hi!
Do you know about a good book that introduces relativistic quantum mechanics with one-forms (diferential forms) like the book a first course of general relativity of Schutz does?
if I apply a force constant F=100N en ##y## direction to a body of mass 2kg, it is elevated 3m and I know that the force of gravity is in ##y## direction too. What is the work that I did?
I think: ##W_{me}=(F_{me}-F_{g})\times 3## it is correct?
and the work made by gravity is "0".
Usually un physics triple or double integrals $$\int \int$$ $$\int \int \int$$
Are changed by only one simbol $$\int$$, so that is the same equation.
The simbol $$\partial v$$ means that the integral Is computed on boundary superfice of $$v$$ or on boundary of $$v$$.
$$\partial v=S$$
i know that
$$S_z |+ >= \frac{\hbar}{2} |+ >$$
also $$ D_z(\phi)= 1+i\frac{S_z}{\hbar}\phi$$ operator of infinitesimal rotations, if we apply this operator $$n$$ times:
$$ (D_z(\phi))^n= (1+i\frac{S_z}{\hbar}\frac{\phi}{n})^n$$ in the limit $$ lim_{n\rightarrow \infty}(D_z(\phi))^n=...
sorry!
for compute:
$$e^{\frac{iS_z\phi}{\hbar}}S_
x e^{\frac{-iS_z\phi}{\hbar}}$$
so, if we use $$S_x=(\frac{\hbar}{2})[(|+><-|)+(|-><+|)]$$
$$e^{\frac{iS_z\phi}{\hbar}}(\frac{\hbar}{2})[(|+><-|)+(|-><+|)] e^{\frac{-iS_z\phi}{\hbar}}$$
so, why that is equal to...
for compute:
$$e^{\frac{iS_z\phi}{\hbar}}S_x e^{\frac{-iS_z\phi}{\hbar}}$$
so, if we use $$S_x=(\frac{\hbar}{2})[(|+><-|)+(|-><+|)]$$
$$e^{\frac{iS_z\phi}{\hbar}}(\frac{\hbar}{2})[(|+><-|)+(|-><+|)] e^{\frac{-iS_z\phi}{\hbar}}$$
so, why that is equal to...
When F(r)= r
One get:
$${A \phi}(r)= r \phi{r}(r)$$ its a eigenvalue equation
when $$f(r)= -i \hbar \partial_r$$
One get:
$${A \phi}(r)= -i \hbar \partial_r \phi{r}(r)$$ it has no sense.
is the why is local or no local
am i right?
Note: i´m using $$A\phi (r)= \int dr´\phi(r) \alpha(r-r´)...
In Hartree-Fock method, I saw the Fock operator has two integrals: Coulomb integral and exchange integral. One can define two operator. "The exchange operator is no local operator" why? Whats de diference: local and no local operator?
And why do the operators have singularities?
thanks
I don't understand why grad(1/r)= (-1/r^2)ê_r. I know its derivate, but in cero the function is not definite and you can't find its derivative in cero.And, why div [grad(1/r)]=-4πδ(r vec) ?