Now in electron microscopy, the situation is different because unlike X-rays, electrons can be focused, and so we gain an advantage by putting a lens behind the sample. And re-focus all of the scattering that emerges from the sample. So let's look at, think about that in more detail. In an electron microscope, here's our sample and there are incident electron waves coming down hitting that sample. And, as it hits the sample, there's millions of scattering centers that introduce phase ripples. And as they propagate forward you see a constructive interference, in a, in a, in a, wave front that could be represented by these waves. And they're produced all over the sample, and remember the action of a lens is to take parallel illumination and focus it to its particular point on the focal plane of that lens. And so all of the straightforward scattering that emerges from the sample is focused to a particular spot behind the lens on the focal plane. And after it's focused into this particular spot, it then propagates and spreads out as it moves further down microscope. And this will be our image plane. In like fashion, there are, is other scattering in different directions. So for instance, all the scattering that comes out in this direction, these wave fronts, they're also collected. Now, a lens will focus any set of parallel radiation into a particular spot of the bo, back focal plane of the lens. But if this radiation is not parallel to the axis of the lens, that spot will also not be on the axis. Instead scattered radiation in this direction is focused to say to a spot over here on the vat, back focal plane of the electron microscope. And from here, it then sails on through that point and spreads out across the image plane. And for completeness, we'll draw one more set, another direction of scattering, say it might be in this direction, is also focused into a different position on the back focal plane of the electron microscope. And from there it sails on through and spreads out across the image plane. Now what you see is that all the radiation scattered in a particular direction, remember, together, this contains the information about how much of a particular Fourier component is present in the sample. In other words, how many scattering centers are a certain distance apart here in the sample. That information is contained in the scattering in this direction. And that's all focused to a particular spot on the back focal plane of the objective lens. And all the possible directions of scattering, each direction is focused to a different spot along the back focal plane of the lens. And so what appears here is actually the Fourier transform of the sample density. Or in other words, a diffraction pattern. On the back focal plane of the electron microscope. Now because this diffraction pattern was formed, by the action of a lens, focusing divergent scattering, if that scattering is allowed to continue propagating through the back focal plane, each of these rays will spread out again across the image plane and will end up interfering with all the other rays that are present here. In other words, a Fourier synthesis occurs here, or an inverse Fourier transform, meaning that these rays which represent one particular sine wave that's needed to reproduce the sample, those spread out here and produce a sine wave across the image. So for instance, the blue scattering might represent a sine wave of a particular frequency, scattering at a different angle, say scattering at a higher angle, here, would be collected at a point further out in reciprocal space away from the origin. But would still carry through, and add its information on the image plane. It might represent a wave that's oscillating much more rapidly. And so we see that the scattering being collected by the lens, produces a Fourier transform and, or a diffraction pattern on the back focal plane. And if that scattering is allowed to continue on, it will re-interfere with itself. And do essentially an inverse Fourier transform, or you might call this a Fourier synthesis to produce an image. And it's just like the ideas that we had when we were studying Fourier transforms. A Fourier transform takes a complicated wave function like the density of a sample and it separates it into its component sine waves. Each component sine wave is represented as a single spot here in the back focal plane of the lens. A Fourier synthesis takes each of these sine waves and adds them back together to reproduce a complicated function. And in this case, this complicated function is a replica of the density of the sample. However, because it's been through a lens system as we explained earlier, this replica of the density can be much higher magnification. It can be very, much larger physically in space than the actual sample was. Now how do we reconcile this with the idea that each electron is a wave? Well if I were to draw again, the electron here as a wave, and remember I, I tried to introduce the idea that it was a plain wave where each pixel had an amplitude in phase, as that plain wave interacts with the sample, the sample delays phases. And scatters so you have phase ripples. So that, as that electron wave is now past the sample it has, embedded within it, a whole series of ripples. And there's so many you can't draw them. All these ripples together-. Are producing the wave fronts that are traveling in, say, this direction, or a higher scattering angle. The lens, bends these scattered rays and recollects them. And so there is a wave function that exists here on the back focal plane of the lens. And that wave function has strong amplitudes in these locations of the Fourier components, and also in all other positions. I've got a drawing now that is some kind of hybrid between a cross section showing the raise and now some kind of oblique view showing the wave. The plane wave as it passes through, but that plane wave carries in it, it is the Fourier transform of the sample density. And then as these rays move on further down the electron microscope. All of these Fourier components interfere with each other to form a complicated wave function. And finally, the wave function that exists here on the image plane is a replica of the density of the sample. But a magnified version, and this wave function is what we record, eventually, on the detector. Now let me try to show the same thing with pictures. If, for instance, the sample is a complex image like the image of the professor. That we had in the last module. Then, the wave function that exists here, on the back focal plane of the objective lens is the Fourier transform of that density, that object. And the wave function that exists on the image plane behind the objective lens is a representation of the density of the original object, only