I know very little about quantum physics, but I just can't wrap my head around the fact that non-spherical orbitals appear to be "oriented", constituting a set of fixed "frame of reference axes".  Does it really work like that? Are there for a single atom given directions in which the orbitals "protrude"? And if so, what are they based on? If I understand it right, the nucleus is seemingly spherical - symmetrical with no unique axis of symmetry. What would then decide the orientation of the orbitals in space (with respect to the surroundings)?  Edit: I understand any orientation works to satisfy the wave equation. I was more curious why the atom would choose one direction, instead of any of the others. Is that fundamentally random? And is this setup retained further, or can it change at any time?
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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?
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I have read , and am still confused about the entire matter.  My understanding is that, using a local coordinate system, if you rotate a model about its X axis then its Y and Z axes experience gimbal lock, and become "locked" together.  However, my understanding is that if you rotate the X axis, then both the Y and Z axes rotate with it, and as such remain separate.  Wikipedia, as always, while a good resource is difficult to understand on this topic.  Could someone please endeavor to explain this to a person who does not have a Master's in physics or mathematics?
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I need to get a nice picture about how electron moves around nucleus? I find concept of probability and orbitals quite difficult to understand?
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Atomic  are used to notate the angular momentum content of the electronic states of an atom, and are normally written down as $$^{2S+1}L_J$$ where the state has total spin $S$, spin multiplicity $2S+1$, total angular momentum $J$, and total orbital angular momentum $L$.  Occasionally, though, one comes across term symbols which cannot fit in this scheme, because they have a half-integral "$L$", such as $$^3[3/2]_{1/2}.$$ (For an example, see  () or the ).   In addition, one can sometimes see even weirder beasts with both an $L$ letter and a half-integer in square brackets, such as () $$^1D[3/2]_{1/2}.$$  Neither of these types of term symbols are covered in normal quantum mechanics textbooks even in passing, even though the states do make an appearance now and then. What do these term symbols mean, and how does one interpret them?
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If I understand correctly, spin is an intrinsic property of particles, which follows the algebra of angular momentum, but has nothing to do with an "orbital angular momentum" in that the particle is not like a small sphere that rotates on itself, to which we could attribute an angular momentum as usual in classical mechanics, cf.  and  Phys.SE posts.   My question is: why does spin combine with orbital angular momentum to give total angular momentum? To me this is surprising, as spin, in fact, has nothing to do with a classical angular momentum. Any insight on that?
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I often hear about subatomic particles having a property called "spin" but also that it doesn't actually relate to spinning about an axis like you would think. Which particles have spin? What does spin mean if not an actual spinning motion?
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I was looking at a plot of hydrogen radial wave functions $r^2|R_{nl}(r)|^2$, and I noticed that for fixed $n$, the states with smaller $l$ were "further out" from the nucleus.  This conflicted with my classical notion that the centrifugal force should push things with high angular momentum outwards.  And indeed, the expectation value of $r$ is  $$\tag{1} \langle r\rangle = \frac{a_0}{2}(3n^2-l(l+1)).$$  Is there a physically intuitive reason why lower $l$ states have higher $\langle r \rangle$, and higher $l$ states have lower $\langle r \rangle$?
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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?
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In (special) relativistic quantum mechanics there is a standard argument that says that the (rigged) Hilbert space of states $H$ should be equipped with a projective unitary representation $U$ of the Poincaré group $P$ that goes something like this:  Suppose we have two observers $O_1$ and $O_2$ and $x\in P$ is the Poincaré transformation that maps $O_1$s coordinate system into $O_2$s coordinate system.  Then, if $O_1$ and $O_2$ attempt to measure the same thing, they will actual find different states, $\left| \psi _1\right&gt;$ and $\left| \psi _2\right&gt;$ respectively, in $H$ (for example, if you first find the vector $(1,0,0)$, rotate your axes by $\pi/2$ and measure the same vector, the numbers you now record are going to be $(0,1,0)$).  In fact, this gives us a map of states:  $\left| \psi _1\right&gt; \mapsto \left| \psi _2\right&gt;$ (it's only going to be well-defined up to phase).  Let us call this map $U(x)$, so that $\left| \psi _2\right&gt; =U(x)\left| \psi _1\right&gt;$.  However, if we accept the principle of relativity that physics should be the same in two reference frames related by a Poincaré transformation, then we better have that $$ \left| \left&lt; \phi _1|\psi _1\right&gt; \right| =\left| \left&lt; \phi _2|\psi _2\right&gt; \right| =\left| \left&lt; U(x)\phi _1|U(x)\psi _1\right&gt; \right| $$ for all (normalized) $\left| \psi _1\right&gt;$ and $\left| \phi _1\right&gt;$ because this represents a probability (of course, $\left| \psi _2\right&gt; :=U(x)\left| \psi _1\right&gt;$ and $\left| \phi _2\right&gt; =U(x)\left| \phi _1\right&gt;$).   then tells us that this gives a projective unitary representation of $P$ on $H$.  (Note:  I am allowing some elements of $P$ to be represented by antiunitaries.)  Now, if you try to do the same argument in curved spacetime, you run into the obvious problem that you will not in general have an analogue of the Poincaré group (I believe there exist spacetimes which possess no Killing fields); however, it naively seems as if the principle of general covariance suggests that we should 'upgrade' the isometry group of spacetime to the entire diffeomorphism group of the spacetime in the above argument.  (In particular, I don't see how this argument makes crucial use of the fact that both observes coordinate bases are orthonormal.)  That then, would imply, that the Hilbert space of states in a quantum theory of curved spacetime should possess a projective unitary representation of the diffeomorphism group of that spacetime.  This, however, at first glance, seems like it would be false.  So, where does the argument break down if you replace the Poincaré group in special relativity with the diffeomorphism group in general relativity?  Or, is it in fact the case that we should obtain a unitary representation of the entire diffeomorphism group.
