitude at a point B of a wave originating from a source at point A and scattered by a potential V is equal to the scattered amplitude at point A originating from the same source at B. Many papers have been written for alternative proofs of this theorem, cf. <jats:xref ref-type="bibr" rid="b3">[4]</jats:xref><jats:xref ref-type="bibr" rid="b4">[5]</jats:xref><jats:xref ref-type="bibr" rid="b5">[6]</jats:xref><jats:xref ref-type="bibr" rid="b6">[7]</jats:xref>, as well as applications of it in the interpretation of TEM images, e.g. in connection with imaging of dislocations, cf. <jats:xref ref-type="bibr" rid="b7">[8]</jats:xref><jats:xref ref-type="bibr" rid="b8">[9]</jats:xref><jats:xref ref-type="bibr" rid="b9">[10]</jats:xref>.</jats:p><jats:p>While some of our results can also be deduced from the reciprocity theorem, like midplane reflection, there are more symmetries in the imaging process which can be proven mathematically by assuming the column approximation and focusing on the Darwin-Howie-Whelan (DHW) equations <jats:xref ref-type="bibr" rid="b1">[2,</jats:xref><jats:xref ref-type="bibr" rid="b10">11]</jats:xref>. Combining the symmetry properties of the imaging process with symmetry properties of the inclusion explains extra symmetries observed in TEM images of strained crystals.</jats:p><jats:p>The DHW equations, which are often simply called Howie-Whelan equations (cf. <jats:xref ref-type="bibr" rid="">[12, &#167;2.3</jats:xref>.2] or <jats:xref ref-type="bibr" rid="">[13, &#167;6.3]</jats:xref>), describe the propagation of electron beams through crystals and can be applied to semiconductor nanostructures, see <jats:xref ref-type="bibr" rid="b13">[14]</jats:xref><jats:xref ref-type="bibr" rid="b14">[15]</jats:xref><jats:xref ref-type="bibr" rid="b15">[16]</jats:xref><jats:xref ref-type="bibr" rid="b16">[17]</jats:xref>. While these equations are typically formulated for infinitely many beams in the dual lattice &#923; * , for all practical purposes, it is sufficient to use only a few important beams, because at high energy and for thin specimens only very few beams are excited by scattering of the incoming beam. A mathematical analysis of the corresponding beam selection is given in <jats:xref ref-type="bibr" rid="b17">[18]</jats:xref>, but this theoretical work is restricted to perfect crystals without inclusions. Here, we stay with finitely many beams, i.e. with so-called m-beam models with wave vectors g &#8712; &#923; * m , but generalize the analysis to crystals with inclusions. The main assumption is however that the crystallographic lattice stays approximately intact and can be modelled as a strained crystal where the positions of the lattice points undergo a displacement u(r). Then, the DHW equation for strained crystals reads Here, &#968; g denotes the wave function of the beam associated with g &#8712; &#923; * m , where g = 0 denotes the incoming beam. The vertical coordinate z &#8712; [0, z * ] gives the depth inside the specimen (z = 0 entry plane and z = z * exit plane), whereas the horizontal coordinates (x, y) are fixed and correspond to the image pixel, see figure 2.</jats:p><jats:p>After a minor transformation, the above system will take the vectorial form</jats:p>&#966; := d dz &#966; = i V + &#931; + F(z) &#966; and &#966;(0) = &#8730; &#961; 0 e 0 &#8712; C m ,(1.2)<jats:p>where &#966; = (&#966; g ) g&#8712;&#923; * m &#8712; C m contains the relevant wave functions. The matrix V corresponds to the electrostatic interaction potential and for elastic scattering it is Hermitian. The diagonal matrix &#931; = diag(s g ) contains the so-called excitation errors and F(z) = diag(g &#8226; (d/dz)u(x, y, z)) &#8712; R d&#215;d contains the projections of the strains to the individual wave vectors g &#8712; &#923; * m . We will call F the strain profile.</jats:p><jats:p>Image symmetries are now easily understood if changing the image pixel (x, y) to another pixel ( x, y) having the same strains throughout the whole thickness, i.e. u(x, y, z) = u( x, y, z) for all z &#8712; [0, z * ], which implies F(z) = F(z). Such a situation is related to a symmetry of the inclusion Downloaded from https://royalsocietypublishing.org/ on 28 February 2023 generating a symmetric strain field. As we will see, additional symmetries may occur in (1.2) in three distinct cases:</jats:p><jats:p>1. if F(z) is replaced by &#8722;F, a so-called sign change symmetry; 2. if F is reflected at the midplane z = z * /2, i.e. F(z) is replaced by F(z * &#8722;z), called midplane reflection symmetry; 3. if &#931; is replaced by &#8722;&#931;, called excitation error symmetry.</jats:p><jats:p>The latter symmetry is relevant when a series of images are done while varying the excitation error s g along the series.</jats:p><jats:p>These symmetries are observed experimentally (cf. <jats:xref ref-type="bibr" rid="b14">[15]</jats:xref>) but occur for the ODE system (1.2) only under additional conditions. Typically, the symmetries are exact only for the case of the twobeam model with &#923; * 2 = {0, g }. Nevertheless, the symmetries are approximately true in m-beam models if the intensities of the two strong beams (bright field and dark field intensities) are much higher than those of the weak beams.</jats:p><jats:p>The structure of our paper is as follows: in &#167;2, we provide the background of TEM imaging, its numerical simulation via the DHW equations, and the modelling of the influence of the strain. In &#167;3, we discuss all issues concerning symmetries in TEM imaging by considering well-chosen examples. In particular, we highlight the relevance of the symmetries for the detection of shapes of inclusions. The mathematical rigorous treatment of the symmetries for the m-beam model (1.2) is given in &#167;4, where the notion of weak and strong symmetries is introduced to provide a coherent structure of the symmetry properties, which also reveals why the two-beam case is different from the m-beam case with m &gt; 2. In the case of absorption, the Hermitian structure of the matrix V is broken and some symmetries do not hold anymore. This case is investigated in &#167;5.</jats:p></jats:sec>
    <jats:sec>Transmission electron microscopy image formation and Darwin-Howie-Whelan equation<jats:p>In TEM, electron beams are transmitted through the specimen to create an image. A parallel electron beam illuminates the specimen. As specimen crystalline materials with a thickness of a few hundred nanometres are considered. Due to the periodic structure of the crystal, the electron beams are diffracted in discrete directions. The diffracted beams leaving the exit plane of the specimen are focused again by the objective. Then, with the objective aperture, the set of beams forming the image can be reduced. This way specific beams can be chosen to create the image. If the image that is created includes the undiffracted beam it is called bright field image, otherwise it is a dark field image. The ray path within the microscope for the creation of a dark field TEM image is illustrated in figure 1a.</jats:p><jats:p>A crystal is a periodic structure created by the repetition of the unit cell across the directions of the direct lattice &#923; &#8834; R 3 . The reciprocal lattice &#923; * is the dual lattice of &#923; defined as &#923; * := g &#8712; R 3 g &#8226; r &#8712; Z for all r &#8712; &#923; . The discrete directions that the beams are diffracted are given by Bragg's Law <jats:xref ref-type="bibr" rid="b18">[19]</jats:xref> (also known as Laue conditions): for an incoming beam with wavevector k 0 a diffracted beam k may occur if the condition k = k 0 + g is satisfied, where g &#8712; &#923; * . For elastic scattering, the energy of the waves is conserved, meaning that the two vectors have the same length. This implies that the wavevectors k 0 and k have to lie on the surface of a sphere, known as the Ewald sphere <jats:xref ref-type="bibr" rid="b19">[20]</jats:xref> and defined as S Ew := g &#8712; R 3 |k 0 | 2 &#8722; |k 0 +g| 2 = 0 . When a reciprocal lattice point g falls on the Ewald sphere, the Bragg condition is satisfied and a diffracted beam in the direction k 0 + g occurs, see figure 1b. However, diffraction can occur even if the condition is not exactly satisfied. The deviation from Bragg's condition is expressed through the excitation error s g defined as where &#945; is the angle between the vector k 0 + g and the foil normal &#957;. The excitation errors are parameters that can easily be controlled by the experimental conditions, like the tilting of the sample. For two-beam approximation, we choose in addition to the incoming beam g = 0 one single reciprocal lattice vector g = 0 satisfying the so-called strong beam conditions, i.e. lies exactly on the Ewald sphere s g = 0. In this situation, the two beams g = 0 and g = g both are strongly excited because of s g = 0. Different choices can give rise to different image contrasts, as seen in figure 1c,d.