magnified. So this again is a Fourier transform of the wave function on the back focal plane. So in a microscope a Fourier transform of the original image is present on the back focal plane. And a Fourier transform of that which is of course a representation of the original function is present on the image plane. Now, this is a description of an ideal microscope. And unfortunately, electron microscopes are not ideal. Instead, some of the scattered rays are delivered to the image more strongly than others. And this is our next subject. In an electron microscope, not all of the components of the image appear equally strong. Some of the Fourier components are represented fully in the final image. Others are only there half as much as they should be. And some are entirely missing. So let's consider. The incident electron as a plane wave traveling down the microscope column. And because it's a plane wave, the amplitude of each pixel here could be a, let's call it A1, and the phase at the beginning would be, let's say 0. Then, as this wave propagates down the microscope, at some point later, we'll have the same plane wave, of, say, amplitude one, but now the phase may have advanced, let's say 45 degrees. At some time later, again the phase will advance. [SOUND] It's still a plane wave of equal amplitude everywhere but now the phase may have advanced to say 90 degrees, etc., etc., until the phase equals a final 360 degrees, at which point the wave is the same as it was originally. Now let's introduce the symbol psi which is commonly used to represent a wave function. And let's call this psi at t equals zero. And this one later psi at t equals sometime t later, let's call this t1and this would be psi at time equals t2. And finally, this would be psi at time equals t3. Now I'd like to introduce another way of representing a wave as it propagates through space. It's called an Argand diagram. Named after a person named Argand. And in an Argand diagram, this would represent the real axis and the imaginary axis and the reason it's called that is beyond what we want to cover today. What I want to use this diagram for though. Is that you can represent a vector quantity with an amplitude and a phase as a line in such an Argand diagram. And the length of the line represents the amplitude of the vector. And the angle away from this real axis represents its phase. So for instance, this wave function psi at time equals 0, I could represent it as this vector here if that represents its amplitude a and its phase here is 0 and that's why I've drawn it coincident with this real axis. Later on, at time t equals 1, I could represent that wave instead as a vector like this. In this case, it's meant to have the same length because that represents the amplitude and that hasn't changed, but now its phase is 45 degrees. And so we'll show this as psi of t1. And over here, this was psi of t0. Later on at time t equals 2, we could represent that wave with a vector like this. Again has the same length because the amplitude hasn't changed. But this would be psi of t, t two. And as the wave propagates forward and the phase advances from 90 degrees to 180 to 270 to 360 this vector would move. Around and around this phase circle, here would be 270 degrees etc., until it came back around to 360 degrees. And so like the hands of a clock, except moving counterclockwise, we can represent the propagation of the wave. Through space, and the phase oscillations that it undergoes on this so-called Argand diagram. Now let's suppose that in the same space there exists not only one wave, but a second wave. And so I'm going to add that in a different color. But there's two waves occupying the same space here, and it has an amplitude, let's call it amplitude 2 and it has its own phase and it propagates with the first wave until, eventually, it reaches some position down here. [SOUND] Okay. Then in our argon diagram, we would have two waves. Here I'll draw the real and the imaginary axes. And in this case now, we have one wave that at some position later for instance at some position here, the first wave might look like this. Let me just call that now psi1. And the second wave that I've drawn in blue might be very different. It might be, say, a wave like that with a different amplitude and a completely different phase. Now let's suppose that we had a detector here. And in this detector, the detect, detector had a number of pixels. [SOUND] And so, each pixel here has some depth to it. I'll try to draw a single pixel within the detector. And the point is that we're going to have one of the waves traveling through that pixel of the detector with a particular amplitude and phase. Now here, I've drawn it oscillating, not that any part of the wave or anything about the wave is moving to the left or to the right, I've just drawn it wavy to help represent the fact that its phase is advancing from 0 to 360 and around and around the circle. And it has a particular wavelength and so, that's represented by the wavelength here. And so whatever that wave is, is oscillating within that pixel of the detector. Now, alongside it, the other wave is also oscillating alongside it and these waves add together in space. Now the way to understand how these waves add together in space, it turns out in the argon diagram, it's relatively easy to do, because you can simply draw one vector where it is, and then draw the second vector starting from the end of the first. And so, I can draw the second vector representing the second wave here. It's the, it's the same vector moving the same amplitude in the same general direction. It's just instead of starting at the origin, I start at the tip of the first vector. And then I can see what the result of both of them together is, because the, the sum of them will be this and I'll call this psi total is now a vector from the origin to the tip of the second. And so the, these two waves, if they have the same frequency when added together, will have an amplitude equal to the distance from here to there and will have a phase equal to that phase. And so the probability of the electron being detected in any particular pixel is equal to this wave function squared, the amplitude of the wave function squared. And so what matters here is the length of these vectors, not the direction. Now, as a result, if two waves are added together and their sum is being measured, it turns out that the angle between these waves becomes incredibly important. That's because if this angle, for instance, was zero and this second wave was headed in the same direction as the first wave, then the sum would be very large indeed. It would, the blue wave would strongly add to the length of the sum vector, so the presence of the blue wave would change the sum significantly. If, on the other hand, the