59688
Why is angular momentum conserved when a planet revolves about sun in an elliptical orbit? Why is linear momentum not conserved in this case?  Please use the minimum amount of equations and try to explain using pure physics logic if possible.
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I got this problem from a journal and curious how they have calculated that.Previously I asked this problem and unfortunately did not get any answer. Hereby I am posting again and hopefully someone can answer as it is a popular platform to understand math. Consider I have a sphere that is equally divided into two different patch(gray and white). The total gray portion can be called as Pt The sphere can rotate and translate(up-down) at the air-water interface.  The angle α defines the position of the sphere (alpha increase means the sphere goes towards the air). The angle β defines the rotation of the sphere. So the surface area of any patch into any phase changes with the position and orientation(example: area of Pt inside air phase will change with α and β angle). Now the calculation of the surface area of Pt part of the sphere in air(Apt-A) is given by this equation.  I understand the first integral limit in polar angle direction but did not find out how the limit comes for azimuth angle direction. Please keep in mind that the picture is only visual representation of a 3D sphere in 2D.
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Let's say I have eigenstates $|x\rangle$ associated with measurement of position. I know that the eigenstates corresponding to their respective eigenvalues form a basis, let's call it $A$. Now let's say I want to expand a generic state vector $|ϕ\rangle$ in that basis formed by eigenstates. Now, what does it mean when physicists say that:     "$ϕ(x)$ is the coordinate of the state $|ϕ\rangle$ in the basis $A$?"    Any explanation or reference to similar kind of questions would be helpful.
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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.
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My intuition has always been that when any sphere is projected into 2D space that the result will always mathematically be an ellipse (or a circle in degenerate cases).  In the past when I was actively doing my own graphics programming and brought this up with other people they were adamant that I was wrong. If I recall correctly they believed the result could be something vaguely "egg shaped".  Who was correct?    Since there is already one answer submitted, I don't wish to totally change my question but I realize I left out important details due to losing familiarity with the field over the years.  I intended to ask specifically about perspective projection where the projection is a linear application.  The other projections are of course interesting for many uses so I wouldn't want them removed at this point. But it would be great if answers could have perspective projection as their most prominent section.
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I read Griffiths EM today and it says something very interesting but a little bothering to me. It states for an atom, the position of center of mass of an electron cloud lies in the center of the proton.  But my question is why? What will happen if this is not true?
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The solutions of the Schrödinger equation for hydrogen are the "electronic orbitals", shown in this picture:  ()  They have the following degeneracy structure:  ()  It is often said that atoms in other elements simply have more electrons "filling up" the orbitals in increasing order of energy, as required by the Pauli exclusion principle.  This is somewhat confirmed via xray spectroscopy:  ()  But why would non-hydrogen atoms have the same orbital degeneracy structure as hydrogen, if the actual Hamiltonian is different from one atom to another?
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I came across a discussion of atomic Rydberg states, which appear to be defined as having their outer electron in a highly excited state. In the same text, it mentioned that these excited electrons could be modeled rather well by the Bohr model.  I don't see why this should be the case. The Bohr model treats the electron as a circularly orbiting body around the nucleus, whereas the electron cloud model seems to produce orbits which are highly anisotropic. In my understanding, the Bohr model also proposes electron orbits with fixed radii.  I don't see how this can be deduced from taking the limit of large $n$. How is Bohr's model related to electron cloud models via the correspondence principle?  On a further side note, I am curious why the different orbital energy splitting as a result of atomic effects (Fine structure, Hyperfine etc) which causes non-degenerate orbitals wrt $\ell$ is ignored for high $n$, where orbital energies are only dependent on the principal quantum number.  References:   Randall G. Hulet and Daniel Kleppner, Rydberg Atoms in "Circular" States, Phys. Rev. Lett. 51 (1983) 1430  R.J. Brecha, G. Raithel, C. Wagner, H. Walther,  Circular Rydberg states with very large n,