</jats:p>s g = &#8722; g &#8226; (2k 0 +g) 2|k 0 +g| cos &#945; = |k 0 | 2 &#8722; |k 0 +g| 2 2(k 0 +g) &#8226; &#957; , (2<jats:p>(a) Multi-beam approach and Darwin-Howie-Whelan equations</jats:p><jats:p>The electron propagation is described by the relativistic Schr&#246;dinger equation, which is a threedimensional problem. However, for computational reasons, the three-dimensional problem is often reduced to a two-dimensional family of one-dimensional problems using the so-called column approximation. TEM uses fast electrons with acceleration voltages in the range 200-400 keV. This means that the angle between the diffracted and undiffracted beam is very small. For thin specimens (thickness in the range 100-200 nm), we can apply the column approximation, which states that an incoming beam will not leave a column centred at the entrance point. The width of the column defines the spatial resolution and is typically in the range of size of a unit cell, e.g. &#8776;0.5-1 nm. It also assumes that electrons are not scattered in neighbouring columns and the propagation can be computed independently, by solving the equations for each column in turn. We divide a rectangular specimen into squares of edge length l c defining the columns (i, j), i = 1, . . . , N x , j = 1, . . . , N y centred around the positions (x i , y j ) &#8712; R 2 , see figure 2. To obtain the simulated TEM image, for every pixel (i, j), the intensity has to be calculated by solving the dynamical diffraction equations for that column. We decomposed the spatial variable r = (x, y, z) into the transversal part (x, y) orthogonal to the thickness variable z &#8712; [0, z * ], see figure 2 Column approximation: an incoming beam is assumed not to leave a column centred at the entering point. For this column, the intensity corresponding to the (i, j) pixel is obtained by propagating the beam along a line scan (blue line) in the z-direction at position (x i , y j ) to the exit plane. (Online version in colour.) numerically, like in pyTEM software <jats:xref ref-type="bibr" rid="b20">[21]</jats:xref>, which in the case of a perfect crystal are</jats:p>d dz &#968; g (z) = i&#960; 2s g &#968; g (z) + 1 &#961; g h&#8712;&#923; * U g-h &#968; h (z) , &#968; g (0) = &#948; 0,g , for g &#8712; &#923; * ,(2.2)<jats:p>where</jats:p>&#961; g = (k 0 + g) &#8226; &#957;,<jats:p>where s g are the excitation errors given in (2.1) and U g are the Fourier coefficients of the periodic electrostatic lattice potential of the crystal. As the dual lattice &#923; * contains infinitely many points, (2.2) is an initial value problem for an infinite system of first-order ordinary differential equations describing the propagation of the electron beam through the specimen from the entry plane z = 0 to the exit plane z = z * . However, in experiments, the set-up is done in such a way that the incoming beam, which will always be given by g = 0 &#8712; &#923; * 0 , is diffracted in a few directions k 0 +g for g lying in a small subset &#923; * m of &#923; * , where m is used to indicate the number of elements in &#923; * m . Replacing &#923; * in (2.2) by &#923; * m , we arrive at an m-beam model, which is an ODE for the vector (&#968; g ) g&#8712;&#923; * m &#8712; C m . Of special importance will be the two-beam model with</jats:p>&#923; * 2 = {0, g'},<jats:p>which is widely used. From now on, we will denote by g the diffracted beam in the two-beam approximation and by g ap the beam chosen by the objective aperture. For a bright field image, we have g ap = 0 and for a dark field image under two-beam approximation we have g ap = g . The problem of finding good subsets &#923; * m , which is the so-called beam selection problem, is discussed from a mathematical point of view in <jats:xref ref-type="bibr" rid="b17">[18]</jats:xref>. There it is argued that the infinite dimensional problem for g &#8712; &#923; * is even ill-posed, and it is shown that under typical assumptions the intensities |&#968; g (z)| 2 decay exponentially like e &#8722;&#945;|g| . With this and further energetic considerations based on the Ewald sphere it was possible to derive rigorous error estimates to justify typical beam selection schemes, like the two-beam approximation or the systematic-row approximation. A simulation example was given, using a set of 30 beams as a reference and comparing the solutions with smaller sets. The simulation was done with parameters for gallium arsenide (GaAs) with a lattice constant a 0 = 0.56503 nm, a specimen thickness z * = 200a 0 &#8776; 113 nm and an acceleration voltage of 400 keV. The comparison showed that the four-beam case gives an accuracy of four significant digits and increasing the number of beams in systematic row did  not improve the accuracy. The accuracy would improve by going beyond the systematic row, where a seven significant digits accuracy was observed in the case of 18 beams. However, for the simulation of TEM images, the four significant digits accuracy is sufficient.</jats:p></jats:sec>
    <jats:sec>(b) Influence of defects and strain<jats:p>TEM imaging is widely used for the study of defects in crystalline materials, see <jats:xref ref-type="bibr" rid="b16">[17,</jats:xref><jats:xref ref-type="bibr" rid="b21">[22]</jats:xref><jats:xref ref-type="bibr" rid="b22">[23]</jats:xref><jats:xref ref-type="bibr" rid="b23">[24]</jats:xref>. Defects are perturbations of the crystal symmetry, in the sense that the atoms are displaced from their original position in the perfect crystal. If an atom was at position r, its new position will be r = r + u(r), where u(r) is the displacement field, see figure 3a. As an elementary example for strained crystals, we consider a spherical particle with radius r 0 and lattice parameter a p inside a matrix with lattice parameter a m , as is done in <jats:xref ref-type="bibr" rid="">[14, ch. 8, p. 479</jats:xref>]. The displacement field is given by</jats:p>u(r) = C(&#948;) min{|r|, r 0 } 3 |r| 3 r,(2.3)<jats:p>where C(&#948;) is a constant that depends on the elastic properties of the isotropic matrix and &#948; the matrix misfit given by &#948; = (a p &#8722; a m )/a m . In this case, the displacement inside the particle is proportional to r = (x, y, z), whereas outside it decays as 1/|r| 2 , see figure 3. The displacement field u is only valid for small isotropic inclusions where the particle diameter is significantly smaller than one extinction distance. An example of such a case can be a spherical InAs QD inside a GaAs matrix, see <jats:xref ref-type="bibr" rid="b15">[16,</jats:xref><jats:xref ref-type="bibr" rid="b24">25]</jats:xref>.</jats:p><jats:p>For small deformations, the displacement will modify the Fourier coefficients of the potential in the DHW equations (2.2) by a phase factor</jats:p>U g &#8594; U g e &#8722;2i&#960;g&#8226;u(r) .<jats:p>Using this and letting &#968; g = &#981; g e &#8722;ig&#8226;u(r) in (2.2), we get the DHW equations for a strained crystal <jats:xref ref-type="bibr" rid="">[14, ch.8</jats:xref></jats:p>] d dz &#981; g (z) = i&#960; 2s g + d dz (g &#8226; u(r)) &#981; g (z) + i&#960; &#961; g h&#8712;&#923; * m U g-h &#981; h (z) (2.4) and &#981; g (0) = &#948; 0,g for g &#8712; &#923; * m . (2.5)<jats:p>To simulate a TEM image with defects the column approximation and the DHW equation as described above can still be used, but now for each horizontal position (x i , y j ), where (i, j) denotes   the image pixel, u(r) in (2.4) is evaluated as u(z; x i , y j ). If this is constant, then the defect will not be visible. Another important fact for the imaging of defects is that the projection of the displacement to the reciprocal lattice vector g is what really matters, see figure 3b. If g &#8226; u(r) is constant, then again the defect is not visible. This means that by choosing different vectors g ap we get different information about the defect. Figure 4 illustrates this for a pyramidal QD. Choosing g ap = (040) will create a TEM image corresponding to the u x component of the displacement, as seen in figure 4b,c. Changing to g ap = (004) will give a TEM image corresponding to the u y component of the displacement, see figure 4d,e. This sensitivity of TEM images to different components of the displacement field is important for the interpretation of images and can be used for the reconstruction or classification of the observed object. In <jats:xref ref-type="bibr" rid="b15">[16]</jats:xref>, this was used to compare DT of four different geometries. It was observed that the projection to the vector g ap = (004) would give better contrast, allowing one to distinguish between pyramidal or lens-shaped dots, while in the g ap = (040) direction all images would show a very similar coffee-bean contrast making it difficult to distinguish among the geometries.</jats:p>u x (z; x 0 , y 0 ) = -u x (z; x 1 , y 1 ) u x (z; x 0 , y 0 ) = -u x (z; x 1 , y 1 ) (b) ( c) ( d) ( e)</jats:sec>
