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In contrast to Dr. Deisenhofer’s beautiful lecture mine concerned as it is
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with methods of crystal emulative structure determination is of necessity highly theoretical.
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However I hope to show by my lecture today that it doesn’t follow that it must be incomprehensible as well.
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The first slide shows in a schematic way the fundamental experiment
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which was done by Friedrich and Knipping in the year 1912 at the suggestion of Max von Laue.
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It shows very briefly that x-rays are scattered by crystals and the scattered x-rays
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if caused to strike a photographic plate will darken the photographic plate at the points
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where the scattered rays strike the plate.
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And the amount of blackening on the photographic plate depends upon the intensity of the corresponding scattered x-ray.
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Because of the consequences of this experiment, because this experiment was the key
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which unlocked during the course of the next seventy-five years, the mystery of molecular structures,
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this experiment must be regarded as a fundamental landmark experiment of this century.
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The slide on the right shows a typical molecular structure.
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It’s the structure of decaborane which consists of ten borane atoms and fourteen hydrogen atoms.
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The borane atoms are located at the vertices of a regular icosahedron.
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I’ve shown these two slides together because I wish to stress the mathematical equivalence between the diffraction pattern,
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which is to say the arrangement and the intensities of the x-rays scattered by crystal and the molecular structure on the right,
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the information content of this defraction pattern and the information content of the molecular structure,
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which is to say the arrangement of the atoms in the molecule,
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the information content of these two slides is precisely the same.
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If one knows the molecular structure shown on the right one
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can calculate unambiguously completely the nature of the defraction pattern shown on the left.
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Which is to say the directions and intensities of the x-rays scattered by the crystal
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which consists of the molecules shown on the right.
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And conversely if one has done the scattering experiment and has measured the directions
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and the intensities of the x-rays scattered by the crystal
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then the molecular structure shown on the right is in fact uniquely determined.
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What I would like to describe next is precisely
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what the relationship is between the structure shown on the right as an example and the defraction pattern shown on the left.
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Here we have an equation which I hope doesn’t frighten you.
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On the left hand side is simply the electron density function
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which is simply a function of the position vector r and it gives us the number of electrons per unit volume.
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And on the right hand side is the formula which enables us to calculate the electron density function Rho(r).
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If we knew all these quantities on the right,
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those of you who are familiar with the elements of x-ray crystallography
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or even with most elementary mathematics,
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know that this function on the right is simply a Fourier series, a triple Fourier series.
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The scaling parameter v is not important for our present purpose.
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This expression on the right is a sum taken over all triples of integers, so called reciprocal lattice vectors.
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And on the right hand side we have simply a Fourier series expressed in pure exponential form.
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We have the magnitudes or the non-negative numbers which are the coefficient of the exponential function.
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We have the reciprocal lattice vector H and a triple of integers.
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We have an arbitrary position vector R, which has also three components.
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This is simply the scale of product.
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And here we have the phases of the structure factors,
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the magnitudes of which are shown here as the co-efficient of the exponential function.
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If we knew everything that we need to know on the right,
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which is to say these magnitudes and these phases, then we could calculate this function.
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This triple Fourier series as a function of the position vector R.
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And therefore we could calculate the electron density function Rho(r),
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read off the positions of the maxima of the electron density function
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and that would give us the positions of the atoms or in other words the crystal structures.
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The problem which was alluded to just a few minutes ago, is
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that although these magnitudes are obtainable directly from the defraction experiment, from the measured intensities.
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The intensity of the x-ray scattered in the direction labelled by the reciprocal lattice vector H,
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although these magnitudes are directly obtainable from the experiment these phases are lost in the defraction experiment.
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And so although from the very earliest years
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because of the known relationship between the fraction patterns and crystal structures it was felt
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that the fraction experiment did in fact unlock the key to the determination of crystal and molecular structures.
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Because these phases were missing, because they were lost in the defraction experiment it was thought
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that after all what could be observed in the defraction experiment was in fact
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not sufficient to determine unique crystal structures.
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The argument that was used was a very simple one and a very compelling one.
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It was simply that we could use for these co-efficient, for these magnitudes the quantities
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which were directly obtainable from the experiment.
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Which is to say the intensities of the scattered x-rays in calculating this function.
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And we could put in for the lost phases, the missing phases arbitrary values.
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And depending upon which values we put in for these phases we would get different electron density functions.
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And therefore different crystal and molecular structures, all however consistent with what could be measured
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which is to say the intensities of the x-rays scattered by the crystal.
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And it was therefore believed for some forty years after this experiment was done, it was therefore believed that
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the fraction experiment could not even on principle lead to unique crystal and molecular structures.
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Now there was a flaw in this argument, as simple as it appears to be
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and as overwhelming as the logic appears to be there was a fatal flaw in it.
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And that was that one could not use arbitrary values for these phases.
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For the simple reason that if one were to do that, one would obtain electron density functions
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which were not consistent with what was known about crystal structures.
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For example one of the properties of the electron density function which must be satisfied by every crystal is
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that the electron density function must be non-negative everywhere.
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After all the electron density function Rho(r) gives us the number of electrons per unit volume.
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And from its very definition therefore it must be non-negative everywhere.
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On the other hand for a given set of known magnitudes F sub H,
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if one used arbitrary values for these phases in general one would obtain electron density functions
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which were negative somewhere, for some values of the position vectors R and therefore would not be permitted.
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So that the known non-negativity of the electron density function restricts the possible values which the phases may have.
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In fact restricts rather severely the possible values which the phases may have.
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And the non-negativity condition alone, the non-negativity restriction on the electron density function is in fact
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sufficient to enable one to solve some rather simple crystal structures.
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However the restrictions on the phases which are obtainable in this way are of a rather complicated nature.
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And therefore the non-negativity conditional law has proven to be not very useful in the actual applications.
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A much more useful restriction may be summarised in the one word – atomicity.