angle between the two waves is, say, 90 degrees, if they're 90 degrees apart, then their sum looks like this. And it's not very much different than the brown wave all by itself in terms of amplitude and because of that, the presence of the blue wave is almost irrelevant. It doesn't really change the length of their sum, of the sum, and therefore, it wouldn't change the probability of an electron being detected there. Finally, for completeness, let's just draw the situation, what if their phases were exactly opposite? So the blue wave was now pointing in the exact opposite direction of the brown wave. Well, in this case, the sum would be much shorter. It would only be this length, much shorter than the original brown wave. And once again, that blue wave, its presence would be felt significantly. It would have a strong impact on the result. Now to understand why this is the case, let's consider three different scenarios for how two electron waves could add together. And what I want you to imagine is that this is all happening within one pixel of a detector in an electron microscope, for instance. And this plot and this plot and this plot are three different scenarios about how two electron waves could be adding together. So, in this first scenario we have an electron wave here shown in blue that is oscillating back and forth. Now, we don't know what about an electron wave oscillates. We understand that in a photon, what oscillates is the electric field, and that's why it's more easy to understand light as waves. With an electron, we don't really know what is oscillating about it. Its electron-ness is oscillating, but whatever it is that oscillates, that gives it its wavelike properties, that's what we're trying to illustrate in blue here. Here's an electron wave moving down through the pixel of the detector with a particular phase and amplitude. Now, if we were to draw this situation with an argon diagram, with the real and imaginary axes, we might represent this blue electron wave at some particular position where it has a particular phase as a vector here. It has an amplitude and it has a particular phase. Now actually, these plots specifically these plots are of 4 sin x and 1.5 sin of x plus a phase factor phi. And so the blue curve here is 4 sin of x, and so its amplitude is 4. And so here, this vector I'm going to represent, it has a length of let's say, arbitrary units of four. Now, in this same pixel of the detector, let's imagine that there's a second electron wave. And this electron wave has the same frequency the same wavelength, you might say, so these, these electrons have the same energy. And they coincide, they coexist inside the detector. But this electron wave. Has an amplitude of just 1.5. So this is the, the red curve here represents the second electron wave. And its amplitude is less, it's only 1.5. And in this first scenario, the phase shift is zero degrees. And so phi is zero. And because the fade shift is zero both the red curve and the blue curve oscillate together. And so they're here in phase. And so if we were to add the blue curve to the red curve in the argon diagram, the way to do it would be to add the amplitude of the red curve starting at the end of the blue wave in the same direction. Because they have the same phase, and together, they would add up to the third curve being shown here which is their sum. The sum of the red and the blue curve is just bigger, and so then their sum is a vector that looks like this. And so its amplitude is even more than the other two curves. The other two curves add together to form their sum. And both of these curves have a strong influence on this one because it increases the amplitude. Okay that's one scenario. Now let's consider this scenario in the middle plot. Okay, here the blue curve is just as it was before 4sin(x). And so at the same position it would look like a vector just like it did over here. But in this case, the phase shift between the two curves is 180 degrees. And therefore, they're exactly out of phase. And so the red curve still has the same amplitude of one point five. So it would rep-, be represented in the argon diagram, as a vector that's shorter. But in this case the vector is moving in exactly the opposite direction of the blue vector, because these two curves are exactly out of phase. And in this case, now their sum has substantially lower amplitude. It's a shorter, vector because here when you add the red curve to the blue curve, the amplitude of their sum is significantly reduced compared to the blue curve. Now if the probability of detecting an electron here is equal, or is proportional to the amplitude of these oscillations squared. Well then you see that the probability of an electron being detected here, is much lower than the probability of the electron being detected here, because here the way, the amplitude of the wave function is very large. Here it's smaller, because the two pieces that were added together to form it were out of phase. Now let's consider the final case, the final scenario. Here again we imagine that the blue wave is as it always was before, and so we'll draw it the same way. Now the second curve is 90 degrees out of phase with the blue curve and so it's shift back. And so it's going to have an amplitude of 1 and a half, but it's direction is 90 degrees away from the direction of the blue vector, because it's phase shifted by that amount. Now, in this situation, the sum of those two vectors is now here. And a remarkable fact is that the length of this vector is approximately equal to the length of the blue vector. And if the probability of detecting an electron is, proportional to the link of this, proportional to the amplitude of that wave squared, then the red component is almost irrelevant, because it doesn't change the amplitude of the resultant wave by much. And that's because it's 90 degrees out of phase. If you look at the sum of these two waves, the brown curve, it has nearly the same amplitude as the blue curve by itself. So the red curve almost disappears in the sum, because all it does is phase shift the sum a little bit. But detection, in a detector, is blind to that phase shift. All that the detector reveals is the square of the amplitude of that wave function. That's why if there are two waves occupying the same space; one large and one smaller. And they have the same frequency. Whether or not the smaller wave is manifest, in the detector, in the sum of those two waves. Depends entirely on the phase shift between the two waves that are being added together. If the phases are the same, the smaller curve becomes visible. If the phases are exactly opposite, the smaller curve, is clearly manifest as a change. But if the phase is 90 degrees. It essentially is lost, it essentially disappears.