Now we're prepared to understand why some Fourier components are strongly represented in an EM image and why some are lost. Let's begin. By drawing the sample and the lens that collects the scattered radiation and below, an image plane. Now let me draw the incident electron [SOUND] and I'll draw it as a plane wave. Coming down, interacting with the sample. Now when that electron wave hits the sample, some of it will be unscattered, and some of it will be scattered. The single electron can experience all of these phenomena, and its contribution to the final image depends on how all of these processes interfere with each other. And so, let's first think about the unscattered part of that electron wave. So the electron wave hits the sample and then from here, let me just draw it as a wavy line. To again remind us that as that wave propagates down through the microscope, the phase of the electron wave is oscillating. In, around from zero to 360, to zero to 360 and so it oscillates. Again I don't mean, to mean, I don't mean to suggest that the wave is moving to the left and then moving to the right and moving to the left or anything like that. It's just whatever the wavelength whatever's oscillating in the electron that gives it. It's wave like properties and its wavelength. It is, it is progressing as it moves down the microscope in this fashion. And it traverses so many wavelengths from the position of the sample down to the position of the image plane. Now remember that when the unscattered part of the electron hits the image plane, it fills that space. I've just represented it as a single oscillating line, as it travels down through the microscope. And when it hits the image plane it will have a particular phase. And so let's draw an argon diagram now, and represent that unscattered part of the electron beam as a vector and let's call it side unscattered. And it has an arbitrary phase, it, it, there was a number of wavelengths as it passed from the sample to the image plane. And so it will hit image plane with some arbitrary phase. Here I've drawn it as a, as approximately 45 degrees but that's not the important thing, so we'll just draw it in that position. Now let's consider another part of the electron that is scattered into some angle. And for now, let's just draw it as an angle like this. So part of the electron will be scattered by the sample into lots of different directions. This is the first direction we're going to consider. And scattering in this direction will be refocused, back on, so that it interferes on the image plane with the unscattered beam. Now there's two key differences between this scattered beam here in blue and the unscattered beam in green. The first difference is that upon scattering the blue curve, suffers a face shift of 90 degrees. It emerges from the sample 90 degrees behind the unscattered being. Now, why is this so? Let's think again for a second about the analogy of a photon being scattered by an atom. So let's imagine an atom with a nucleus and an electron cloud around that nucleus. And so we have a single electron in the, in the cloud around the atom. And let's suppose we have an incoming photon. Now in the case of a photon, we know what it is that's oscillating, it's an electric field. Photons are oscillating electric fields. And when this electric field interacts with and electron in an atom, it we can imagine that it causes the electron to move up and down and up and down. Because electrons experience forces in, in electric fields. And as that electron is moving up and down, we also know that moving electric charges, create electric fields. And so a charge that's moving up and down causes an electric field disturbance that will [SOUND] move outwards away from that movement. And propagate in little circles away from that up and down trajectory. In a driven oscillator, I want you to imagine that there is a, a lever arm that's moving up and down and up and down just oscillating. And on that arm there's a spring with a mass below it so as the lever arm moves up and down, the mass on the spring may go with it like this. It turns out that each spring as a resonant frequency and in, under some conditions, if, if the oscillation of the driver is lower than the resonant frequency of that spring. Then the mass and the lever arm will move up and down together. If, however, the oscillations of the driver are faster than the resonant frequency of the spring, then what will happen is they'll be opposite of each other. The driver and the mass will move up and down like this. If, in the third case, the driver is moving up and down at the resonance frequency of the spring. Then what will happen is, as it moves down, it compresses the spring which then pushes the mass down. And then as the driver starts rising, the mass will stop and then start following it up. Then the driver will turn around and start pushing down, and that will pus, push the mass down. And, I, I can't do it very well with my hands, but what happens is that the driver and the mass are exactly 90 degrees out of phase. And this is what I want you to imagine might happen when a photon drives an electron up and down, that the scattered radiation that emerged from that electron bouncing up and down will be emitted exactly 90 degrees behind the driver oscillations of the incoming photon. And so, it could be that what happens with electrons is similar. That whatever is oscillating up and down in an electron, when it hits an atom in the sample, something in that atom is caused to oscillate and scatter the radiation in turn. And the scattering emerges 90 degrees behind the incoming electron wave. And so after this part of the electron that's scattered, this wave emerges 90 degrees out of phase with the unscattered wave. So that's the first big difference between the scattered beam and the unscattered beam. The second difference is that they travel different path links. So the distance that the unscattered beam travels