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Since molecules consist of atoms it follows that the electron density function is not only non-negative everywhere
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but must take on rather large positive values at the positions of the atoms.
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And must drop down to very small values at positions in between the atoms.
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And this requirement of atomicity, this property of the electron density function turns out to be a severely restrictive one.
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And in general at least for small molecules, say in molecules consisting of a hundred or a hundred and fifty non-hydrogen atoms,
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this requirement is sufficiently restrictive
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that the measured intensities in the x-ray defraction experiment is in general enough.
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In fact, in general far more than enough to determine unique crystal structures.
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I should also mention before I leave this slide is that we should carry with us the fact that if we know these magnitudes
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which as I said are obtainable directly from the measured intensities in the defraction experiment and
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if somehow or other we can find these phases then by calculating this Fourier series on the right
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we can calculate the electron density function Rho(r) and therefore determine the crystal and molecular structure.
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In the next slide I want to show that not only do the crystals structure factors,
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which is to say magnitudes and phases of the crystal structure factors, determine crystal structures,
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but that the converse is also true.
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In order to exploit the atomicity property of real crystal structures, it turns out we have to make a small change in these F’s.
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We replace theses structure factors by what is called the normalised crystal factors E,
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shown on this slide, and defined in this way.
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Again we have a magnitude E sub H which is directly obtainable from the measured intensities in the diffraction experiment.
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We have the missing phases Phi(H), and this complex number may be represented in polar from in this way.
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The product of the magnitude times the pure exponential function e ^i * Phi(H).
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Where this Phi(H) is of course the phase of the normalised structured factor E(H).
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And what this equation tells us is that if we know the atomic position vector r(J),
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the r(J) now represents an atomic position vector labelled by the index J.
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We have here a sum of a linear combination of exponential functions taken over all the N atoms,
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in the unit cell of the crystal.
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On the right hand side we have the atomic number of the atom labelled J.
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We have the atomic position vector r(J) in the atom labelled J.
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H is a fixed reciprocal lattice vector and ordered triple of integers.
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Sigma sub 2 is not very important.
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For our present purpose it is simply the sum
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of the squares of the atomic numbers of all the atoms in the unit cell of the crystal.
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What this equation tells us then is that
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if we know atomic position vectors we can calculate magnitudes and phases of the normalised structured factors E(H).
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This slide tells us that the converse is true.
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If we know magnitudes and phases by calculating the Fourier series we can get the electron density function
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and therefore the crystal structure.
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This tells us that the converse of that statement is also true.
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If we know atomic position vectors we can calculate essentially the co-efficient of this Fourier series.
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However, I’ve already suggested that because of the requirement of atomicity
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that measured magnitudes alone provide a very strong restriction on the values of the phases
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and in fact require that the phases have unique values.
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But what that means of course is that if we have measured a large number of intensities,
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therefore magnitudes E(H), somehow or other these phases are determined.
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And now our problem is, in fact the solution of the phase problem requires
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that using only known magnitudes E(H) how does one calculate the unknown phases Phi(H)?
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Now, this equation tells us actually that right away if we examine it closely we see that we have a complication.
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And the complication comes from the fact that the position vectors r(J) are not uniquely determined by the crystal structure.
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Because if we have a given crystal, then the position vectors or the atomic position vectors r(J) depend
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not only on the crystal structure but depend also on the choice of origin.
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If we move the origin around in the unit cell of the crystal and in this way do not change the crystal structure,
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we change the value of this function and therefore
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we change the value of the normalised structure factor E(H) on the left hand side.
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What this suggests then is that these normalised structured factors,
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which is to say these magnitudes and these phases depend not only on the crystal structure but also on the choice of origin.
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And this of course causes a complication.
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As it turns out the crystal structure does determine unique values for these magnitudes no matter where the origin may be chosen.
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But the values of the individual phases do in fact depend not only on the crystal structure but also on the choice of origin.
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As you can see that complicates our problem.
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Because if the phases are not uniquely determined by the crystal structure,
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if the phases are not uniquely determined by the crystal structure
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then certainly they are not uniquely determined by measured intensities alone.
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Or by the known values of these magnitudes.
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Because we have somehow or other to find unique values for the individual phases
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we have to have a mechanism for specifying the origin.
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So what's called for before we can even hope to solve the phase problem,
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to calculate the values of the phases for given values of these magnitudes.
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Before we can even hope to do this we have to, in the process which leads from known magnitudes to unknown phases,
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we have to incorporate a recipe or a mechanism for origin fixing.
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Now that as I say introduces a complication which is not too difficult to resolve.
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The way to resolve it is to separate out from the contributions to the value of a given phase.
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There are as I indicated two kinds of contributions to the value of an individual phase.
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The contribution which comes from the crystal structure.
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And the contribution which comes from the choice of origin.
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And the first thing that has to be done is to separate out these two contributions.
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So we can decide once and for all what part of the value of the phase depends upon the crystal structure
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and what part comes from the choice of origin.
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And the best way to do that is to observe something that I don’t,
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which is not possible for me to show where without causing a lot of confusion.
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The best way to do that is to introduce the idea of what is called the structure and variant.
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Which is to say certain special linear combinations of the phases
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which have the remarkable property that their values are in fact uniquely determined by the crystal structure,
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no matter what the origin may be.
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So the first thing to do then is of course is to identify these very special linear combinations of the phases.
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The so-called structure invariance
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and I would like to show on the next slide a typical example of such a special linear combination of the phases.
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The three phase structure invariant
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the so-called triplet is simply a linear combination of three phases, Phi(H) + Phi(K) + Phi(L) where H + K + L = 0.
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If this condition is satisfied this linear combination of three phases as a structure invariant and it has the property
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that its value is uniquely determined by the crystal structure no matter where the origin may be chosen.
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Now, you can see the fundamental importance of these structure invariance
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because it’s only linear combinations of this kind whose values we can hope to estimate in terms of measured intensities alone.
