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Sep
8
comment Term symbol - how do we know the number of electrons $e^-$?
But in the book A. Beiser, Concepts of modern physics it says that multiplicity is $2S+1$ only in cases when $L>S$ while in cases when $L<S$ the multiplicity is $2L+1$. This is why we can't just state $4=2S+1$ as it could be $4=2L+1$... so can this even be solved?
Sep
8
asked Term symbol - how do we know the number of electrons $e^-$?
Sep
8
comment LS Coupling - weird image in the book
So i did this like i should but forgot that there is no such thing as negative magnitude... so from all the solutions for $L=3\hbar, 2\hbar, 1\hbar, 0, -1\hbar, -2\hbar, -3\hbar$ i have to simply remove negative ones and 0.
Sep
8
comment LS Coupling - weird image in the book
Why do you all use $l$ (orbital quantum number) in place of $L$ (orbital angular momentum)? In the book it only says that $L$ can be 3 2 or 1 and not -3, -2, -1...
Sep
8
revised LS Coupling - weird image in the book
added 42 characters in body
Sep
8
asked LS Coupling - weird image in the book
Sep
8
accepted Total angular momentum - single electron
Sep
8
comment Total angular momentum - single electron
This means that the $z$ axis must be defined in the same direction as the total angular momentum $\mathbf{J}$.
Sep
8
comment Total angular momentum - single electron
Spin magnitude of an electron is allways the same $S=\sqrt{s(s+1)}\hbar=\frac{\sqrt{3}}{2}\hbar$ where $s$ is the spin quantum number which for electron equals $1/2$. By $s$ you probably meant the $z$ component of the spin $S_z=m_s\hbar$ where $m_s= \pm s = \pm\frac{1}{2}$? So in one case $S_z = \frac{1}{2}\hbar$ while in the other case $S_z = -\frac{1}{2}\hbar$?
Sep
8
asked Total angular momentum - single electron
Sep
2
comment positron / electron annihilation - where is the invariant?
Thank you. So as long as I stay in the same coordinate system there is no need to use the equation $E^2={E_0}^2+(pc)^2$. But I have allso been dealing with systems where we used this equation within the same coordinate system to connect the momentum $p$ and energies - in this case it comes quite handy. Is my thinking ok now?
Sep
2
revised positron / electron annihilation - where is the invariant?
added 2 characters in body
Sep
2
comment positron / electron annihilation - where is the invariant?
I am sorry but I don't understand. Is it different fi I say the Lorentz invariance is: $E^2={E_0}^2+(pc)^2$ or if I say: ${E_0}^2 =E^2-(pc)^2$? It is the same equation. Is above example solved correctly?
Sep
1
asked positron / electron annihilation - where is the invariant?
Sep
1
revised Kaon spontaneously splits into two pions (just need an confirmation)
deleted 2 characters in body
Sep
1
comment Kaon spontaneously splits into two pions (just need an confirmation)
Sorry - it was a mistake which has now been fixed :)
Sep
1
revised Kaon spontaneously splits into two pions (just need an confirmation)
added 275 characters in body
Sep
1
comment Kaon spontaneously splits into two pions (just need an confirmation)
Thank gosh! I am so happy now, because i think i understand this :). What about if the kaon would spolit into two different particles with different mass and speeds. Look at the EDIT.
Sep
1
asked Kaon spontaneously splits into two pions (just need an confirmation)
Aug
23
accepted Transition integral from 1-D cartesian into 3-D polar coordinate system