is from here to the image plane. But the distance that the scattered beam travels is over to this position. Then it's bent and brought back to the image plane. And there's a difference in the path length, delta l. And because of that, the scattered wave will be additionally phase shifted compared to the unscattered wave. Now, if that path length difference is, for instance, a fourth of a wavelength. Then this will correspond to an additional 90 degrees phase shift, compared to the unscattered wave. And the third thing you need to know about the scattered wave is that its amplitude is very small compared to the unscattered wave for the kinds of samples we'll image in cryo-EM. So if we were to add a vector representing this scattered wave to our argon diagram, the unscattered wave is headed in this direction. The scattered wave would be short in comparison because the amplitude of the scattered component is less than the unscattered component by a lot. So it's short. In addition, it's 90 degrees phase shifted because of the scattering process. But in addition to that, it's a further 90 degrees phase shifted because it has to travel an additional path length equivalent in this imagined case of a quarter of a wavelength. And so, the total phase shift is the full 180 degrees. And as a result, this scattered wave will add to the unscattered wave in the opposite direction. And the sum of those two waves together will be, could be represented by a vector like this, which is significantly shorter than the unscattered vector itself. So, down here in the detector I'm going to draw a particular pixel of this detector. And as these two waves interfere with each other as they pass through that detector, you'll see that because the small, blue wave is a full 180 degrees out of phase with the green wave. The sum of those two waves is now going to have a much lower amplitude than the unscattered beam itself. Let me add some labels. The blue curve is the scattered component, and the red curve is the total, the sum of the two. Now I'm going to draw a graph. And the horizontal axis is going to be spatial frequency. And the vertical actis, axis is going to be called contrast. And contrast is going to vary from 1 to negative 1. And what we're going to do is, as a function of spatial frequency, we're going to plot how much a wave at that spatial frequency can be seen. Or how much of that wave has an impact in the sum within the detector. Now this scattered wave is scattering at some particular spatial frequency. Let's just say that it, it, it corresponds to a spatial frequency here along this axis. And we can see here that the blue curve contributes maximally in that it, it is 180 degrees out of phase with the unscattered curve. And so it shortens the sum maximally. And so I'm going to put a single spot here for that blue curve here, corresponding that its full amplitude is seen in the sum. And it contributes negatively, for instance, like that. Okay, some of this will be clearer when we start drawing additional curves. Now let's consider the part of the electron that's scattered to an even higher angle. 'Kay, so it's scattered out here. And then from here, the lens focuses it back in here to interfere with the other, with the, the other waves within the detector. Now in this case, if we were to draw the argon diagram, it would again, it would look like this. The green wave, again we'll represent it as a vector, that's the unscattered component of the electron. Now, this magenta scattered component here will have a much smaller amplitude. And it will be phase shifted with respect to the unscattered beam for two reasons. First of all when it's scattered, it, it suffers the same phase shift of 90 degrees. And in addition, its path length is, is quite different than the path length of the unscattered beam. In fact, if the path link difference of the blue curve was say a quarter of a wave length, the difference in path link here might be say, a half of the wave length or correspond to a full 180 degree phase shift. In that case, if we were to plot a vector that represented this scattered wave on the argon diagram, first of all it would be phase shifted relative to the unscattered wave by 90 degrees because of the scattering process. And then an additional 180 degrees because of the additional path length that it travels, and as a result, we would draw it in that position. So that's the scattered wave, and the sum of those two waves is now this. And the key thing to recognize is that this magenta wave scattered at this spatial frequency is essentially invisible to the detector. Because what the detector is going to detect is the amplitude of that electron wave squared. And so, because the amplitude of the total wave here is essentially the same as the amplitude of the unscattered wave, the magenta wave is essentially invisible. It contributes nothing. It, it doesn't change the probability of detecting an electron there. And so we're going to mark its contrast as zero. So I'll make that point magenta. Now the reason I'm drawing it further along this axis, which is spatial frequency, is because it was scattered to higher angle. And remember that the ang, the scattering angle is proportional to spatial frequency. Remember that at the back focal plane of the lens exists the Fourier transform of the sample density. And so the unscattered beam. Represents the zero spatial frequency component of the image, which is here at the beginning of this spatial frequency axis. And as we proceed to higher and higher scattering angles, we're moving to higher and higher spatial frequency components of the image. Close to the origin here, or here. These are low resolution details, far from the optical axis at high scattering angles, are high spatial frequencies. So, as we move to higher and higher angles, I'm going to plot the points further, and further out. So, let's plot the next one. [SOUND] Let's consider the part of the wave that scattered