    <jats:sec>Symmetries in transmission electron microscopy images<jats:p>In this section, we study observed symmetries in TEM images of strained crystals and discuss their interpretation. To this end, we introduce selected examples demonstrating different kinds of symmetries, e.g. images that are pixelwise symmetric, like (c) and (e) in figure 4. This kind of symmetry occurs when there is a sign change in the displacement component, see &#167;3a, or when the inclusion is shifted from the centre, see &#167;3b. In &#167;3c, symmetries of a series of TEM images for varying excitation errors s g and varying positions are discussed. These examples show the importance of distinguishing different kinds of symmetries that can occur and to examine which ones are connected to specific properties of the displacement or strain profile and which are independent of it.</jats:p><jats:p>To understand the origin of these symmetries, we performed an analysis on the symmetry properties of solutions of the DHW equations. The main results are explained in &#167;3d, while the formal proofs are given in &#167;4. This analysis revealed three important symmetry principles, stated in &#167;3(iii). By combining these principles with specific properties of the strain profiles we can explain all the observed symmetries introduced in &#167;3a-c. The capability of our approach to explain symmetries in TEM images beyond these examples is demonstrated in &#167;3f, where the developed theory is applied to a more complex problem featuring general displacement profiles.  In figure 4c,e, we see two simulated TEM images for different choices of the vectors g ap . Each image is pixelwise symmetric, in the sense that for two different pixels (x 0 , y 0 ) and (x 1 , y 1 ), we have the same intensities:</jats:p>I g ap (x 0 , y 0 ) = |&#981; g ap (z * ; x 0 , y 0 )| 2 = |&#981; g ap (z * ; x 1 , y 1 )| 2 = I g ap (x 1 , y 1 )<jats:p>. For image 4e, this in not surprising since the profile of the vertical component of the displacement along the column related to pixel (x 0 , y 0 ) is the same as the one for pixel (x 1 , y 1 ), namely u y (z; x 0 , y 0 ) = u y (z; x 1 , y 1 ) for z &#8712; [0, z * ], due to the symmetry of the pyramid. However, the pixelwise symmetry in image 4c is interesting: the profiles of the horizontal displacement component, which are responsible for the image contrast, have opposite values u x (z; x 0 , y 0 ) = &#8722;u x (z; x 1 , y 1 ). This indicates that there might be some symmetry in TEM images with respect to the sign of the displacement.</jats:p><jats:p>In (2.4), we see that it is the product of the strain (d/dz)u with the reciprocal lattice vector g that enters the equations. This term will from now on be expressed as F g (z) = (d/dz)(g &#8226; u(z; x i , y j )) and the influence of the strain to an m-beam system will be represented by the matrix-valued function</jats:p>F(z) = diag(F g ) g&#8712;&#923; *<jats:p>m . So we want to know if the transformation F(z) &#8722;F(z) gives the same intensity. If it does, then the question that arises is whether it is for a specific shape of the strain profile F(z) or if it is independent of it and applies to general strain profiles.</jats:p><jats:p>(b) Symmetry with respect to the centre of the sample Our next example is inspired by images provided in <jats:xref ref-type="bibr" rid="b14">[15]</jats:xref>, where TEM imaging of an inclined strained semiconductor quantum well, like the one in figure 5a, has been studied. A quantum well is a planar heterostructure consisting of a thin film, forming the quantum well, sandwiched between barrier material layers forming the matrix. Due to the lattice mismatch between the materials the lattice of the quantum well is deformed. For pseudomorphically grown quantum wells with perfect interfaces, it can be assumed that the displacement grows linearly within the quantum well region and has a constant value outside, resulting in a strain profile similar to an indicator function, see figure 5b.</jats:p><jats:p>The intensity values of the dark field for such a structure are shown in figure 5c, for different values of the excitation error and for different positions. Due to the incline angle between the planar interface and the imaging direction, the different positions correspond to different depths  of the quantum well as measured from the surface of the specimen, see figure 5a. An interesting first observation here is that the intensity seems to be symmetric with respect to a shift in the position from the centre of the sample and for every excitation error s g ap . A natural question that occurs is whether this shifting symmetry is a general property of TEM imaging. The answer is negative and this can be seen in figure 6c which shows the dark field intensities for a spherical QD (figure 6a) again for different excitation errors and different positions. Shifting the QD from the centre for an excitation error s g ap = 0 does not give the same intensity. However, if we choose s g ap = 0 then we observe again a symmetry with respect to shifting.</jats:p><jats:p>To analyse these observations, we take a closer look into the shape of the strain in each case. For the quantum well in figure 5, the strain profile is an even function (figure 5b) while for the QD in figure 6, it is an odd function (figure 6b). The latter is due to the symmetry of the sphere, cf. displacement field for spherical inclusion (2.3). Shifting the inclusion would correspond to shifting the strain in both cases as seen in figures 5d and 6d, respectively. The questions to be answered here are (i) what is special in the case s g ap = 0 that makes shifting a symmetry, (ii) how does shifting an even or odd strain profile affect the intensities and (iii) what happens for a general strain profile? (c) Symmetry with respect to the sign of s g</jats:p><jats:p>In the previous examples, we considered pixelwise symmetry for one specific image. This was expressed as I g ap (x 0 , y 0 ) = I g ap (x 1 , y 1 ). In this section, we talk about pixelwise symmetry between images. This means that if I g ap corresponds to the intensity of an image and I g ap to the intensity of another image then the two images are pixelwise symmetric if I g ap (x i , y j ) = I g ap (x i , y j ) for every pixel (i, j).</jats:p><jats:p>In figure 7, we have TEM images, adapted from <jats:xref ref-type="bibr" rid="b24">[25]</jats:xref>, of a spherical QD at different positions and for different excitation errors. The observations we made for shifting at the previous section apply here as well. Shifting the QD for an excitation error s g ap = 0 creates images that are pixelwise symmetric with each other (red boxes) while for s g ap = 0 they are not symmetric (&#945; and &#946; blue boxes). Shifting for an s g ap = 0 however seems to create mirrored images, in the sense that the image &#945; is a mirrored version of image &#946; with respect to the symmetry axis of the sphere. Interestingly though we see that if we shift the QD from the centre and additionally change the sign of the excitation error s g ap then the two images are pixelwise symmetric (green boxes or blue &#946; and &#947; boxes). Again the question that arises here is whether these observations are connected to a specific property of the strain profile or is there a symmetry connected to shifting and sign change of s g ap that occurs for general profiles?</jats:p></jats:sec>
    <jats:sec>(d) Symmetries explained via Darwin-Howie-Whelan equations<jats:p>To understand the symmetries in TEM images we described above, we studied the properties of the beam propagation through the specimen using the DHW equations. It turned out that the intensity at the exit plane is invariant under specific transformations of the strain field. In the following, we give an introduction to our approach and an overview of the different types of symmetries formally defined and proved in mathematically rigorous terms in &#167;4. We conclude the section with an explanation of the observed symmetries using the theory we developed.</jats:p></jats:sec>
