Conversation started May 10, 2016 at 10:55.
user116211
May 10, 2016 10:55
Okay, isn't Hamiltonian self-adjoint operator?
user116211
At-least I know so... it has to follow unitarity... isn't it?
user116211
But while googling, I stumbled across this paper cited by over 250:
user116211
Must a Hamiltonian be Hermitian?
Slereah
Slereah
It is.
user116211
Okay, did I learn wrong?
Slereah
Slereah
May 10, 2016 10:58
Otherwise you will have non-real energies
yuggib
yuggib
May 10, 2016 11:10
@MAFIA36790 self-adjoint and symmetric(hermitian) are not the same
for an operator to have real numerical range it is sufficient that it is symmetric
for it to be the generator of a unitary group of evolution (dynamics) it has to be self-adjoint
user116211
sorry, got offline... @yuggib o/
yuggib
yuggib
\o
user116211
@yuggib ooh.
yuggib
yuggib
@MAFIA36790 and you should take things said by people that do not know the difference between symmetric and self-adjoint with a grain of salt
and very carefully
user116211
I have @yuggib ;D
yuggib
yuggib
May 10, 2016 11:17
quantum mechanics is not a theory of matrices
with accidentally an infinite number of entries
user116211
@yuggib o.O
yuggib
yuggib
it is a wonderful and rich theory with plenty of counterintuitive stuff
that is nevertheless very important
user116211
Thanks @yuggib for the insight o/
user116211
Ah!
user116211
15
Q: Distinguishing between symmetric, Hermitian and self-adjoint operators

Josef K.I am permanently confused about the distinction between Hermitian and self-adjoint operators in an infinite-dimensional space. The preceding statement may even be ill-defined. My confusion is due to consulting Wikipedia, upon which action I have the following notion. Let $H$ be a pre-Hilbert spa...

functional-analysis
yuggib
yuggib
May 10, 2016 11:20
it is just that I find the sloppiness of physicists really stupid in this context
user116211
is hermitian different from symmetruc ;(
user116211
investigating
yuggib
yuggib
because they often end up overlooking some very relevant features of the theory
user116211
@yuggib ohh.
yuggib
yuggib
that distinction between hermitian and symmetric is not used in mathematical physics/analysis
at least as far as I know
simply the word "hermitian" does not appear
ever
Slereah
Slereah
May 10, 2016 11:23
"T is Hermitian if it is symmetric and bounded."
HINT THOUGH
Most operators are bounded in QM
Although I guess momentum wouldn't be hermitian then
yuggib
yuggib
@Slereah never read such a definition
Slereah
Slereah
that's what the math dude said
yuggib
yuggib
apart from the urreferenced math.se answer
Slereah
Slereah
What are the definitions you heard
yuggib
yuggib
I know the definitions of books like Kato, Yosida, Reed Simon, etc.
Slereah
Slereah
May 10, 2016 11:25
I do recall working with operators that were self-adjoints but not hermitians, though
yuggib
yuggib
and they never use the word hermitian
Slereah
Slereah
Like the momentum operator in a curved space
$p + \Gamma$ or something
user image
that's the one
$-i\hbar(\partial_a- \frac{1}{2} \Gamma_a)$
 
Conversation ended May 10, 2016 at 11:32.