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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