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We’ve already seen that measured intensities alone do not determine unique values for the individual phases,
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because the values of the phases depend also on the choice of origin.
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But measured intensities alone do determine the values of these special linear combinations of the phases.
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So the phase problem then is really broken down into two parts.
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First to use the measured intensities, to provide estimates of this structure invariance,
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these special linear combinations of the phases, and once the values of a sufficiently large number of these structure
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and variance are known, then we can hope to calculate the values of the individual phases.
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Provided that in the process leading from the estimated values of a large number of these structure invariance
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to the values of the individual phases we incorporate a mechanism for origin fixing.
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So these structure invariance therefore play a fundamental role in the solution of the phase problem.
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They serve to link the observed magnitudes, these quantities here with the desired values of the individual phases.
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Because we can hope to estimate these linear combinations of the phases in terms of these measured magnitudes.
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And once we have estimated a sufficiently large number of these we can hope to calculate the values of the individual phases.
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Now, I have to indicate briefly how one estimates the values of these, not only this structure invariance but others as well.
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In order to do this the method which was introduced is a probabilistic one.
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Because of the large number of intensities which are available from experiment a probabilistic approach to this problem,
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to the solution of the phase problem is strongly suggested.
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And the strategy, the device which is used is simply to replace these position vectors R
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or the atomic position vectors r(J) replace them by random variables
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which are assumed to be uniformly and independently distributed.
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This is using the language of mathematical probability.
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In every day terms what we are doing is assuming
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that all positions of the atoms in the crystal are equally likely that no positions are preferred over any other.
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And that amounts the same then that the atomic positions vectors r(J) are assumed
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to be a primitive random variables uniformly and independently distributed.
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Now, once we do that then the, (could I have the previous slide on the right hand side please).
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If we assume these atomic position vectors are random variables, uniformly and independently distributed
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then the right hand side becomes a function of random variables.
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The left hand side is also a function of random variables and is therefore itself a random variable.
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And we can calculate by standard techniques its probability distribution, if we choose to do that.
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However, its probability distribution will not be useful to us
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but what will be more useful to us is the probability distribution of the structure invariance.
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These linear combinations of the phases.
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(If I could have the next slide on the right hand side again, no,
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the other direction you are going the wrong direction, the one before this).
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Okay this is a structure invariant.
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What we are asking for now is the probability distribution of this structure invariant
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because we know from the discussion that I’ve already given
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that it’s only the values of these special linear combinations of the phases
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which we can hope to estimate in terms of measured magnitudes alone.
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Therefore, what we are looking for is the probability distribution of a structure invariant
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in the hope that the probability distribution will give us some information about its value.
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In particular we not only are looking for the probability distribution of this structure invariant
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but we are looking for the conditional probability distribution of this structure invariant
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assuming as known a certain set of magnitudes.
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Because after all the magnitudes or intensities are known.
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This is what is given to us from the defraction experiment.
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And we want to use that information in order to estimate the values of these structure invariance.
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What that calls for then is the conditional probability distribution of a structure invariant given a certain set of magnitudes.
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And on this slide if I can have the next slide we’ll see the formula
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which tells us what is the probability distribution of a structure invariant.
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Here we have a three phase structure invariant, Phi(H) + Phi(K) + Phi(HK).
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I’ve written it in this form rather than this form
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where we clearly show explicitly that the sum of the three indices, H+K-H-K adds up to zero.
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So that this condition is satisfied.
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This triplet then is a structure invariant
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and we can ask for its conditional probability distribution assuming as known these three magnitudes.
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And these three magnitudes are of course known from the defraction experiment.
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I have written down the formula only in the case that all the atoms are identical and that we have N of them in the unit cell.
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It isn’t necessary to specialise it in this way but I’ve done so in order to simplify the formulas.
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This gives us the conditional probability distribution then of the three phase structure invariant,
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the triplet, assuming as known three magnitudes.
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And this is the analytic formula and in a few seconds I’ll show you what it looks like by means of the next slide.
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Right now what I would like to emphasise, is that we can calculate any parameters of this distribution that we chose
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and in particular we can calculate the expected value or the average value of the cosine of this triplet.
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This is the formula for it, it turns out to be a ratio of these two vessel functions.
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It’s not important for us to know what they look like at the moment, it’s something,
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these functions are known functions and I’ve abbreviated it by writing T(H).
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And the important thing, the only thing we should carry away with us is that the average value of the cosine,
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of the triplet can be calculated from the distribution.
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This is what it’s equal to, it depends only on known quantities measured magnitudes
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and the number of atoms and in the unit cell.
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And it turns out always to be greater than zero.
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The next slide, on this side shows us as picture of what that distribution looks like.
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And we can clearly see when the parameter A shown on the previous slide is about seven tenths the distribution looks like this.
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It goes from -180° to +180° and what the distribution tells us is that the values of the triplet,
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of the three phase structure invariant tends to cluster around zero.
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There are more values of this triplet in the neighbourhood of zero than there are let’s say in the neighbourhood of 180°.
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So the distribution then, the known distribution
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which we can calculate carries information about the possible values of these triplets.
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And in fact it enables us to estimate the triplet, the estimate in this simple case would be
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that this triplet is probably approximately equal to zero.
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But in this case when the parameter A is only about 7/10ths the estimate is not a very good one
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because values near 180° were much, well not very frequent are still possible.
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It’s still possible to get a substantial number of values of the triplet in the neighbourhood of 180°,
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when the parameter A is only about 7/10ths.
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However, when the parameter A is larger as shown on this next slide,
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when the parameter A is 2.3 or so the distribution looks like this.
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Again values of the structure invariance in the neighbourhood of zero are much more common now than in the neighbourhood of 180°.
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So the estimate of the triplet in this favourable case
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when the parameter A is about 2.3 the zero estimate of the triplet is a particularly good one in this favourable case.