to yet, higher angle. And imagine that it is focused by the lens, my drawing wasn't big enough. But it was focused by the lens, and so it comes back to interfere with the unscattered beam, and the others. Now, the path length here, now is quite a bit longer than the path length of the unscattered beam. And if we were to represent the situation in an argand diagram, the unscattered beam would be represented again like this, a single vector. So, [SOUND] that's the unscattered beam. And now, in this case the scattered wave has a small amplitude because it's a scattered beam, and it's phase shifted by 90 degrees because of the scattering event. In addition, let's imagine that the path length difference here corresponds to three quarters of the wavelength, or a 270 degree phase shift. In this case, there's another 270 degree phase shift, phase delay. Because of the additional path length that it traversed. In this case, the vector ends up pointing in the same direction as the unscattered beam. And so, to find their sum, we would add it to the tip, pointing in the same direction. So, that's the scattered beam, and the sum of these two beams now, is this full length, so that's the total beam. As a result, the beam that scattered at this spatial frequency, adds up with the unscattered beam with a phase shift that makes it fully visible. It's full extent is manifest in the amplitude of the total wave. And because of that, we would graph, if this is the spatial frequency of the orange scattered wave that we're considering right now, we would give its contrast at 1, because it's fully visible. It has the same phase as the unscattered wave, and becomes fully visible in the total wave. And so, we would plot it here. [NOISE] Now, you're probably starting to see the pattern. There will be a part of the scattered wave that's scattered to even high angle, and that will be focused by the lens to interfere with the unscattered wave. And in this case, the phase shift will not only be 90 degrees, but if this additional path length corresponds to a full one single full wavelength, then its phase shift will be an additional 360 degrees. And so, it's additional phase shift because of the path length will bring it all the way back into pointing in this direction. And as a result, it will add with the unscattered beam like this. And the, [SOUND] that's the scattered component we're considering, and so the total will be here. And the amplitude of the sum is essentially, the same as the amplitude of the unscattered beam. And so this Fourier component, is essentially invisible to the image. [SOUND] And so, if we plot that here at this spatial frequency. It contributes zero. It's contrast is zero. [NOISE] Now, the last beam I want to consider [NOISE] is, a beam that is scattered by the sample. But to very low angle. And so I'm going to draw it right along side the unscattered beam. It was scattered, but to very low angle. And because of that, when we draw the argand diagram to represent the situation. The unscattered beam is as it always has been. Now, the scattered beam, to very low spatial frequency, now suffers a 90 degree phase shift. But no additional phase shift because it, it follows almost the same path as the unscattered beam. So, there's no additional phase shift due to a different path length. And so, the phase shift is simply the 90 degrees due to scattering. And so, we would draw it here, that's the scattered beam, and the total. Is now this. And you can see that the scattered beam to very low spatial frequency is almost invisible in the image because its phase shift of 90 degrees. Causes it not to impact the total amplitude of the sum. And so, if we plot it on our plot here, it's at zero spacial frequency, and its contrast is zero. It, it, it doe, it fails to change the image. Now, if we were to consider all kinds of scattering directions in-between, you would see that each one, because of the different path length, its contribution would vary from no contribution in zero spatial frequency to a full negative contribution. Here at this position would vary like that. And then, past that it would move around into a position where again it had no contribution to the image, and that was this point, this magenta curve. In between here and the orange curve, these spatial frequencies would contribute more, and more, and more until eventually, at the orange curve, they were contributing fully. Because they were in-phase with the unscattered beam, and so they rise to this fully contributing situation. [NOISE] Then between the orange curve, and the purple curve, these spatial frequencies suffer path length differences that cause them to roll around the circle in this way, until they are once again, invisible. And that was the situation with the purple curve. And if we were to think about spatial frequencies at higher, and higher, and higher scattering angle. We would see that they contribute more, or less in an oscillatory fashion, as a function of spatial frequency. And this curve is what we call [SOUND] the contrast [SOUND] transfer. Function or CTF of the electron microscope. Now bore, more specifically what we've just drawn out is the phase contrast transfer function. Meaning that, it tells us how much contrast is transferred into the image, as a function of spatial frequency. For each of these spatial frequencies, it answers the question. How much of that Fourier component is actually transferred into the final image? And, as we can see, some, some of the, the spatial frequencies, like this one, is transferred fully at a value of 1. Others are entirely invisible, where these spatial frequencies where the contrast transfer function crosses 0. These are spatial frequencies that are essentially missing in the image. And, there are spatial frequencies, for instance, this spatial frequency, where it's Fourier component is fully present in the image, but the contrast it contributes is actually in opposite direction of these other