9

For a BEC, you want atoms to be in the same quantum state, not necessarily at the same position. For a BEC, the temperature is low enough so that the de Broglie wavelength $\lambda_{\mathrm{dB}} \propto 1/\sqrt{T}$ is larger than the interatomic spacing $\propto n^{-1/3}$, $n$ being the density. This means that the wave nature of the atoms is large enough ...


4

Nobody knows the answer. All we know is that quantum mechanics is in perfect agreement with experiment. All you can do is to critically investigate any intuitive concepts that you may have that are incompatible with quantum mechanics.


4

The frequencies of atoms in a molecule will be different. First of all, the binding will results in splitting of frequencies in proportion to the rate with which electrons hop between the two atoms. Secondly, the electron-electron interaction between the electrons of different atoms will introduce corrections. Finally, there will be vibrational modes due to ...


4

Short story: it's zero, but with interesting physics to explore along the way. However, since it is in general not a product of single-electron wavefunctions the angular momenta of the two individual electrons are not defined (which makes it difficult for me to exploit symmetries for the integration of the two coordinates). In principle, yes, it's true ...


2

Nuclei, like molecules, have vibrational states, but the vibrational character of these states is not as pure for nuclei. You can model them as shape oscillations (usually ellipsoidal deformations about a spherical or ellipsoidal equilibrium shape), but in reality they're usually more like a slightly coherent superposition of a few different particle-hole ...


2

ESPECIALLY I WOULD LIKE for a given n (for example n =3) a graphics comparison of the different splitting for each case. Unless you specify which atom you want, and provide spectroscopic data for the constants before the $\mathbf{L}\cdot\mathbf{S}$ and $\mathbf{I}\cdot\mathbf{J}$ terms for that specific atom & isotope, we cannot help you. Definition ...


2

Your confusion arises from a misunderstanding of how the mass of composite systems works in special relativity. When you have a composite system where objects interact, the energy of that interaction counts towards the total energy of the system at rest (in other words, the mass). This is a totally separate, standalone contribution to the mass of the system. ...


2

(a) The title of your question. The concept of coefficient of restitution is useful for macroscopic bodies, but not for microscopic bodies like atoms. The exception is those head-on collisions between atoms when no kinetic energy is lost (elastic collisions). The relative velocity of separation is then equal and opposite to the relative velocity of approach, ...


2

Yes, the energy (and therefore the photon wavelength) required for the transition between two energy levels is independent of the "direction". One way to see this is observing absorption and emmision spectra of the hydrogen, where the position of the lines for both types is the same.


2

I'll add a couple notes to the nice answer by @SuperCiocia. Interactions Regarding attractive vs. repulsive interactions. Your original intuition that you would want attractive interactions for a BEC is understandable. You want the atoms to be very cold and densely formed so that they bose condense. Surely attractive interactions would bring the atoms closer?...


2

Let's say the hydrogen atom started in the $2p$ state. Suppose we keep it in a vacuum so it doesn't interact with any other particles. Then the $2p$ state is an eigenstate of the Hamiltonian, and therefore the energy does not change and the only change in time of the state is to pick up a phase $e^{i \omega t}$, where $\hbar \omega=E_{2p}$, and $E_{2p}$ is ...


2

The energy levels of the Hydrogen atom have specific energy levels because the solution of the S-equation says so. Other than this no one knows why things are described by the S-eqn. It just the way things are. The S-eqn provides a mathematical explanation of the way things work. It has been checked against known results and it agrees perfectly so there is ...


2

I believe that when Richard Feynman was asked what the Schrödinger equation (SE) was, he replied (paraphrasing from memory): 'it's something that happened in Schrödinger's mind!' In truth we don't know why the SE works, only that it does tremendously well! I understand that electrons do not orbit the nucleus, instead they have a higher probability to be ...


2

I'd like to elaborate some more on my comment. As you probably know, quantum mechanics take place in a Hilbert space $\mathcal H$, and every physical state is represented by a unit vector in $\mathcal H$. The time evolution of a given state $\vert\psi\rangle$ is given by the Schrödinger equation $i\hbar\vert\dot\psi\rangle=H\vert\psi\rangle$. Using some math ...


1

The question that you have shared does deal with this point, but I thought I'd elaborate on it a little more. When we solve a quantum mechanical system, we would like to work with an eigenbasis that is labelled by quantum numbers that completely specify the state of the system. Essentially, this comes down to finding something called a Complete Set of ...


1

It is very desirable for $\psi$ to be an eigenfunction of as many operators as possible In fact, we would probably like it to be an eigenfunction of all angular momenta, but they do not commute, so we can't make it be an eigenfunction of all momenta, so we choose only one (usually $L_z$) There are many reasons, but the main ones are: They are measurable ...


1

Your expression for the angular momentum of the system is wrong. It assumes that both the nucleus and the electron are traveling in a a circle of radius $r$; but in reality, the electron travels in a circle of radius $r_e$ and the nucleus travels in a circle of radius $r_n$.


1

If I have a single hydrogen atom with an electron in the ground state, can light of the correct energy excite this electron? The answer to this question is yes. To bring in your discussion of equilibrium statistical mechanics phenomena, it is best to consider an ensemble of hydrogen atoms at zero temperature. Then your question may be recast as: can light ...


Only top voted, non community-wiki answers of a minimum length are eligible