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When the parameter A is large, bigger than two or so, then we get a very reliable estimate of the triplet.
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And if we can estimate a sufficiently large number of them
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as I’ve already indicated we can then hope to calculate the values of the individual phases provided once again
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that in the process leading from estimated values of the structure invariance to the values of the individual phases
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we incorporate a mechanism for origin fixing.
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What I would like to do next is show another class of structure invariance.
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The so called quartets which are linear combinations of four phases now, Phi(H)+Phi(K)+Phi(L)+Phi(M) where H+K+L+M is equal to zero.
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This is very analogous to the triplet that I showed on an earlier slide.
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It’s a linear combination of four phases now, instead of just three phases.
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Just as we did with the triplets so we can do with the quartets.
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We can find the conditional probability distribution of the quartet assuming as known certain magnitudes.
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But there is an important difference between the quartet and the triplet which I showed earlier.
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The distribution actually has a very similar functional form.
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It’s exactly the same as for the triplet but the parameter BLMN is an abbreviation for this,
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well I see I didn’t write the quartet on this slide.
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I suppose because there wasn’t enough room.
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But BLMN is simply an abbreviation, no it’s not an abbreviation.
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BLMN is given by this, Phi represents the quartet.
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What this shows us is that here too we can calculate the conditional probability distribution of the quartet now.
294
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Assuming as known not three magnitudes as we had in the case of the triplet but seven magnitudes - EL, EM, EN and this.
295
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These are the magnitudes corresponding to these indices and three other magnitudes, so-called cross terms.
296
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It’s not important to know what these magnitudes are it’s sufficient to know that the single parameter
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on which the distribution depends can be calculated from seven known magnitudes.
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Magnitudes obtained from the defraction experiment.
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The important difference though between the quartet distribution and the triplet distribution is
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that the parameter B now on which the distribution depends may be positive or negative
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depending upon the sine of this expression embraces.
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If these three cross stems are large then this term embraces will be positive and the parameter B will be positive.
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And the distribution will have a maximum around zero, as we had for the case of the triplets.
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But if these three cross terms are small the expression embraces is negative
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and this distribution instead of having a maximum at zero will have a maximum at 180°,
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so that the estimate of the quartet in that case and it’s a case which can be calculated in advance,
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the estimate of the quartet becomes not zero but 180°.
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However just as in the case of the triplet we can calculate again the expected value of the cosine of the quartet,
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again it turns out to be the ratio of vessel functions
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because it has the same functional form as the distribution for the triplet.
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And we call it for abbreviation T(LMN) but now T may be positive or maybe negative
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depending upon whether this parameter B is positive or negative.
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And we know in advance which it will be.
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So the next slide on this side will show us what the distribution looks like
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in the case that the parameter B is negative.
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I’ve shown it for the case -7/10ths.
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Now in sharp contrast to what the situation is for the triplets the distribution has a maximum at 180°.
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So that the estimate for the quartet instead of zero will now be 180°.
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But it will not be a very reliable estimate in the case that B has such a small value
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because as you can see values of the quartet in the neighbourhood of zero while less likely
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than values in the neighbourhood of 180° still will occur.
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What's needed then is a distribution which is sharper than the one shown here and that will happen
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when the value of the parameter B is say -1.2.
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In that case we again have a peak at 180° so that the estimate of 180° is rather reliable
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but certainly not as reliable as we would like it to be.
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Now the traditional techniques of direct methods which have proven to be useful in the case
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that we are determining structures of so-called small molecules,
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molecules of less than a 100 or 150 non-hydrogen atoms in the molecule,
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those can be solved in a rather routine way using estimated values of the structure invariance.
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The reason that the methods eventually fail when the structure becomes very large is
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that we can no longer obtain distributions which give us reliable estimates of the structure invariance.
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As the structures become more and more complex there are very few distributions
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which have a sharp peak, whether at zero or 180°.
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And therefore there are very few structure invariance, whether they are triplets or quartets,
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whose values we can reliably estimate.
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And therefore eventually the methods fail.
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The one point which should be emphasised however, and which I have emphasised on the next slide on the right hand side,
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is what I’ve called the fundamental principle of direct methods.
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And this simply states that the structure invariance link the observed magnitudes E with the desired phases Phi.
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By this I mean, this is what the traditional direct methods tell us, the direct methods for solving the phase problem,
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is that if we can estimate from measured intensities alone a sufficiently large number of these structure invariance
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whether they are triplets or quartets or whatever.
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Then we can hope to use those estimates to go from, which are after all determined by the measured magnitudes,
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we can use those estimates to derive a value or to calculate the values of the individual phases provided
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that in the process leading from estimates of the structure invariance to the values of the individual phases
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we incorporate a mechanism for origin fixing.
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For this reason the structure invariance serve to link measured magnitudes, known magnitudes with unknown phases.
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But they require that we estimate fairly reliably the values of a large number of structure invariance.
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Well we can’t do that for very complex structures,
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for very complex structures we don’t get a sufficiently large number of probability distributions
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which yield reliable estimates for the structure invariance.
352
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So we have to do something else,
353
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when we try to strengthen the traditional direct methods to be useful for much more complicated structures.
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Say structures in the neighbourhood of three or four or five hundred or even more non-hydrogen atoms in the molecule.
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We have to do better than we have done in the past.
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But again we use the fundamental principle of direct methods.
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We use again the fact that it is the structure invariance which link these measured magnitudes with unknown phases,
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even though we can no longer estimate reliably the values of a large number of these structure invariance
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in the case of very complex molecular structures.
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We can always calculate reliably these conditional probability distributions.
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So just as for the traditional direct methods, the structure invariance link known magnitudes E with unknown phases Phi.
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Now they all again link these magnitudes with these phases but the property of these structure invariance
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which we surely know is their conditional probability distributions. That we surely know.
364
00:38:07.222 --> 00:38:12.031
And so we can try to solve the following problem.
