In literature on an introduction to quantum mechanics which I am working through, there is a section which explains that a vector has different representations based on the basis you choose. It then makes a statement that
The same is true for the state of a system in quantum mechanics. It is...
@Obliv Not as far as I know. I didn't get anything for reaching 100K except a congratulations e-mail from the SE guys. Basicaly once you're at 20K that's as as regards extra stuff.
@ACuriousMind no-one believes I could be that handsome and distinguished? :-)
@Obliv that is me in the picture, though the picture is twenty years old and I have since lost the face fuzz.
@Obliv I used to do amateur operatics back then, and that was me in costume for Die Lustige Witwe by Franz Lehar. The operetta is set at around the beginning of the 20th century, so yes it was in the heyday of classical physics.
@0celo7 assume $v$ is a normalized eigenvector with real eigenvalue $\lambda$. We have $\lambda=v^t A v$. Therefore, $\lambda=\lambda^t=v^t A^t v=-\lambda$.
for a "physicist proof", we could try to think what would happen if $F^{\mu\nu}$ had non-zero eigenvalues
you know how the prime factorization of $a$ and $b$ is $a = {p_1}^{\alpha_1}{p_2}^{\alpha_2}...{p_t}^{\alpha_t}$ and $b = {p_1}^{\beta_1}{p_2}^{\beta_2}...{p_t}^{\beta_s}$ and the G.C.D of them is $(a,b) = {p_1}^{min({\alpha_1},{\beta_1})}{p_2}^{min({\alpha_2},{\beta_2})}...{p_s}^{min({\alpha_s},{\beta_s})}$? How come it says minimum g.c.d?
The reason it's not obvious is that in number theory $(a,b)$ is also used for lcm or gcd or something so it can be confusing as all of number theory is
I don't know what you mean by ''algebraically''. I'll show you a proof.
I write $(m,n)$ for $gcd$ and $[m,n]$ for $lcm$. If $(m,n) = 1$, then $m$ and $n$ both divide some integer $r$ if and only if $mn$ divides it (easy consequence of the Euclidean algorithm); it means that $[m,n] = mn$. Otherw...
maybe you got into a horrific accident without knowing.. or you just didn't sleep for a year straight. I don't know many causes for brain damage admittedly
This is a snippet of code from an array library I'm using. This runs fine on windows, but when I compile with gcc on linux if crashes in this function. when trying to narrow down the problem, I added a printf statement to it, and the code stopped crashing.
void _arrayCreateSize( void ***array, i...
Possible Duplicate:
Common reasons for bugs in release version not present in debug mode
Sometimes I encouter such strange situations that the program run incorrectly while running normally and it will pop-up the termination dialog,but correctly while debugging.This do make me frustrat...
@BernardMeurer Yes I read that earlier. I'm sure your recursive stack size is limited. Consider a global variable that increments when you call a recursion and decrements on return. See how deeply you're nesting your stacks.
--num-callers=<number> [default: 12] By default, Valgrind shows twelve levels of function call names to help you identify program locations. You can change that number with this option. This can help in determining the program's location in deeply-nested call chains. Note that errors are commoned up using only the top four function locations (the place in the current function, and that of its three immediate callers). So this doesn't affect the total number of errors reported.
The maximum value for this is 50. Note that higher settings will
Also look at the --main-stacksize option and consider changing your ulimit.
espaco@Arch ~> ulimit -s unlimited espaco@Arch ~> ulimit -a Maximum size of core files created (kB, -c) unlimited Maximum size of a process’s data segment (kB, -d) unlimited Maximum size of files created by the shell (kB, -f) unlimited Maximum size that may be locked into memory (kB, -l) unlimited Maximum resident set size (kB, -m) unlimited Maximum number of open file descriptors (-n) 1024
In literature on an introduction to quantum mechanics which I am working through, there is a section which explains that a vector has different representations based on the basis you choose. It then makes a statement that
The same is true for the state of a system in quantum mechanics. It is...
I don't know what you mean by ''algebraically''. I'll show you a proof.
I write $(m,n)$ for $gcd$ and $[m,n]$ for $lcm$. If $(m,n) = 1$, then $m$ and $n$ both divide some integer $r$ if and only if $mn$ divides it (easy consequence of the Euclidean algorithm); it means that $[m,n] = mn$. Otherw...
Possible Duplicate:
Common reasons for bugs in release version not present in debug mode
Sometimes I encouter such strange situations that the program run incorrectly while running normally and it will pop-up the termination dialog,but correctly while debugging.This do make me frustrat...