    <jats:sec>(i) Transformation to Hermitian form<jats:p>To begin with, it is essential to use the self-adjoint structure that is somehow hidden in the DHW equations. This can either be done as in <jats:xref ref-type="bibr" rid="b17">[18]</jats:xref>, where C m is equipped with the scalar product a, b = g&#8712;&#923; * m &#961; g a g b g , or by the simple transformation</jats:p>&#966; g = &#8730; &#961; g &#981; g for g &#8712; &#923; * m ,<jats:p>which will be used in this paper. This has the advantage that C m is equipped with the standard (complex) Euclidean scalar product, but the intensities take the form</jats:p>I g (x i , y j ) = |&#981; g (z * ; x i , y j )| 2 = |&#966; g (z * ; x i , y j )| 2 /&#961; g .<jats:p>In terms of &#966; = (&#966; g ) g&#8712;&#923; * m &#8712; C m , the system (2.4) is rewritten in matrix form as follows: Subsequently, we will omit the normalizing factor &#8730; &#961; 0 in the initial condition &#966;(0), because it is not relevant in TEM imaging, where grey-scale pictures are created using relative intensities only.</jats:p>&#966; := d dz &#966; = i A + F(z) &#966; and &#966;(0) = &#8730; &#961; 0 e 0 &#8712; C m . (3<jats:p>The system matrix A = V + &#931; and the influence F(z) of the strain are given via</jats:p>V = &#960; U g&#8722;h &#8730; &#961; g &#961; h g,h&#8712;&#923; * m , &#931; = diag(2&#960; s g ) g&#8712;&#923; * m F(z) = diag d dz (g &#8226; u(z)) g&#8712;&#923; * m , &#9131; &#9130; &#9130; &#9130; &#9132; &#9130; &#9130; &#9130; &#9133; (3.1b)<jats:p>where V describes the interaction of the beams via the scattering potential and &#931; is related to the excitation conditions. As the Fourier coefficients of the scattering potential satisfy U &#8722;g = U g , we see that V &#8712; C m&#215;m is indeed a Hermitian matrix, while &#931; and F(z) are real-valued diagonal matrices.</jats:p><jats:p>What is important in TEM imaging is the intensity of the strongly excited beams at the exit plane z = z * and not all components of &#966;. Our theory is developed in such a way that it focuses on the amplitude of the undiffracted beam, |&#966; 0 (z * )|, which corresponds to a bright-field image. The point is that this generates a potential reflection symmetry z z * &#8722; z, because the initial condition &#966;(0) = e 0 and the exit measurement &#966; 0 (z * ) = &#966;(z * ) &#8226; e 0 use the same vector e 0 .</jats:p><jats:p>Intensities of solutions for different choices of the pair (A, F(z)) are compared to see which replacements of (A, F) by ( A, F(z)) lead to the same (measurement) results. Such transformations are then called symmetries. Changes in F(z) correspond to transformations in the strain, while changes in the matrix A can correspond to transformations in the excitation errors (given by &#931;) or the potential (given by U).</jats:p></jats:sec>
    <jats:sec>(ii) Strong and weak symmetries<jats:p>Two kind of symmetries are defined in &#167;4.1: strong and weak symmetries. For strong symmetry, the intensity of the beam along the whole column [0, z * ] is invariant under the transformation (A, F) &#8594; ( A, F), that means the corresponding solutions &#966; and &#966; satisfy |&#966; 0 (z)| = | &#966; 0 (z)| for all z &#8712; [0, z * ]. For weak symmetry, this invariance holds for the intensity of the beam at the exit plane only, namely |&#966; 0 (z * )| = | &#966; 0 (z * )|. In TEM imaging, this distinction is not visible since we only see the intensity at the exit plane. For the mathematical analysis, however, this distinction is highly relevant because of the different underlying mechanisms. Of course, any composition of weak and strong symmetries provides a weak symmetry again.</jats:p><jats:p>We illustrate strong and weak symmetries by numerical simulations of the DHW equations for a strained GaAs layer using four beams and a displacement field as given by (2.3), see figure 8. In this example, we also observe a dark field symmetry, namely |&#966; g ap (z)| = | &#966; g ap (z)| (strong symmetry) or |&#966; g ap (z * )| = | &#966; g ap (z * )| (weak symmetry) with g ap = (1, 0). In table 1, we see the intensities for all four beams at the exit plane. While figure 8 suggests an exact symmetry, table 1 reveals that the symmetry is only approximate with an error up to 1%. The reason is that the four-beam model does not enjoy the symmetries, however, the solutions stay close to the solutions of the two-beam model, see table 2, which has the desired symmetries. This simple example demonstrates the importance of the two-beam approximation in the study of symmetries for both bright field and dark field.</jats:p></jats:sec>
    <jats:sec>(iii) Three important symmetry facts<jats:p>Here, we give an overview of the necessary results from &#167;4 that help us explain the symmetries in TEM imaging observed at the beginning of the section. The results are stated as facts and put into physics words, while the formal version of them and the proofs can be found in the next section.</jats:p><jats:p>The first fact concerns the change in the sign of the strain, which corresponds to changing the sign of F(z), and is proved in corollary 4.3. The next fact concerns reflections at the midplane of the specimen given by the transformation F(z) F(z * &#8722;z) and is proved in corollary 4.5 part (W3). Here, it is important to note that fact 3.2 does not require strong beam conditions, so it can be applied for excitation errors s g = 0. This result is equivalent to the Type II symmetry in <jats:xref ref-type="bibr" rid="b5">[6]</jats:xref> or to <jats:xref ref-type="bibr" rid="b1">[2]</jats:xref> who showed this symmetry for bright field images. Part (W2) of corollary 4.5 shows that, in the general m-beam case, the midplane reflection symmetry holds under the assumption that all relevant U g are real. However, even in the case of absorption, this midplane reflection symmetry still holds, as seen in theorem 5.1.</jats:p><jats:p>In the next fact, we combine the first two facts with an additional sign change of the excitation error s g , proved in corollary 4.6. </jats:p></jats:sec>
    <jats:sec>Table 1.<jats:p>Comparison of intensities at the exit plane for the four-beam model in figure 8. For both bright field (g = (0, 0)) and dark field (g = (1, 0)), we observe an approximate symmetry with an error of about 1%.  8 and under two-beam approximation. Both bright and dark field show a perfect symmetry in this case (up to some numerical error).  The Type I symmetry in <jats:xref ref-type="bibr" rid="b5">[6]</jats:xref> is a special case of this result for s g = 0. All results are derived for a general strain profile. The strain profiles in the examples we discussed before have an additional symmetry, namely they are even or odd functions which are shifted relative to the centre of the specimen, see figures 5d and 6d. In the next subsection, we will show how the above observations interact with the parity of the strain profile z &#8594; F(z).</jats:p>g (A, F(z)) (A, &#8722;F(z)) (A, F(z * &#8722; z)) (A, &#8722;F(z * &#8722; z)) (&#8722;g (A, F(z)) (A, &#8722;F(z)) (A, F(z * &#8722; z)) (A, &#8722;F(z * &#8722; z)) (0,</jats:sec>
    <jats:sec>(iv) Explanation of observed symmetries<jats:p>With the symmetries that we have in hand now we are able to answer all the questions that occurred from the observations we made before. We start with the symmetry with respect to the sign of the strain (F(z) &#8722;F(z)), that was discussed in &#167;3a using the example of the pyramidal QD in figure 4. We can now say that this is a direct application of fact 3.1 to every pair of pixels (i, j) and (i , j ) such that F(z; x i , x j ) = &#8722;F(z; x i , y i ) and F(z) being a general strain profile.</jats:p><jats:p>For the symmetry with respect to the centre of the sample discussed in &#167;3b a combination of the facts 3.1 and 3.2 with the parity of the strain profile can explain the observations. We take each case separately. For the inclined quantum well the strain has an even profile, as in figure 9a. From fact 3.2, we know that we can apply midplane reflection (F(z) F(z * &#8722; z)) and get the same pixel intensity. For an even profile midplane reflection and shifting coincide, see figure 9a. This is the reason why the image shows a pixelwise symmetry with respect to shifting. In the case of the spherical QD, the strain has an odd profile, as in figure 9b. Applying midplane reflection, we do not get the same result as shifting, see figure 9b. We would need to apply the sign change as well, as stated in fact 3.1. This however cannot be done unless we have strong beam conditions (meaning s g ap = 0). This is the reason why, for s g ap = 0, we observe a symmetry with respect to shifting while for s g a p = 0 we do not.</jats:p><jats:p>The observations concerning the sign change of the s g made in &#167;3c, e.g. see green and red boxes in figure 7, can be explained from fact 3.3: it says that a midplane reflection combined with a sign change in the strain (F(z) &#8722;F(z * &#8722;z) ) is a symmetry if we also change the sign of the excitation error (&#931; &#8722;&#931;). In this case, strong beam condition (s g ap = 0) is not a requirement. This means that we can apply midplane reflection plus sign change of the strain, which for the odd strain profile in figure 7 would correspond to shifting the strain profile with respect to the centre, and then change the sign of the excitation error. This explains the symmetric images in  </jats:p>z = z * z = 0 z = z * 2 z = z * z = 0 z = z * 2 F(z) F(z * &#8722; z) F(z) z z (a) ( b) F(z * &#8722; z) -F(z * &#8722; z)</jats:sec>