365
00:38:12.181 --> 00:38:21.511
We can try to estimate the values of a large number of individual phases, say several hundred,
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three-hundred, four-hundred or five-hundred individual phases in one block, at one stroke.
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By requiring that the values have the property that when we construct from those phases,
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several hundred phases all the structure invariance which we can construct.
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Let’s say all the triplets and all the quartets that those structure invariance have a distribution of values
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then which agrees with theoretical distributions.
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We know their theoretical distributions and we require that the individual phases have such values
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that when we generate all the triplets and all the quartets which we can that their distributions,
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their conditional distributions assuming as known certain magnitudes, agree with the known theoretical distributions.
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The one thing we know for sure is that even for complex structures
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we know the probability distributions of the structure invariance.
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We may not be able to use these distributions to give us reliable estimates of the structure invariance
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but we know their distributions.
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And we have from this point of view a tremendous amount of over-determination
379
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because from a set of say three-hundred phases or so we can generate
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in any given case some tens of thousands of triplets and hundreds of thousands of quartets.
381
00:39:53.471 --> 00:39:58.471
And we know of course the distributions of all these triplets and all these quartets.
382
00:39:58.591 --> 00:40:05.142
And we can ask the question, whether we can answer it or not is another question, but we can certainly ask the question.
383
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What must be the values of the individual phases so that when we generate these enormous numbers of structure invariance,
384
00:40:13.952 --> 00:40:21.033
perhaps millions of them in any give case, that they have distributions of values
385
00:40:21.123 --> 00:40:28.232
which agree with their known theoretical distribution? If I may use that term.
386
00:40:28.332 --> 00:40:31.632
So that’s the problem that we try to answer now
387
00:40:31.712 --> 00:40:38.012
and I hope in the next few minutes to tell you what the answer to that question is.
388
00:40:38.383 --> 00:40:44.222
On this slide I just have just a brief summary of what I’ve already shown.
389
00:40:44.352 --> 00:40:50.272
I’ve already shown that for the triplets, Phi(HK) and for the quartets,
390
00:40:50.352 --> 00:40:54.853
Phi(LMN) we can calculate these parameters of the distribution.
391
00:40:54.853 --> 00:41:00.642
For example the expected value of the cosine for the triplet, I already showed you the formula for that.
392
00:41:00.742 --> 00:41:07.133
We can also calculate what I’ve called the weight which is the reciprocal of the variance for the cosine.
393
00:41:07.273 --> 00:41:12.893
I haven’t shown you the formula for that but it’s easily calculated once we know the distribution.
394
00:41:12.993 --> 00:41:17.472
And we can do exactly the same thing for the quartet, we can calculate as I’ve already shown you
395
00:41:17.591 --> 00:41:26.333
what the expected value of the cosine of the quartet is and we can also calculate the variance of the cosine of the quartet.
396
00:41:26.433 --> 00:41:32.403
So we can assume that these are known parameters of the distributions that we are concerned with.
397
00:41:32.513 --> 00:41:36.432
I should mention one other thing that I haven’t stressed.
398
00:41:36.521 --> 00:41:41.492
That is because from a set of phases, let’s say three or four hundred phases
399
00:41:41.581 --> 00:41:45.512
we can generate hundreds of thousands of invariance it follows
400
00:41:45.652 --> 00:41:55.013
that their must exist a very large number of identities which the invariance must satisfy.
401
00:41:55.262 --> 00:41:57.943
The very fact of the redundancy here,
402
00:41:58.022 --> 00:42:03.512
the fact that we can generate hundreds of thousands of invariance from just a few hundred phases means
403
00:42:03.612 --> 00:42:11.112
that the invariance must of necessity satisfy a very large number of identities.
404
00:42:11.222 --> 00:42:18.192
We shall make important use of that over determination property of this method.
405
00:42:18.352 --> 00:42:28.042
On this slide I’ve shown you what the mathematical formulation is of the requirement that the structure invariance,
406
00:42:28.212 --> 00:42:34.013
these hundreds and thousands of them which are generated by a set of several hundred phases,
407
00:42:34.143 --> 00:42:41.892
the requirement that those structure invariance obey their known theoretical probability distribution.
408
00:42:41.982 --> 00:42:47.073
The requirement is very simple, here we have the triplets.
409
00:42:47.203 --> 00:42:52.623
Here we have the quartets, incidentally in this work it’s absolutely essential
410
00:42:52.713 --> 00:42:55.773
that we use the quartets in addition to the triplets.
411
00:42:55.863 --> 00:43:02.212
Although the traditional direct method depends mostly on the triplets and very little on the quartets if at all.
412
00:43:02.332 --> 00:43:07.042
For the present formulation we need to have both triplets and quartets,
413
00:43:07.122 --> 00:43:12.043
because of the fact that with the triplets the only estimates of the triplets
414
00:43:12.133 --> 00:43:17.193
that we can obtain are the zero estimates where the cosines are positive.
415
00:43:17.302 --> 00:43:27.793
But for the quartets where the quartets may have the value, most probable values may be 180° the cosines are negative
416
00:43:27.913 --> 00:43:32.002
and we need to use those quartets.
417
00:43:32.062 --> 00:43:38.663
The fact that we have one or two orders of magnitude, more of these so-called negative quartets.
418
00:43:38.782 --> 00:43:49.903
Quartets whose probable cosines are, the expected values of these cosines is negative we need to make very strong use of those.
419
00:43:50.013 --> 00:43:56.103
Well I’ve already told you that these parameters, this T is determined from the known distributions.
420
00:43:56.203 --> 00:43:58.932
It’s simply the expected value of the cosine of the triplet.
421
00:43:59.082 --> 00:44:02.493
This is the expected value of the cosine of the quartet.
422
00:44:02.622 --> 00:44:06.363
These are simply weights which I already described before
423
00:44:06.453 --> 00:44:11.183
and I relate it to the variances of the cosines of the quartets and triplets.
