Timeline for Toy Models of Quantum Mechanics
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2011 at 9:33 | comment | added | Roland Bacher | The easiest example of a non-diagonalizable matrix is perhaps the non-zero nilpotent matrix $\left(\begin{array}{cc}1&1\\1&1\end{array}\right)$ over the field of two elements. | |
| Mar 20, 2011 at 20:18 | comment | added | Chris Heunen | I'm not sure this is a problem. In standard QM the spectral theorem means that observables just happen to coincide with selfadjoint operators. This is of course mathematically very handy, but not essential to modelling observables mathematically in the first place. I'd rather conclude that "nonstandard toy models of QM" might need to model observables by adapting a different formulation, such as POVMs, instead of selfadjoint operators. | |
| Mar 20, 2011 at 16:17 | comment | added | Chris Godsil | The fact that $\mathbb{Z}_2$ is not algebraically closed is irrelevant - a symmetric matrix over the algebraic closure of $\mathbb{Z}_2$ need not be diagonalizable. For example if $$ A= \left(\begin{array}{ccc}0&1&0\\ 1&0&1 \\0&1&0\\ \end{array}\right) $$ then the minimal polynomial of $A$ in characteristic two is $x^3$, and so $A$ cannot be diagonalizable. | |
| Mar 20, 2011 at 7:38 | comment | added | Robert Israel | Also a 2 x 2. The point is that Z_2 is not algebraically complete: the polynomial z^2 + z + 1 has no roots in it. So e.g. the matrix [ 1 1; 1 0] has no eigenvalues in Z_2. | |
| Mar 20, 2011 at 5:43 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 2.5 |