    <jats:sec>(e) Mirrored transmission electron microscopy images induced by strain<jats:p>Here, we will focus on explaining the images in figure 7 that are indicated by the blue boxes. First, we start with the TEM images &#945; and &#946;, see also figure 10. This means we have two images of a spherical QD using the same excitation error, here s g ap = 12(1/&#956;m), but placed in different positions, symmetrical to the centre of the sample, see figure 10a,b.</jats:p><jats:p>To analyse the images pixelwise, we make two line scans in the z-direction, A and B. The corresponding pixels for each image are indicated in figure 10a (image &#945;) and (b) (image &#946;), using the same notation A and B. We can see in figure 10a that the pixel intensities corresponding to the line scans A and B in image &#945; are not the same. So the image itself does not have pixelwise symmetry. Comparing the pixels between the images &#945; and &#946; though shows that the pixel in the image &#945; that corresponds to the line scan A (or B) is the same as the pixel in the image &#946; that corresponds to the line scan B (or A).</jats:p><jats:p>To understand these properties using the theory we developed we study the strain profile for each line scan, shown in figure 10c. First, we focus on why the image itself is not pixelwise symmetric. For the QD in image &#945;, the strain profile across the two line scans is shown in figure 10c by the solid black and red lines. We see that the difference between these two profiles is the sign. Changing the sign of the strain though is a symmetry only under strong beam conditions (fact 3.1) but in this case, we have s g ap = 0. The same exact argument applies to image &#946;.</jats:p><jats:p>Next, we compare the two images with each other. The strain profile for the spherical QD in image &#946; is given in figure 10c by the dotted black and red lines. The reason that the pixel corresponding to the line scan A in image &#945; is equal to the one that corresponds to line scan B in image &#946; is fact 3.2, since the strain profile for the first case (solid black line in figure 10c) is a midplane reflection of the strain profile in the second case (dotted red line in figure 10c). This is due to the fact that for an odd function shifting the strain (black dotted line) plus sign change correspond to midplane reflection (see also figure 9b).  Additionally, from fact 3.3, we know that image &#946; is symmetric to the image &#947; in figure 7. Combining all the above, we can see why also the images corresponding to the same position but with opposite excitation errors are mirrored images of each other, see figure 7. This mirror-like symmetry is induced by the parity of the strain profile.</jats:p></jats:sec>
    <jats:sec>(f) Symmetries for general profiles<jats:p>The examples discussed until now were for a strain profile with odd or even parity. However, this parity is not the essential cause for the symmetries observed between two pixels. What is important is the symmetry between the strain profiles with respect to sign change and midplane reflection. To make this clear we consider the case of a general strain profile without a specific parity. For this purpose, we examine TEM images of a pyramidal QD with a rhomboid as a base instead of a square. We assume that the QD is placed at the centre of the sample. To create these TEM images, we used the computational method described in <jats:xref ref-type="bibr" rid="b15">[16]</jats:xref> and the tool chain employed therein. First, a three-dimensional mesh is generated to represent the geometry of the QD using TetGen <jats:xref ref-type="bibr" rid="b25">[26]</jats:xref>, see figure 11a. Then the generated mesh enters the FEM-based solver WIAS-pdelib <jats:xref ref-type="bibr" rid="b26">[27]</jats:xref>, in order to find the displacement u, see figure 11c. Finally, the relevant displacement component enters the DHW solver PyTEM <jats:xref ref-type="bibr" rid="b20">[21]</jats:xref> in order to simulate the corresponding TEM image, see figure 11b. For this set up, two TEM images are computed, corresponding to different vectors g ap using strong beam excitation conditions.</jats:p><jats:p>For an excitation corresponding to g ap = (004), the TEM image is shown in figure 11b. The (projected) component of the displacement, which is responsible for the image contrast in this case, is shown in figure 11c. This was taken in a cross-section parallel to the base of the pyramid, as indicated by the white dotted lines in figure 11a. Next, we analyse the displacement profile along the two line scans A and B evolving in the z-direction, as indicated in figure 11a,c by the black and red dotted lines. The displacement profiles across these line scans are shown in figure 11d, where we can see that they are not even or odd. However, we observe a pixelwise symmetry in the TEM image. The displacement u A (z) across line A is a midplane reflection of the displacement u B (z) across line B: u A (z * &#8722; z) = u B (z). This means that the strain across A differs with the strain  For an excitation corresponding to g ap = (040), the TEM image is shown in figure 12b. Under this excitation, the imaging is sensitive to a different component of the displacement field as in the case before. The corresponding displacement field in the cross-section is shown in figure 12c. We can see that the displacement obeys the sign change symmetry with respect to the centre of the structure, as also observed for the pyramidal QD with square base, see figure 4. As in the example before, we plot the displacement profile across lines A and B as shown in figure 12d. Here again we see that it is not even or odd. However, midplane reflection and sign change of the displacement profile in A equals the displacement in B: u B (z) = &#8722;u A (z * &#8722; z). This gives for the corresponding strain that (d/dz)u B (z) = (d/dz)u A (z * &#8722; z). Then fact 3.2 explains the pixelwise symmetry we observe. These two examples demonstrate that the results from &#167;(iii) are indeed valid for a general displacement profile. </jats:p>A, F) = (A+&#948;I, F) (S2) complex conjugation: ( A, F) = (&#8722;A, &#8722;F) (S3) constant phase factors: ( A, F) = (Q &#968; AQ * &#968; , F) with Q &#968; = diag(1, e i&#968; 2 , . . . , e i&#968; m )<jats:p>, where &#948;, &#968; j &#8712; R.</jats:p><jats:p>Proof. In all three cases, the result follows easily by writing down the corresponding evolution operators.</jats:p><jats:p>(S1) U &#948;I+A+F (z) = e i&#948;z U A+F (z) giving &#966; 0 (z) = e i&#948;z &#966; 0 (z). (S2) By complex conjugation of (3.1a), we easily obtain U &#8722;A&#8722;F (z) = U A+F (z). As the initial condition &#966;(0) = e 0 is real, we conclude &#966;(z) = &#966;(z) and hence &#966; 0 (z) = &#966; 0 (z). (S3) For this case, we use the transformation rule (4.2) with Q = Q &#968; and observe that</jats:p>Q &#968; F(z)Q * &#968; = F(z) because F is diagonal. Hence we have &#966; 0 (z) = &#966; 0 (z).<jats:p>As a first non-trivial result, we now reduce to the case m = 2 with the additional restriction A 00 = A g g . Indeed, the condition  </jats:p>A 00 = A g'g' , which means U 0 &#961; 0 + 2s 0 = U 0 &#961; g' +2s|&#951; 0 (z)| 2 m=2 = |&#951;(z)| 2 &#8722; |&#951; g' (z)| 2 unit. = |e 0 | 2 &#8722; |&#951; g' (z)| 2 .<jats:p>Hence, we obtain | &#966; g (z)| = |&#966; g (z)| from the corresponding result for g = 0.</jats:p><jats:p>To study the midplane reflection, we introduce the</jats:p>flip operator R(z) = z * &#8722;z acting on C 0 ([0, z * ]; D m ) via (F&#8226;R)(z) = F(R(z)) = F(z * &#8722;z).<jats:p>The following identity will be crucial for the understanding of the flip symmetry as a weak symmetry. Of course, one cannot expect that flipping gives rise to a strong symmetry. To see this, we consider a non-trivial strain profile F with F(z) = 0 for z &#8712; [z * /2, z * ], i.e. the perturbation acts only in the upper half of the specimen. The flipped case F = F&#8226;R then corresponds to a perturbation acting only in the lower half of the specimen. In such a case one cannot expect that the bright-field intensities |&#966; 0 (z)| 2 and | &#966; 0 (z)| 2 are the same inside the specimen. However, because of the double occurrence of the vector e 0 there is some chance that the intensities match for z = z * only. According to this theorem, midplane reflection symmetry will survive absorption. In figure 13, a numerical simulation of the DHW equations is shown, using four beams and the displacement field given by (2.3). The simulation is done for the same experimental conditions as the ones in figure 8. The electrostatic lattice potential has an imaginary part of magnitude 10% of the real part, i.e. U &#8594; U + 0.1iU. The simulation is done under strong beam conditions, which means that s (0,0) = s (1,0) = 0. Comparing the values in table 3, which shows the intensities at the exit plane, we can make the following observations:</jats:p><jats:p>-midplane reflection for the bright field seems to hold.</jats:p><jats:p>-midplane reflection plus sign change seems to hold for the dark field.</jats:p><jats:p>The first observation is justified by theorem 5.1. For the second observation, which also holds in the case of elastic scattering as seen in table 1, further analysis is required. This symmetry of the dark field will break if strong beam conditions do not hold, i.e. s g ap = 0. The reason that strong beam conditions make dark field images symmetric with respect to sign change plus midplane reflection is not clear yet and will be investigated in future work.  </jats:p></jats:sec>
