Therefore, it must be true that $g_n = 0$, so that $f(\xi) = 0$. However, it is clear from this expression that, unless $g_n = 0$, $f(\xi)$ will be a polynomial of degree $n$, which would contradict the earlier observation that it can only be of order $n-1$ at most. Select true when prompted to set the value and click OK. In the Enter the preference name field, enter and click OK. Where in the last line $g_n$ must be a constant because we have finished factoring. Right-click Hold down the control key while you click in the list of preferences, select New, and then select Boolean. Suppose $\xi$ is a real (self-adjoint) linear operator that satisfies the "algebraic equation" The argument leading up to this is as follows. I am hoping that someone can explain whatever it is that I am missing. 34), there is a sentence that is very confusing to me. At the end of Section 9 in Dirac's Principles of Quantum Mechanics (p.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |