Electrons appear in electron clouds of atoms but positrons are found in cosmic rays and radioactive elements. Why do electrons seem more common then positrons?  Edit: Is this an example of asymmetry? I haven't studied it much yet.
120991
Baryon asymmetry refers to the observation that apparently there is matter in the Universe but not much antimatter. We don't see galaxies made of antimatter or observe gamma rays that would be produced if large chunks of antimatter would annihilate with matter. Hence at early times, when both were present, there must have been a little bit more matter than antimatter. This is quantified using the asymmetry parameter  $\eta = \frac{n_{baryon} - n_{antibaryon}}{n_{photon}}$  From cosmological measurements such as WMAP,  $\eta \approx (6 \pm 0.25) \times 10^{-10}$  However, the source of baryon asymmetry is said to be one of the Big Problems of Physics.  What is currently the state of the art regarding this puzzle? What's the best fit we can get from the Standard Model? What do we get from lattice simulations?
40574
How does  work? I'm wondering why matter and antimatter actually annihilates if they come into contact. What exactly happens? Is that a known process? Is it just because of their different charges? Then what about neutrons and anti-neutrons? I guess that would be about their quarks, but wouldn't it be very unlikely what all of their individual quarks would actually touch the anti-quarks of the other (anti-)neutron at exactly the same time?
117671
Is there any physical matter which won't appear on x-ray i.e. invisible matter? I think I learnt at university that all phsyical matter appears on x-ray and there is no matter that can be invisible. Is this true? For instance, does dark matter and antimatter appear on x-ray or what technology is used to prove that there is something instead of nothing?
4258
Suppose we have an atom. It is commonly said that because of the PEP, two electrons can't be in the ground state unless they have opposite spins, because no two electrons can have the same wavefunction.  What bugs me is that spin up and spin down aren't the only possible spin states. There's a whole continuum of linear combinations of them, and as far as I can tell the PEP wouldn't exclude the possibility of having lots of electrons, all sharing the same spatial wavefunction but with different combinations of $\mid\uparrow \rangle$ and $\mid\downarrow\rangle$. Why doesn't this happen?
4862
While studying the semi-empirical mass formula for nuclei, I came across an "asymmetry term" whose function, as far as I understand, is to build in the fact that nuclei "prefer" to have equal numbers of protons and neutrons. This is explained by the Pauli exclusion principle; the neutrons and protons are distinguishable, and hence have fill separate energy levels. Hence, if the number of neutrons and protons is close, the nuclei will have a smaller energy.  Yet, it is observed that large stable nuclei have more neutrons than protons. Why is this so? If nuclei prefer to have equal protons and neutrons, shouldn't the stable nuclei lie along the N=Z line? (in the image)
57443
In an hydrogen-like atoms the orbitals are solutions to the Schrodinger equation suitable for the problem.  They describe the regions where an electron can be found.  So, why don't they have spherical symmetric?  As in this picture:    Why, in a certain distance from the nucleus, are there points where an electron can be found, and points where it can't?  EDIT: Is there any intuation about the symmetry breaking?
28991
Until yesterday I thought they have. Googling a little bit around the measured values, I've found nothing.  As I've heard on the , the nuclei don't have an EDM. Although it would be possible by the symmetry violations of the weak interaction, it doesn't happen.  But I am asking not for the EDM due to weak asymmetry. I am asking for the lack of the EDM due to the asymmetric charge distribution in them.  Nuclei have a lot of elementary particles, interacting in quite complex ways (strong, EM and weakly). Why should the charges be distributed in them always to a 0 EDM? What is the mechanism, or what is the reason, to zero out any EDM in them?
123102
Given the wavefunction, $\psi$ is explained as the flow of probabilities or in other words probability density over a certain region of space. In the case of electrons, say in $s$ orbital, the probability of its existence is distributed over the whole spherical region. Now, when we talk about orbits is it true that the electrons are literally orbiting around the nucleus or are they just chaotically whizzing "around" with no sense of direction unlike we'd classically understand an electron to "orbit" around a positive nucleus.
30216
When we first discovered the proton and neutron, I'm sure scientists didn't think that it was made up of quark arrangements, but then we figured they could be and experiments proved that they were.  So, what is it about the electron that leads us to believe that it isn't a composite particle? What evidence do we have to suggest that it it isn't?
51988
Ordinary matter and antimatter have the same physical properties when it comes to, for example, spectroscopy. Hydrogen and antihydrogen atoms produce the same spectroscopy when excited, and adsorb the same frequencies. The charge does not make a difference in the potential (regardless if it's generated by a proton or an antiproton) nor in how the positron behaves in this potential (being its mass equal to the mass of an electron)  How can astronomy evaluate if a far galaxy is made of matter or antimatter, given that from the spectroscopy point of view, they behave in the same way? In other words, how do we know that an asymmetry exists between matter and antimatter in the universe?
53679
It's an old-new question (I found only one similar question with unsatisfactory (for me) answer: )  It's strange for me how all books simply pass by such an important question, and mentioning strange and mathematically unsupported reasons such as:   orbits are stationary (while as I know this is just idealization, there is no stationary orbits in reality even for Hydrogen)  electrons are actually not localized due to uncertainty principle, thus they have no acceleration (while obviously in a non-spherically symmetric orbits a kind of "charge acceleration distribution" always exist) vacuum fluctuations play a major role (according to QED).   I'm not interested in how Bohr or Schroedinger explained it, I want to see a rigorous proof with QM, QED or maybe even the standard model as whole. I would like to see how this question was closed.
121361
It is well established that:   'In quantum mechanics, the behavior of an electron in an atom is described by an orbital, which is a probability distribution rather than an orbit.   There are also many graphs describing this fact: : (hydrogen atomic orbital - one electron) In the figure, the shading indicates the relative probability to "find" the electron, having the energy corresponding to the given quantum numbers, at that point.  My question is: How do these symmetries shown in the above article occur?                 What about the 'preferable' axis of symmetries? Why these?
56706
Quark doesn't constitutes more fundamental particle and proton and neutron consist of quarks. Now come to beta decay.      $n \rightarrow p + e^{-} + \bar{\nu}_e $    How can  an electron  emit from an neutron whereas electron is a fundamental particle? Is the electron created from gluon?
121848
Experiments used to observe particle spin properties (such as ) rely on a varied magnetic field and a dipole-like reaction in the particle, deflecting it in one direction or another.  In the case of a point-particle such as an electron, how is it explained that a dipole with spatially separated poles exists in a single point?
104175
I am a bit confused about the terminology concerning identical fermions. In quantum mechanics, identical fermions need to obey certain anticommutation relations i.e. have an antisymmetric total wavefunction (product of spatial and spin parts) under the exchange of two particles. But when exactly are two fermions considered to be 'identical'? As far as I understand, two electrons are identical in the sense that they for example have the same electric charge, mass and spin quantum number $s=\frac{1}{2}$. (These are shared by all electrons.) How about the secondary spin (or spin projection) quantum number $m_s=\pm \frac{1}{2}$? If in an atom we have two electrons with the same principal quantum number, orbital quantum number and magnetic quantum number, but different $m_s$, are these two electrons 'identical' or not? The electrons are definitely identical if they additionally have the same $m_s$, but the Pauli exclusion principle forbids that. How does the consideration of the quantum numbers fit with the electrons having the same electric charge, mass, and spin $s$, which are independent of atomic orbitals? Furthermore, are 'identical' and 'indistinguishable' fermions different things or not? If 'identical' particles cannot be told apart by an experiment, how about the Stern-Gerlach experiment? This links back to the above question whether $m_s=\pm \frac{1}{2}$ electrons are identical. My partial answer is that if we have electrons labelled 1 and 2, we cannot say which of the electrons has $m_s=-\frac{1}{2}$ and which has $m_s=\frac{1}{2}$. So in an experiment, you cannot say that 'The electron with $m_s=-\frac{1}{2}$ is electron 1' even if you were able to separate the $m_s=\pm \frac{1}{2}$ states. Or something along those lines. In other words, electrons are identical regardless of the quantum numbers. Further confusion arises when we go beyond electrons. I have learned that atoms with an odd number of neutrons in an 'optical lattice' can be used to mimic electrons in a metal. Then the role of $m_s=\pm \frac{1}{2}$ is played by different hyperfine states of the atom as far as I can tell. If I now have atoms of the same mass (same species) in two different hyperfine states corresponding to spin-up and spin-down electrons, how are these atoms classified? Are they 'identical', 'indistinguishable', 'distinguishable', or what? Do these atoms obey the fermionic anticommutation relations? I thought I understood what's going on i.e. two atoms of the same species are identical but can be in different internal states representing the spin projections and thus the atoms obey the usual fermionic anticommutation relations of spin-$\frac{1}{2}$ electrons. But then I found this paper () which says that atoms in the $| F=\frac{1}{2}, m_F=\pm \frac{1}{2} \rangle$ states of $^6$Li are distinguishable. How can distinguishable atoms play the role of electrons? How can they obey the anticommutation relations of spin-$\frac{1}{2}$ electrons? If I have understood correctly, there are no symmetry requirements for distinguishable particles in quantum mechanics. Or are the authors just possibly using sloppy language by calling two identical atoms in different internal states distinguishable? I would call two atoms of different species, e.g. $^6$Li and $^{40}$K, distinguishable. Sorry for the long text, but hopefully I made my confusion sufficiently clear. Any clarifications are appreciated.
53576
The mass of a neutron is greater than mass of a proton so how is it possible in positron emission for a proton to form a neutron and a positron?
53276
I am a beginner in QED and QFT.  What is known (or expected to be) about bound states in QED? As far as I understand, in non-relativistic QM electron and positron can form a bound state. Should it be true in QED? Most of the basic text books on QFT I known treat QED with perturbative methods which do not help to study bound states. Is there a literature to read about it?  UPDATE: According to , electron and positron form a positronium. It is an unstable particle which can annihilate to two photons. In particular electron and positron cannot form a bound state. But still there is a logical possibility that several electrons and positrons can form a bound state (though probably once one has at least one electron and one positron, they will necessarily annihilate. Is it true?)  On the other hand, if one considers QED with both electrons and muons, then electron and antimuon form muonium. Clearly they cannot annihilate to photons. However in  in wikipedia it is claimed that muonium is unstable. Is it due to some effects of QED or due to actual presence of other, say weak, interactions?