    <jats:sec>Conclusion<jats:p>The symmetry properties of the TEM imaging process were analysed via the DHW equations. This analysis showed that the imaging process in elastic scattering is invariant under special transformations. The most important symmetries are the sign change of the strain field and the midplane reflection as well as a symmetry related to the sign change of the excitation error. The latter can be of particular importance in experiments, since modern TEM can easily create series of images by changing the excitation error. Combining these results with specific properties of the strain profile of the inclusion explains extra symmetries observed in TEM images. For inelastic scattering including absorption, the midplane reflection symmetry was proved to hold. However, the absorption case needs further investigation to explain symmetries observed in dark field images.</jats:p><jats:p>The distinction between symmetries of the imaging process and symmetries of the strain field can be used to extract information for the inclusion, e.g. shape or size. The approach can also be applied to the imaging of dislocations, since the TEM images are sensitive to the strain field they induce. However, the practitioner should always first check the validity of the fewbeam approximation and absorption effects for their application by suitable simulations. This will yield some thickness regimes where the approximations are valid and thus also the findings of the paper are valid in this regime. In addition, the theory was developed for a homogeneous material assuming a spatially constant scattering potential. If one considers thickness and composition variations inside a heterostructure the potential coefficients in the DHW equations become position-dependent. This generalization goes beyond the scope of this paper and will be considered in future work. Data accessibility. Data and code for our paper can be found at: https://doi.org/10.5281/zenodo.7248879 [28]. Authors' contributions. T.K.: conceptualization, formal analysis, investigation, software, visualization, writingoriginal draft, writing-review and editing; A.Ma.: conceptualization, formal analysis, investigation, software, visualization, writing-original draft, writing-review and editing; A.Mi.: conceptualization, formal analysis, investigation, software, visualization, writing-original draft, writing-review and editing.</jats:p><jats:p>All authors gave final approval for publication and agreed to be held accountable for the work performed therein.</jats:p><jats:p>Conflict of interest declaration. We declare that we have no competing interests. </jats:p></jats:sec>
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        <jats:p>d dz &#981; g (z) = i&#960; 2s g + g &#8226; d dz u(x, y, z) &#981; g (z) + i&#960; &#961; g h&#8712;&#923; * m U g-h &#981; h (z) for g &#8712; &#923; * m .(1.1)</jats:p>
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      <jats:label>. 1 )Figure 1 .</jats:label>
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        <jats:p>Figure 1. Image formation in TEM: (a) ray paths in TEM for dark field imaging, where the objective aperture allows only selected diffracted beams (red) to pass to the detector. The incoming beam with wavevector k 0 enters the specimen, is partially transmitted, and generates beams with nearby wave vectors k 0 + g. The red beam fulfils strong beam conditions on the Ewald sphere (b). Experimental dark field TEM images of InAs quantum dots for g = (040) in (c), sensitive to [010] component of strain, and g = (004) in (d), sensitive to [001] component of strain, both for an acceleration voltage of 300 keV. The whole figure is adapted from [16, figs. 1 and 2] used under CC-BY. (Online version in colour.)</jats:p>
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      <jats:caption>
        <jats:p>Figure 2. Column approximation: an incoming beam is assumed not to leave a column centred at the entering point. For this column, the intensity corresponding to the (i, j) pixel is obtained by propagating the beam along a line scan (blue line) in the z-direction at position (x i , y j ) to the exit plane. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Figure 3 .</jats:label>
      <jats:caption>
        <jats:p>Figure 3. Crystal lattice with spherical inclusion: (a) deformation of the lattice (b) variation of the projection of the displacement u on the g vector along the line scan in z-direction (red) in the crystal. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Figure 4 .</jats:label>
      <jats:caption>
        <jats:p>Figure 4. Simulation of TEM images for pyramidal InGaAs (green) QD (indium content 80%) embedded in a GaAs (red) matrix for an acceleration voltage of 300 keV and sample thickness 150 nm: (a) QD geometry indicating the crystallographic directions, (b) u x component and (d) u y component of displacement field along a cross-section in the centre of the structure. (c,e) corresponding TEM images for strong beam conditions as indicated by the direction of the chosen vectors g. The images in the figure are adapted from [16, fig. 5] used under CC-BY. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>FFigure 5 .</jats:label>
      <jats:caption>
        <jats:p>Figure 5. TEM imaging of inclined InGaAs quantum well (indium content 25%): (a) illustration of a specimen for an inclined quantum well (pink) embedded in a GaAs matrix (green) showing the z-direction and two line scans (purple dotted lines). (b) The displacement (red) and strain (black) profiles projected to the reciprocal vector g ap . (c) Intensity values for different positions and different excitation errors s g ap for a beam propagating in z-direction and acceleration voltage of 300 keV . Adapted from [15, fig. 5] used under CC-BY. (d) Strain profile for the different positions corresponding to a shift of the strain across the z-direction. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>FFigure 6 .</jats:label>
      <jats:caption>
        <jats:p>Figure 6. TEM imaging of spherical InGaAs quantum dot (indium content 80%) embedded in a GaAs matrix: (a) illustration of the specimen in (x, y) projection showing the direction of the chosen beam g ap . (b) Propagating the beam across z for a chosen (x 0 , y 0 ) gives the intensity at the corresponding pixel. The displacement (red) and strain (black) profiles projected to the reciprocal vector g ap . (c) Intensity values for different positions and different excitation errors s g ap , for acceleration voltage of 300 keV. Adapted from [25, fig. 5.13] used under CC-BY. (d) Strain profile for the different positions corresponding to a shift of the strain across the z-direction. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Figure 7 .</jats:label>
      <jats:caption>
        <jats:p>Figure 7. Series of TEM images for a spherical InGaAs quantum dot (indium content 80%) embedded in a GaAs matrix and acceleration voltage 300 keV: (a) schematics of the position of quantum dot in the sample. (b) Simulated TEM images for different depths of the quantum dot and for different excitation errors s g ap . For s g ap = 0, the TEM images show a pixelwise symmetry with respect to the centre (red boxes). For s g ap = 0, the TEM images are symmetric with respect to the centre if in addition the sign of the excitation error is changed (green boxes). The images are mirrored to each other with respect to the centre for the same excitation error (&#945; and &#946; blue boxes) or with respect to the sign of the excitation error for a fixed position (&#945; and &#947; blue boxes). Adapted from [25, fig. 5.12] used under CC-BY. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Fact 3 . 1 .Figure 8 .</jats:label>
      <jats:caption>
        <jats:p>Figure 8. Strong and weak symmetry in four-beam approximation: (a) intensities for system (A, F(z)) used as reference, (b) intensities for system ( A, F(z)) = (A, &#8722;F(z)). The solution &#966; has a strong symmetry compared to the solution &#966; of (a). (c) Intensities for system ( &#194;, F (z)) = (A, F(z * &#8722; z)). The solution &#966; has a weak symmetry compared to the solution &#966; of (a). (d) Intensities for system ( &#195;, F (z)) = (A, &#8722;F(z * &#8722; z)) showing a strong symmetry to case (c) and weak symmetry to (a,b). For the simulation material, parameters for GaAs have been used. The lattice constant was a 0 = 0.56503 nm and the specimen thickness z * = 200a 0 &#8776; 113 nm. The acceleration voltage was considered to be 400 keV. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Fact 3 . 2 .</jats:label>
      <jats:caption>