424
00:44:11.283 --> 00:44:13.683
So all these parameters are known.
425
00:44:13.792 --> 00:44:20.263
Phi(HK) is an abbreviation for this triplet. Phi(LMN) is an abbreviation for this quartet.
426
00:44:20.403 --> 00:44:26.542
The condition which has to be satisfied if we are to find an answer to the question
427
00:44:26.642 --> 00:44:33.292
that I raised a few minutes ago, is that the cosines of the triplets must,
428
00:44:33.383 --> 00:44:39.723
well the value of this function of the invariance, Phi(HK) and Phi(LMN),
429
00:44:39.853 --> 00:44:45.843
this function of these invariance of which there are maybe hundreds of thousands of them.
430
00:44:45.923 --> 00:44:50.632
So this is a sum over several hundred thousand of terms.
431
00:44:50.742 --> 00:44:56.822
The value of this function, of these invariance, this one and this one must be a minimum.
432
00:44:56.952 --> 00:45:04.523
When this function is a minimum then we can be sure that we have answered our question which I raised before.
433
00:45:04.613 --> 00:45:08.703
That is to say - what must be the values of the individual phases so
434
00:45:08.783 --> 00:45:12.662
that when we generate triplets and quartets we get distributions of values for these
435
00:45:12.743 --> 00:45:17.463
which agree with their known theoretical distribution?
436
00:45:17.543 --> 00:45:26.763
The answer to the question is to minimise this function of invariance, Phi(HKK) and Phi(LMN) subject to the constraint
437
00:45:26.862 --> 00:45:33.132
that all the identities which the invariance must satisfy are in fact satisfied.
438
00:45:33.472 --> 00:45:39.113
Now that requirement that the identities which must exist among the invariance simply
439
00:45:39.183 --> 00:45:43.413
because there are so many of them and there are relatively few phases,
440
00:45:43.503 --> 00:45:48.092
that requirement of course is a tremendously restrictive requirement.
441
00:45:48.202 --> 00:45:52.462
So our problem then is formulated in a very simple way.
442
00:45:52.552 --> 00:45:56.122
Here is a known function of several hundred thousand invariance.
443
00:45:56.293 --> 00:46:03.763
We have to find the values of the phases which minimise that function of several hundred thousand of invariance.
444
00:46:03.863 --> 00:46:10.532
Subject to the condition that all identities which must hold among the invariance are in fact satisfied.
445
00:46:10.632 --> 00:46:17.282
The answer is very simple. However we still have a major problem. How do we find the answer?
446
00:46:17.372 --> 00:46:25.283
How do we determine the phases which will make this function a minimum, considered as a function of these invariance?
447
00:46:25.413 --> 00:46:30.622
And the first step to the answer to that question is shown on the next slide,
448
00:46:30.742 --> 00:46:34.583
on the right hand side which looks very similar to this.
449
00:46:34.723 --> 00:46:38.993
Except now, and I’ve called this the minimal principle.
450
00:46:39.103 --> 00:46:42.853
It’s the minimal principle for the individual phases.
451
00:46:42.972 --> 00:46:46.873
This is a function of invariance, Phi(HK), Phi(LMN),
452
00:46:47.013 --> 00:46:52.813
but the invariance themselves are explicitly expressed in terms of individual phases.
453
00:46:52.913 --> 00:46:59.712
So this defines implicitly a function of phases of which there may only be a few hundred.
454
00:46:59.822 --> 00:47:04.983
Here we have several hundred thousand invariance, here on the right hand side
455
00:47:05.083 --> 00:47:11.962
when we consider this function to be a function of phases, we have only three or four or five hundred phases.
456
00:47:12.102 --> 00:47:16.622
So this is a function of a relatively small number of phases.
457
00:47:16.712 --> 00:47:25.643
And the minimal principle says that that set of phases is correct which minimises this function of the phases.
458
00:47:25.752 --> 00:47:33.113
So the answer to the question that I previously raised is in fact formulated in a very simple way.
459
00:47:33.173 --> 00:47:39.422
It’s formulated as this minimal principle. But there still remains a major problem.
460
00:47:39.522 --> 00:47:44.942
Even a function of three or four or five hundred phases is a function for
461
00:47:45.002 --> 00:47:51.242
which it is very difficult to find the global minimum, especially if as in this case there are many local minimum.
462
00:47:51.312 --> 00:47:55.932
In the case like this with several hundred phases
463
00:47:56.002 --> 00:48:02.842
there may be something of the order of ten to the one-hundredth power local minima.
464
00:48:02.972 --> 00:48:09.803
From this enormous number how are we to select the one global minimum which is the answer to our question?
465
00:48:09.893 --> 00:48:15.532
Well, it would be very nice of course, if this function were very well behaved in the sense
466
00:48:15.612 --> 00:48:23.082
that we could start with a random set of values for the phases. Just choose phases at random.
467
00:48:23.172 --> 00:48:28.412
And then use standard techniques to find the minimum nearby that.
468
00:48:28.503 --> 00:48:35.292
There is several ways of doing that one is the least squares technique which however has the disadvantage
469
00:48:35.372 --> 00:48:38.753
that it will get the local minimum which is near to the starting point,
470
00:48:38.863 --> 00:48:44.572
will be trapped in a local minimum far away from the global minimum that we are looking for.
471
00:48:44.732 --> 00:48:48.773
So that’s a method that in general will not give us the answer.
472
00:48:48.933 --> 00:48:56.133
Or we could use a different method, a method called parameter shift method in which we vary the phases one at a time,
473
00:48:56.263 --> 00:49:03.612
look for the minimum as a function of a single phase and that way escape the trap of being caught in the local minimum.
474
00:49:03.752 --> 00:49:13.293
We may get an answer; a minimum far removed from the stating set but in general still a local minimum as it turns out.
475
00:49:13.432 --> 00:49:16.142
Not the global minimum that we are looking for.
