Thanks for the great explanation! If you don't mind, I had a question about '3.' I get how it's used in making your larger argument, but as a thing in itself I'm kinda lost. Part of what I'm wondering is how it's okay to say 'L(W) = 0' when earlier you described L(W) like 'L(W: wavefunction) => X: wavefunction'; so is the zero in 'L(W) = 0' just shorthand for something like 'L(W) = (f(x,t) = 0)'?(I'm also curious why mapping to the zero function ends up being the central criterion for physical possibility—but I'm guessing that's a rather deep subject ;))Edit: NVM I see what's going on with 'L(W(x,t)) = 0' now: the equation is equating the full evaluation of the nested functions on the left (using the x,t params) with zero. When written like 'L(W) = 0' it looks like it's equating the function returned by L(W) with zero.

 >it looks like it's equating the function returned by L(W) with zeroThere is a deeper point to be made here that I'm glad you brought up. Functions form a vector space (because they satisfy the axioms of vector behavior, basically because they can be added to each other and scaled by constant multiples). In linear algebra the symbol 0 often does double-duty as the zero vector, which is defined as the vector that doesn't change other vectors when it's added to them. So, here, when I write L(W) = 0 I'm implicitly invoking 0 = f_zero(x,t) = 0.As for why "mapping to zero" has a physical basis, well, it's really more of a thing we're always guaranteed to be able to do. You can always subtract everything from the right-hand side of an equation! For example, Wikipedia introduces the one-dimensional wave equation as D_t^2 u = a^2 D_x^2 u. I can also write that as L[u] = D_t^2 u - q^2 * D_x^2 u = 0, so L[u] = 0. (In my notation, D_x is the derivative with respect to x, and D_x^2 is the second derivative with respect to x.)The real question is why the addition thing works; if I had to try explaining it I would just say it's just fundamental that Maxwell's equations are linear, and when dealing with things that aren't, we usually approximate them with linear functions anyways[0]. That's how gravitational waves emerge from GR, by the way: at low energies the nonlinear equations behave nearly linear, and in that approximation the familiar wave equation falls out.[0] If you zoom in to a small enough range in the graph of all but the most esoteric functions, the thing on your screen will look like a line. Try it, it's a good intuition to have.
 Okay, that makes a lot more sense. Thanks. I missed this in your original phrasing, "We can SET UP operators L(W) so that they..."—but I think I see now: equating L(W) with the zero function is algebraically convenient, since we want to solve for an unknown function/vector, and using zero there allows us to be 'agnostic' about what that unknown function/vector is (any other value would 'say something' about the unknown).> The real question is why the addition thing worksUnfortunately I'm still at the point where I can't see why it should be surprising that it works. I'm assuming that by the 'addition thing' you are referring to the fact that adding two wave functions always produces another wave function—or maybe it's something about the characteristics of the wave function produced through adding? I'm not sure how linearity plays into things here. Maybe it's surprising that it's possible to form a linear operator (I'm assuming the "operator" you mentioned is this: https://en.wikipedia.org/wiki/Linear_map) for wave functions? I guess not though since it's probably just using the structure of those functions as vectors and it doesn't matter what they're 'about'. Nope, not sure :)
 While the wavefunctions are vectors by virtue of being functions, it still remains to be seen that the set of "physically possible" wavefunctions is also a vector space. For example, if the physics were such that we were equating the local kinetic energy of a string to, I don't know, it's position or something, we would end up with an equation looking like (D_t f(x,t))^2 - f(x,t) = 0. It still can be written as an operator (not all operators are linear), but the addition thing wouldn't work anymore. It's a remarkable pattern in physics that those squared terms tend not to appear in reality.

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