        <jats:p>In the two-beam approximation &#923; * 2 = {0, g } a midplane reflection of the strain (F(z) F(z * &#8722;z)) is a weak symmetry.</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Fact 3 . 3 .</jats:label>
      <jats:caption>
        <jats:p>In the two-beam approximation &#923; * 2 = {0, g } combining the sign change of the strain with a midplane reflection (F(z) &#8722;F(z * &#8722; z)) and changing the sign of the excitation error s g &#8722;s g is a weak symmetry.</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Figure 9 .</jats:label>
      <jats:caption>
        <jats:p>Figure 9. Plot of a shifted even function F(z) (black) and the midplane reflection of it F(z * &#8722; z) (red) illustrating that the midplane reflection corresponds to shifting F(z) (a). Plot of a shifted odd function F(z) (black) and the midplane reflection of it (red) illustrating that shifting (black dotted) needs an additional sign change to correspond to midplane reflection (b). (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Figure 10 .</jats:label>
      <jats:caption>
        <jats:p>Figure 10. Mirrored images from figure 7: (a) spherical quantum dot positioned at z = 3 8 z * and the corresponding TEM image (&#945;). Solving the equations across the line scans A and B gives the corresponding pixels in &#945;, denoted also by A and B. (b) Spherical quantum dot positioned at z = 5 8 z * and the corresponding TEM image (&#946;), showing again the pixels A and B that correspond to the same line scans. (c) Strain profiles across the line scans A and B. The solid black and red profiles correspond to the TEM image &#945;, while the dotted black and red profiles to TEM image &#946;. The TEM images are adapted from [25, fig. 5.12] used under CC-BY, cf. figure 7. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Figure 11 .Figure 12 .</jats:label>
      <jats:caption>
        <jats:p>Figure 11. Pyramidal quantum dot with a rhomboidal base: (a) 3D geometry showing the z-direction and two line scans A and B. (b) Simulated TEM image for g ap = (004) showing a pixelwise symmetry. (c) Displacement component responsible for the image contrast at the cut (white dotted lines) in (a) and the line scans A and B. (d) Displacement profiles across the line scans A (black solid) and B (red solid). We see that the displacement profile is not even nor odd and that the displacement in B is the midplane reflection of the displacement in A (black dashed). An InGaAs quantum dot (indium content 80%) embedded in a GaAs matrix was used with an acceleration voltage of 300 keV and a sample thickness of 150 nm, as the material in figure 4. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (S1) simultaneous linear phase factor: (</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>g' , is typically satisfied (in high enough accuracy) in the case of the strong two-beam conditions, because one usually chooses s g = s 0 = 0 and one has &#961; 0 = k 0 &#8226; &#957; &#8776; &#961; g . This holds automatically if g &#8226; &#957; = 0 or it is approximately true in the case of high-energy electrons, i.e. |k 0 | |g |.</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Corollary 4 . 3 (</jats:label>
      <jats:caption>
        <jats:p>Sign change using m = 2 and A 00 = A g g ). In the caseA = V + &#931; &#8712; C 2&#215;2 Herm with A 00 = A g g , the transformation ( A, F) = (A, &#8722;F) is a strong symmetry, i.e. | &#966; g (z)| = |&#966; g (z)| for z &#8712; [0, z * ] and g &#8712; &#923; * 2 = {0, g }.Proof. The result follows by combining the three strong symmetries (S1)-(S3). We writeA= a b b a with a &#8712; R and b = |b|e i&#946; . Applying first (S2), we find a strong symmetry with (A 1 , F 1 ) = (&#8722;A, &#8722;F). Next, we apply (S1) with &#948; = 2a such that (A 2 , F 2 ) = (2aI &#8722; A, &#8722;F) is again a strong symmetry. Finally, we apply (S3) with &#968; 2 = &#960; &#8722; 2&#946; and observe that e i&#968; 2 = &#8722;e &#8722;i2&#946; , which gives diag(1, &#8722;e i2&#946; ) 2aI&#8722;A diag(1, &#8722;e &#8722;i2&#946; ) = A. Hence, (A 3 , F 3 ) = (A, &#8722;F) is a strong symmetry giving | &#966; 0 (z)| = |&#966; 0 (z)| for all z &#8712; [0, z * ]. Finally, the assumption m = 2 and the unitarity (4.1) give, for &#951; = &#966; or &#966;, the relation</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>Figure 13 .</jats:label>
      <jats:caption>
        <jats:p>Figure 13. Strong and weak symmetry in four-beam approximation, for wide-angle absorption with imaginary part of U of 10% of the real part: (a) intensities for system (A, F(z)) used as reference, (b) intensities for system ( A, F(z)) = (A, &#8722;F(z)), (c) intensities for system ( &#194;, F (z)) = (A, F(z * &#8722; z)) and (d) intensities for system ( &#195;, F (z)) = (A, &#8722;F(z * &#8722; z)). Material parameters and experimental conditions as the ones in figure 8. (Online version in colour.)</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label>.</jats:label>
      <jats:caption>
        <jats:p>. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p>Funding. The research was partially supported by the DFG through the Berlin Mathematics Research Center MATH+ (EXC-2046/1, project ID: 390685689) via the subproject EF3-1 Model-based geometry reconstruction from TEM images.</jats:p>
      </jats:caption>
    </jats:fig>
    <jats:fig>
      <jats:label/>
      <jats:caption>
        <jats:p/>
      </jats:caption>
    </jats:fig>
    <jats:table-wrap id="tab_0">
      <jats:label/>
      <jats:caption>
        <jats:title/>
        <jats:p>.1a) royalsocietypublishing.org/journal/rspa Proc. R. Soc. A 478: 20220317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
      <jats:table>
        <jats:thead>
          <jats:tr/>
        </jats:thead>
        <jats:tbody/>
      </jats:table>
    </jats:table-wrap>
    <jats:table-wrap id="tab_1">
      <jats:label>Table 2 .</jats:label>
      <jats:caption>
        <jats:title>Table 2 .</jats:title>
        <jats:p>. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Comparison of intensities at the exit plane for the systems in figure</jats:p>
      </jats:caption>
      <jats:table>
        <jats:thead>
          <jats:tr>
            <jats:th>1, 0)</jats:th>
            <jats:th>0.00012040357</jats:th>
            <jats:th>0.00330035539</jats:th>
            <jats:th>0.00004461419</jats:th>
            <jats:th>0.00230899563</jats:th>
          </jats:tr>
        </jats:thead>
        <jats:tbody>
          <jats:tr>
            <jats:td>(0, 0)</jats:td>
            <jats:td>0.15359073146</jats:td>
            <jats:td>0.15209371434</jats:td>
            <jats:td>0.15359073146</jats:td>
            <jats:td>0.15209371434</jats:td>
          </jats:tr>
          <jats:tr>
            <jats:td>(1, 0)</jats:td>
            <jats:td>0.84539398729</jats:td>
            <jats:td>0.84447759832</jats:td>
            <jats:td>0.84447759832</jats:td>
            <jats:td>0.84539398729</jats:td>
          </jats:tr>
          <jats:tr>
            <jats:td>(2, 0)</jats:td>
            <jats:td>0.00089487769</jats:td>
            <jats:td>0.00012833195</jats:td>
            <jats:td>0.00188705604</jats:td>
            <jats:td>0.00020330274</jats:td>
          </jats:tr>
        </jats:tbody>
      </jats:table>
    </jats:table-wrap>
    <jats:table-wrap id="tab_2">
      <jats:label/>
      <jats:caption>
        <jats:title/>
        <jats:p>. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
      <jats:table>
        <jats:thead>
          <jats:tr>
            <jats:th>0)</jats:th>
            <jats:th>0.15309988945</jats:th>
            <jats:th>0.15309988945</jats:th>
            <jats:th>0.15309988945</jats:th>
            <jats:th>0.15309988945</jats:th>
          </jats:tr>
        </jats:thead>
        <jats:tbody>
          <jats:tr>
            <jats:td>(1, 0)</jats:td>
            <jats:td>0.84690011055</jats:td>
            <jats:td>0.84690011055</jats:td>
            <jats:td>0.84690011055</jats:td>
            <jats:td>0.84690011055</jats:td>
          </jats:tr>
        </jats:tbody>
      </jats:table>
    </jats:table-wrap>
    <jats:table-wrap id="tab_3">
      <jats:label>Table 3 .</jats:label>
      <jats:caption>
        <jats:title>Table 3 .</jats:title>
        <jats:p>Comparison of intensities at the exit plane for the four-beam model in figure 13. Midplane reflection for bright field still holds. Midplane reflection plus sign change holds for dark field. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .</jats:p>
      </jats:caption>
      <jats:table>
        <jats:thead>
          <jats:tr>
            <jats:th>g</jats:th>
            <jats:th>(A, F(z))</jats:th>
            <jats:th>(A, &#8722;F(z))</jats:th>
            <jats:th>(A, F(z  *  &#8722; z))</jats:th>
            <jats:th>(A, &#8722;F(z  *  &#8722; z))</jats:th>
          </jats:tr>
        </jats:thead>
        <jats:tbody>
          <jats:tr>
            <jats:td>(&#8722;1, 0)</jats:td>