476
00:49:16.242 --> 00:49:23.502
So it looks as if we have traded one very difficult problem for another problem just as difficult.
477
00:49:23.612 --> 00:49:33.672
But I would like to describe in the remaining few minutes that I have what we have done in order to try to solve this problem.
478
00:49:33.762 --> 00:49:39.463
And to show in fact that at least for a small molecule we have been able to resolve this problem.
479
00:49:39.542 --> 00:49:48.693
We have in fact found the unique global minimum chosen from this set of maybe ten to the one-hundredth power local minima
480
00:49:48.813 --> 00:49:51.733
we have in fact gotten the global minimum.
481
00:49:51.833 --> 00:49:56.372
I would like to describe in the next few minutes how we have done this.
482
00:49:56.512 --> 00:50:04.902
We have taken a small molecule, a molecule consisting of twenty-nine atoms, non-hydrogen atoms in the molecule.
483
00:50:05.022 --> 00:50:10.303
And we’ve constructed this function, this RFV function, and we calculated that function.
484
00:50:10.393 --> 00:50:16.003
First when we put in, since we know the answer beforehand, we know the values of the phases.
485
00:50:16.083 --> 00:50:21.893
And when we put in those values, the value of this function turns out to be approximately four tenths.
486
00:50:22.013 --> 00:50:30.193
And then we also have put in seven other randomly chosen values for the phases
487
00:50:30.283 --> 00:50:36.553
and in each case as you can see the values of the function is bigger than when we put in the true values of the phases.
488
00:50:36.643 --> 00:50:40.813
Which of course is in agreement with the property that I’ve already stated.
489
00:50:40.923 --> 00:50:44.083
That it is for the true phases that this function has a minimum.
490
00:50:44.162 --> 00:50:48.413
And has the minimum of approximately four tenths compared to random phases
491
00:50:48.493 --> 00:50:52.962
which give minima running around .67 or .68 or so on.
492
00:50:53.082 --> 00:51:01.142
Incidentally in this case we have calculated not merely the values of the function for seven randomly chosen phases
493
00:51:01.202 --> 00:51:02.923
but for thousands of them.
494
00:51:03.013 --> 00:51:06.612
And in all cases the value of the function is much larger than four tenths.
495
00:51:06.682 --> 00:51:11.183
It runs from about .66 to .69 or so.
496
00:51:11.263 --> 00:51:18.303
So there is no doubt that we have in fact confirmation of the theoretical result
497
00:51:18.393 --> 00:51:22.563
that the function as a minimum when the phases are equal to their true values.
498
00:51:22.663 --> 00:51:31.513
Well, starting with the true values, we went through two methods for getting the local minimum near to the starting set.
499
00:51:31.623 --> 00:51:36.552
One method was the least squares method, we went through a number of cycles of least squares
500
00:51:36.632 --> 00:51:42.103
and we ended up with values from the phases near to the starting set, not exactly the same.
501
00:51:42.223 --> 00:51:45.963
And it gives us a minimum of .366.
502
00:51:46.073 --> 00:51:50.682
The set of phases incidentally corresponding to this global minimum now
503
00:51:50.832 --> 00:51:55.863
gives us by means of the Fourier synthesis essentially the whole structure.
504
00:51:55.963 --> 00:52:00.313
The whole 29 atoms appear in the Fourier map when the phases
505
00:52:00.423 --> 00:52:06.773
which are put in are the phases which correspond to the global minimum of this function which is .366.
506
00:52:06.893 --> 00:52:12.293
If we use a parameter shift method for getting the minimum near to the starting set
507
00:52:12.403 --> 00:52:15.693
we get the same minimum which is not too surprising.
508
00:52:15.813 --> 00:52:21.452
But what happens when we put in a random set of phases and we go through both processes
509
00:52:21.573 --> 00:52:25.813
we get a local minimum, .44 here and .46 here.
510
00:52:25.963 --> 00:52:30.612
It’s not a global minimum clearly, this is the global minimum so we get a local minimum.
511
00:52:30.732 --> 00:52:36.193
And the same thing happens with each of these other random starts; we get local minima
512
00:52:36.303 --> 00:52:41.102
which however are not the global minimum.
513
00:52:41.212 --> 00:52:48.852
Well of all these minima we have chosen two to be of particular interest, 1.4125
514
00:52:48.952 --> 00:52:51.553
which is the smallest one in this column.
515
00:52:51.693 --> 00:52:58.322
And the other .43 which is the smallest one here except for the true global minimum.
516
00:52:58.482 --> 00:53:08.172
And we have made the assumption that because .41 and .43 are both less than the other local minima
517
00:53:08.242 --> 00:53:17.892
which run about .45 or .46, that the phases which give us these minima, these local minima now,
518
00:53:18.033 --> 00:53:23.833
somehow or other carry some structural information in them.
519
00:53:23.942 --> 00:53:30.233
They are not, certainly they are not the correct phases, we know that, the correct phases give us the global minimum.
520
00:53:30.343 --> 00:53:35.193
But the assumption is made that they carry some structural information.
521
00:53:35.283 --> 00:53:42.072
If they are to carry structural information the question is how do we find what that structural information is?
522
00:53:42.162 --> 00:53:49.872
And the answer of course is very simple, all we do is use the phases that we get let’s say from this local minimum,
523
00:53:49.962 --> 00:53:52.933
calculate the Fourier series and have a look at it.
524
00:53:53.003 --> 00:53:55.763
See if in fact the structure is in there.
525
00:53:55.883 --> 00:53:59.843
Well we’ve done that, the next slide shows what happens.
526
00:53:59.952 --> 00:54:06.863
We’ve done that for that minimum, this was the random start, after minimisation we get .4125.
527
00:54:06.963 --> 00:54:16.443
We construct the Fourier series with co-efficients using these phases and known magnitudes and we take a look at it.
