And where I'm going to integrate over all the transverse components. And of course, I need to give them some different amplitude to which one. So I'm thinking like Fourier again. You can decompose anything into things that oscillate. I just need to find what they are. So I'm going to put something here. Depends on these transverse things here. And I'm going to integrate only over D, new X, and D, new Y. Why not the C component? Why only X and Y? Because the C1 is determined by the X and Y. But if I write this again, substituting this form, then this is the same as the integral. And it's from minus infinity to infinity, A of nu, e to the minus i 2 pi nu dot X. And this part that I can write as e to the i, let me write this part as h of nu. The nu X, the nu Y, where h of nu is that exponential, e to the i, 2 pi. And then I can bring that lambda instead of the square root. So this gives me 1 over lambda squared minus nu X squared minus nu Y squared. Is that OK? And it is assumed that when this is bigger than 1, this is positive. When this is smaller than 1, the square root is positive imaginary. So that this whole thing, the case, times C, times C. Now, how do I find this A? Four minutes. How I find this A? So this h depends on nu and depends on z. What happens if I let z equal 0? What is the exponential equal to? 1. So notice that h of nu comma 0 is equal to e to the i, 0 equal to 1. So if I evaluate the field at the initial plane, u at X, Y, let me say z equal 0. And then what I have is the integral from minus infinity to infinity of A of nu e to the minus i, 2 pi nu dot X, the nu X, the nu Y, the h goes away. And what is this formula? I will see in this before. Is the inverse Fourier transform? The inverse Fourier transform of A is equal to the field at the first plane. So I can call this, let me call this the initial plane of X, Y. That means that A is the Fourier transform of the initial plane. So therefore, this A of nu, which is called the angular spectrum, is the Fourier transform. As X goes to nu X, let's say Y goes to nu Y of the initial field. So if I have a field and I know what it is at the initial plane, I Fourier transform that. And that tells me what the weight of the plane waves is. From there to propagate, I just let this plane waves travel. In other words, then the field at X, Y, Z is now, this is the inverse Fourier transform of what? Of this? This whole expression looks now like an inverse Fourier transform of the product of this times h. So that is, and this is the Fourier transform of the initial field. So this is the inverse Fourier transform. As nu X goes to X, nu Y goes to Y of the transfer function times the Fourier transform. As X goes to nu X, and Y goes to nu Y of the initial field. So to propagate the field from a known initial plane, the plane waves that the professor tells me, I just have to do the Fourier transform, multiplied by this simple transfer function, and the inverse Fourier transform. So it's again our signal and response. The signal is now what? It is the initial field, and the response is the final field after propagation. And transfer function is this phase that accumulates in the propagation or the exponential decay. So the plane waves give us this connection between solving the Hamilton's equation and Fourier theory. So I'm going to start from this tomorrow, and try to interpret again the uncertainty relation and all these properties in terms of this, and show you on the computer how this can be used to propagate and all the properties, and then we'll go into measuring the resolution of an imaging system, et cetera. I just wanted to get to the punch line before we leave. All right, so coffee break, and at 4, please be here for Professor. Sorry? It's about being free for us in protein. OK, so about that. Green Flores in proteins. Very good.