spatial frequencies. Now remember that each of the scattered waves represents a particular Fourier component of the density present in the original sample. Remember when we took a complex function and took its Fourier transform that separated into all of the Fourier components. There were some low frequency waves. For instance, a wave that might have oscillated once across the, the unit size of the sample. There is another higher frequency component that oscillated twice. There were others that oscillated four, five times, and the, the most extreme spatial frequency that can be present in a digital image is a wave that oscillates up and down every other pixel. And so each of these scattered waves represents one of those sine waves that is needed to produce a magnified image of the sample in the image plane. Now, do not confuse the spatial frequency of the information that those waves carry with the frequency of their oscillation as they traverse through the microscope, these are different frequencies we're talking about. It's a single electron that comes through the column one at a time and hits the sample. That single electron is scattered into many different directions. Some of it remains unscattered and un phase shifted. Other parts of it are scattered into each of these possible different directions, and in fact all of the directions around the full circle. And each of these components has exactly the same wavelength as the unscattered beam, because it's all part of the same electron. And that electron has a wavelength set by its energy. For instance, in a 300 kilovolt microscope the wavelength is approximately two picometers. And so as it travels down the microscope column whatever it is about an electron wave that's oscillating, the wavelength there is about two picometers. And so the wavelength of all of these beams is the same. They are phase shifted relative to each other, because of the scattering event. And because they're different path lengths. By, they're brought together by the lens to re-interfere at the image plane. Because they have the same wavelength, they add up. Just like in the previous slide where we had two different wave functions that had the same wavelength. But they were phase shifted with respect to each other slightly. And because of that, some of them added significantly to the sum and others did not add significantly to the sum. Whether they add or not, this is dependent on the phase shifts that are created by the electron microscope through the scattering event and the difference in path links. So, all of these have the same wave length because they're part of the same electron, and when they recombine in all of the individual pixels of the detector. In each pixel there is a probability of that single electron being detected. The probability is proportional to the amplitude of the wave function at that position, squared. And whether or not each of these scattered components. Contribute significantly to that sum depends on the scattering angle and the phase shift that they,that, that they suffer as they travel through the microscope. Some of them contribute fully. Others are totally lost to the image. And others contribute in a negative sense or at least in opposite sense compared to what the others are contributing. But each one is carrying a separate Fourier component of the ultimate image that needs to be present. Now to try to help make this a little clearer let me compare it to, say, an orchestra. In an orchestra you have all kinds of different instruments. You have the violins playing the high notes. You have cellos playing medium notes. And you have basses playing the low notes. And you can be in the audience and you can be listening to the orchestra. And you hear all the high, and the medium, and the low notes all combined. That would be like a perfect microscope that delivered all of the spatial frequencies at their full volume. The contrast transfer function of a perfect microscope would look like this. Meaning that all the spatial frequencies were delivered at full contrast, one. And so for instance, the violins are here at high frequency, the cellos are at some intermediate frequency, and the basses are at low frequency playing the low notes. And when you listen to the orchestra, they all come to you and so you hear them all. Unfortunately in an electron microscope, not all of the components of the signal are transferred or delivered into the final image. Some of them are fully transferred, others are silenced, and others are transferred with opposite contrast. In the orchestra analogy, this would be like putting a microphone above say, the cellos so that you hear the cellos fully. But putting another microphone by the basses and the sound that they create actually instead of adding that sound to the total so now subtracting it away so that it comes in a negative sense. And finally not microphoning at all, say the violins. The violins, without a microphone, would be like a position here in spatial frequency where you don't hear them at all. So contrast transfer function tells us how much of each of these components are going to arrive at the image. Just like placing the microphones in different positions in an orchestra tells you which of the instruments we hear strongly and which will be silenced. Now the exact form of the contrast transfer function can be derived exactly considering issues like the additional path length suffered by each of the scattered beams. And you can find that in books. And its derived form is that the contrast transfer function turns out to be the sine of pi times a term defocus times the wavelength of the radiation, in this case the incident electron. Times spacial frequency squared. That's this spatial frequency plus pi times the coefficient of spherical aberration of the objective lens times wavelength cubed times spatial frequency to the fourth power divided by 2. This is the derived form of the contrast transfer function.