528
00:54:16.533 --> 00:54:21.123
Well it doesn’t look very good, it doesn’t seem to have any structural information in it.
529
00:54:21.243 --> 00:54:27.773
But we expect there will be some structural information in it and the way that we have chosen to extract
530
00:54:27.853 --> 00:54:36.163
that structural information is to assume that the information is contained in the largest peaks of that Fourier series.
531
00:54:37.266 --> 00:54:46.157
So we’ve taken the top six peaks of that Fourier series, that gives us what we hope is a fragment of the structure.
532
00:54:46.436 --> 00:54:53.916
Using those presumed atomic position vectors we can now calculate normalised structure factors E,
533
00:54:54.005 --> 00:54:56.986
which is to say both magnitudes and phases.
534
00:54:57.066 --> 00:54:59.817
In this way we get a new set of phases.
535
00:54:59.947 --> 00:55:07.027
Different from the random set we started with and certainly different from the set which gave us that local minimum.
536
00:55:07.107 --> 00:55:09.147
We get a new set of phases.
537
00:55:09.267 --> 00:55:18.097
We use the known magnitudes of the normalised structure factors with this new set of phases in our minimal function again.
538
00:55:18.197 --> 00:55:24.126
Well it turns out that the value of the function is now less than what happened when we had random start
539
00:55:24.246 --> 00:55:27.686
but more than the local minimum which we got before.
540
00:55:27.776 --> 00:55:34.576
And that's not surprising because we are using only six peaks among the total of maybe several hundred peaks.
541
00:55:34.676 --> 00:55:36.717
We are using the six strongest peaks.
542
00:55:36.807 --> 00:55:43.247
But when we go through the minimisation process again we find that we get a smaller minimum than we had before.
543
00:55:43.316 --> 00:55:51.397
Another local minimum .39, smaller than before and so we expect that the phases
544
00:55:51.487 --> 00:55:58.417
which give rise to this local minimum carry still more structural information than this set of phases.
545
00:55:58.527 --> 00:56:05.167
Well it turns out although we might have difficulty doing this if we didn’t know the structure that the full structure,
546
00:56:05.247 --> 00:56:11.196
all 29 atoms do in fact appear among the strongest 135 peaks.
547
00:56:11.306 --> 00:56:16.606
That may not seem like a very useful result of course because it may be difficult
548
00:56:16.696 --> 00:56:24.067
in the case that we didn’t know the structure to see it, to see the 29 atoms in the 135 strongest peaks.
549
00:56:24.167 --> 00:56:26.856
Well we don’t assume that we’ve done that.
550
00:56:26.946 --> 00:56:34.367
Instead from this Fourier series, the Fourier series calculated with the phases which give us this local minimum.
551
00:56:34.467 --> 00:56:41.557
From that Fourier series we take the top twelve peaks now, again under the presumption
552
00:56:41.647 --> 00:56:48.887
that most or all of these peaks do in fact correspond to true atomic positions.
553
00:56:48.996 --> 00:56:50.736
We go through the process once more,
554
00:56:50.796 --> 00:56:56.567
we calculate the value of this function for the set of phases calculated on the basis of these 12 peaks.
555
00:56:56.667 --> 00:57:01.307
And we now find the value of this minimum function to be .439.
556
00:57:01.397 --> 00:57:04.836
Smaller than each of these but bigger than what we got before.
557
00:57:04.946 --> 00:57:11.537
Again we are not surprised at that because we are using here only 12 peaks among maybe 135 peaks.
558
00:57:11.637 --> 00:57:20.227
But we go through the minimisation process again, and now the local minimum turns out to be .37.
559
00:57:20.397 --> 00:57:26.176
By doing this process then is among these enormous numbers of local minima
560
00:57:26.276 --> 00:57:30.907
we have been able to find the unique global minimum or something very close to it.
561
00:57:31.037 --> 00:57:35.606
Sufficiently close that it’s trivial to pick out the structure.
562
00:57:35.697 --> 00:57:42.216
Now I see that my time is up, so I can’t describe the second application which however is very similar to this,
563
00:57:42.306 --> 00:57:51.687
instead of using the local minima of .41 as the next slide shows we used the next local minimum which was .43.
564
00:57:51.767 --> 00:57:57.566
We go through a rather similar process and we end up with the same result, essentially the same results.
565
00:57:57.666 --> 00:58:07.486
After two cycles 28 of the 29 atoms appear among the strongest 31 peaks and the 29th atom appears at the peak number 44.
566
00:58:07.586 --> 00:58:13.247
For this starting point as well as the starting point shown on the previous slide
567
00:58:13.377 --> 00:58:21.437
we are able to find essentially the global minimum or something very close to the global minimum
568
00:58:21.547 --> 00:58:23.867
and in both cases to solve this structure.
569
00:58:23.987 --> 00:58:30.517
What remains to be seen is whether we can do the same thing for a much more complicated structure.
570
00:58:30.607 --> 00:58:36.446
Say a structure with several hundred atoms where the calculations then become much greater than they are now.
571
00:58:36.556 --> 00:58:40.847
Because instead of using only 300 phases as we’ve done in this case.
572
00:58:40.937 --> 00:58:50.237
300 phases in the base with something like 139,000 negative quartets and 7,000 triplets and a few thousand positive quartets.
573
00:58:50.397 --> 00:58:57.337
We may need to use for a much more complicated structure instead of 300 phases maybe 1,000 phases.
574
00:58:57.447 --> 00:59:04.837
And instead of a couple hundred thousand invariance we may need to use a couple of million.
575
00:59:04.977 --> 00:59:07.186
So the calculations become much greater.
576
00:59:07.266 --> 00:59:13.027
But if the only problem is complexity of calculation then we have made a big advance
577
00:59:13.107 --> 00:59:18.597
because even existing computers are capable of handling that kind of calculation.
578
00:59:18.687 --> 00:59:21.556
Thank you.