            <jats:td>0.00002047358</jats:td>
            <jats:td>0.00095395052</jats:td>
            <jats:td>0.00000913034</jats:td>
            <jats:td>0.00059545046</jats:td>
          </jats:tr>
          <jats:tr>
            <jats:td>(0, 0)</jats:td>
            <jats:td>0.08282721744</jats:td>
            <jats:td>0.02081205071</jats:td>
            <jats:td>0.08282721744</jats:td>
            <jats:td>0.02081205071</jats:td>
          </jats:tr>
          <jats:tr>
            <jats:td>(1, 0)</jats:td>
            <jats:td>0.26635428539</jats:td>
            <jats:td>0.25940647707</jats:td>
            <jats:td>0.25940647707</jats:td>
            <jats:td>0.26635428539</jats:td>
          </jats:tr>
        </jats:tbody>
      </jats:table>
    </jats:table-wrap>
  </jats:body>
  <jats:back>
    <jats:sec>
      <jats:sec>
        <jats:p>Acknowledgements. The authors are grateful to Tore and Laura Niermann for helpful discussions and to Timo Streckenbach for his support in creating the 3D graphics, cross-sections and the line scans using WIAS-gltools, which have been used in figures 11 and 12.</jats:p>
      </jats:sec>
    </jats:sec>
    <jats:sec>
      <jats:sec>Mathematical treatment of the symmetries<jats:p>We now provide the mathematics underlying the symmetry considerations for the solutions of the DHW equations. For this, we use the general m-beam model in the Hermitian form derived in <jats:xref ref-type="bibr" rid="">(3.1)</jats:xref>. To study the symmetries, we consider the system matrix A = V + &#931; and the strain function F as data specified to lie in the following spaces:</jats:p><jats:p>The typical measurements for generating TEM images do not involve all components of &#966;(z * ) &#8712; C m at the exit plane, but only the intensity of beam g ap &#8712; &#923; * m selected by the objective aperture, see figure 1a, namely I ap (x i , y j ) = |&#966; g ap (z * ; x i , y j )| 2 . As mentioned above a special mathematical role is played by the so-called bright field which is given by the choice g ap = 0. The reason for this is the double appearance of the vector e 0 , namely (i) in the initial condition &#966;(0) = e 0 and (ii) in the exit measurement</jats:p><jats:p>The double appearance of e 0 can even be used for symmetries in the dark field where g ap = 0 under the assumption that we have a two-beam model, i.e. &#923; * 2 = {0, g } and g ap = g . In this case, we can exploit the Hermitian structure of ( 3 To analyse these two symmetries and their joint effect for both m-beam models and the twobeam model, we consider more general classes of transformations involving also changes of A = V + &#931; and not only of the strain-related part F. This will uncover the proper mathematical structure of the symmetries and show why the case m = 2 is special. For this, we define two types of symmetries. We call the replacement a weak symmetry if we have</jats:p><jats:p>Throughout we will denote by U A+F (z) &#8712; C m&#215;m the evolution operator solving .</jats:p><jats:p>As A + F(z) is Hermitian for all z, the evolution operators U A+F are unitary, i.e.</jats:p><jats:p>In particular, this implies that the Euclidean norm |&#966;| = ( g&#8712;&#923; * m |&#966; g | 2 ) 1/2 is preserved for solutions &#966;(z) of (3.1). Of course, we have a general transformation rule for arbitrary unitary matrices</jats:p><jats:p>The first result concerns the set of all strong symmetries.  </jats:p><jats:p>Thus, U satisfies the same ODE as U &#8722;A&#8722;F&#8226;R , but the initial conditions are different. This observation, U(z * ) = I, and the unitarity relation (4.1) imply</jats:p><jats:p>Restricting to the case z = z * gives the desired assertion.</jats:p><jats:p>Of course, all compositions of a strong symmetry with a weak symmetry again provides a weak symmetry. Hence, combining the above lemma with proposition 4.2 gives the following result that relies on the double occurrence of e 0 in the definition of weak symmetries. </jats:p><jats:p>Proof. We use that weak symmetry is defined in terms of</jats:p><jats:p>where e 0 occurs as initial condition as well as test vector at z = z * .</jats:p><jats:p>For (W1), we exploit the relation (4.3) from the previous lemma, which gives This immediately implies | &#966; 0 (z * )| = |&#966; 0 (z * )| as desired. For (W2), we simply apply the complex conjugation (S2) and use that F is real-valued. For (W3), we start from (W2) and use m = 2 to replace A by A using (S3) as for corollary 4.3.</jats:p><jats:p>From symmetry (W2) follows that under the assumption that all relevant Fourier coefficients of the scattering potential U g are real the midplane reflection symmetry is also valid for the general m-beam model and not only for the two-beam approximation. This property may be satisfied for specific crystal structures. One example is centrosymmetric materials, such as Al, Cu and Au obeying a face-centred cubic lattice, see <jats:xref ref-type="bibr" rid="b13">[14,</jats:xref><jats:xref ref-type="bibr" rid="">Ch. 6.5]</jats:xref>.</jats:p><jats:p>Our last result concerns a symmetry in the two-beam model when one changes the sign of the excitation error s g . This is relevant in experimental observations, where s g can easily be varied, cf. <jats:xref ref-type="bibr" rid="b24">[25]</jats:xref>. In particular, we refer to the figures 5-7.</jats:p></jats:sec>
      <jats:sec>Corollary 4.6 (Excitation-error symmetry for<jats:p>Herm with V 00 = V g g . Then, the transformation</jats:p><jats:p>Proof. The result follows by combining corollary 4.3 and part (W3) of corollary 4.5. More precisely, we first observe U A+F = U V+(&#931;+F) . Applying corollary 4.3 with (A, F) replaced by (V, &#931;+F) yields that (A 1 , F 1 ) = (V, &#8722;(&#931;+F)) is a strong symmetry. Combining this with part (W3) of corollary 4.5 shows that (A 2 , F 2 ) = (V, &#8722;(&#931;+F)&#8226;R) is a weak symmetry.</jats:p><jats:p>To conclude we observe that </jats:p></jats:sec>
      <jats:sec>Symmetries in the presence of absorption<jats:p>To expand the theory developed in this paper to the case of absorption, one needs to add an imaginary part to the Fourier coefficients of the potential. This would make the matrix V &#8712; C m&#215;m herm non-Hermitian. This Hermitian structure was important for two reasons. First, due to this structure, we had conservation of the Euclidean norm. This allowed us to conclude that if |&#966; 0 (z * )| is preserved by a symmetry, then so is |&#966; g (z * )|, as stated in &#167;4. Since the theory is developed for the bright field, we can say that even if we expand it to the non-Hermitian case, we cannot make conclusions about the dark field as well. The second reason the Hermitian structure is important is the fact that it made the evolution operator U A+F unitary, see <jats:xref ref-type="bibr" rid="">(4.1)</jats:xref>. Without this unitarity some of the symmetries will not survive. In this section, we examine which symmetries survive absorption.</jats:p><jats:p>To model absorption, one introduces a positive semidefinite Hermitian matrix D &#8712; C m&#215;m such that the m-beam DHW equation takes the form</jats:p><jats:p>(</jats:p><jats:p>The associated evolutionary operator is again denoted by U A+iD+F (z) &#8712; C m&#215;m , which is no longer unitary because the generator i(A+iD+F) is no longer skew adjoint.</jats:p><jats:p>If the matrix D = diag(d) is diagonal, with the same value on all components, it would influence all beams in the same way. This is also known as small angle absorption. In the end, this would simply reduce the overall intensity of the resulting images, and would only play a role when experimental and simulated images are compared on an absolute scale. In this case, the theory developed in &#167;4 still holds, since we can transform the system (5.1) to a Hermitian form by using &#966;(z) = e dz &#966;(z). If the matrix D does not have this special form, but is a full matrix, which is the case with the so known wide-angle absorption, then not all symmetries survive.</jats:p><jats:p>The generalization of the unitary property (4.1) is the identity</jats:p><jats:p>where we used our standing assumption A = A * and D = D * in the last identity. To show the first identity, simply take the z-derivative on both sides and observe that both satisfy the same equation U = &#8722;iU(A&#8722;iD+F) and U(0) = I. Applying complex conjugation and the flip operation R(z) = z * &#8722; z, we obtain the following result.</jats:p></jats:sec>
      <jats:sec>Theorem 5.1. (Symmetry with absorption).<jats:p>For A = A * and D = D * , we have the weak symmetries:</jats:p><jats:p>Proof. We first observe that complex conjugation (as in (S2)) leads to the identity</jats:p></jats:sec>
      <jats:sec>.3)<jats:p>For understanding the flip F&#8226;R, we generalize (4.3) by following the proof of lemma 4.4:</jats:p><jats:p>where we used (5.2) for the middle identity and (5. </jats:p></jats:sec>
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