# Tag Info

## Hot answers tagged quantum-mechanics

28

Yes, nucleus is composed of subatomic particles that have probability cloud. Protons and neutrons fill orbitals in the nucleus just like electrons in the atom do. What's more, every proton or neutron is a complex particle itself and the quarks inside have their very own probability cloud. (Quarks are simple objects that have no internal structure as far as ...

17

Very often indeed the nucleus is assumed motionless. It is then assumed that the motion of the nuclei and the electrons can be treated separately. This is known as the Born-Oppenheimer approximation. The reason is that solving the equations simultaneously is very difficult and would not be very efficient. Note that for the hydrogen atom this approximation ...

11

The nucleus does have a probability cloud. As the simplest example, consider the hydrogen-1 atom. Conservation of momentum requires that the center of mass of the electron and proton remain fixed. Therefore we have $$|\Psi_p(\textbf{x})|=|\Psi_e(-\alpha\textbf{ x})|,$$ where $\Psi_p$ is the wavefunction of the proton, $\Psi_e$ is the wavefunction of the ...

10

To build upon Bruce's answer, what happens according to Schrödinger's equation is the following: The initial quantum state of the system (after putting the cat in the box) is $$|\text{cat alive}⟩ |\text{you don't see the cat}⟩$$ As the Geiger counter measures whether the radioactive atom decayed or not, and kills or spares the cat accordingly, this ...

8

The answer is that the protons in the nucleus are quantum particles and don't have a well-defined position, but the uncertainty isn't a big factor in determining the potential experienced by the orbiting electrons, so we can just treat them as a fixed source of potential. That does, also, make the calculations much, much easier.

6

You use the phrase phosphorescent molecule but phosphorescence is normally seen in solids where there are many interacting molecules. When you excite electrons in a solid they will generally decay in a radiationless manner, usually by transferring their energy to vibrations of the lattice (i.e. heat). It is unusual for excited states in solids to decay by ...

4

Hint: $$x^2+y^2+xy = \frac34 (x+y)^2 + \frac14 (x-y)^2.$$ This means that if you transform to the new variables \begin{align} \xi & = \frac1{\sqrt{2}}(x+y) \\ \eta & = \frac1{\sqrt{2}}(x-y), \end{align} and with a similar transformation from $p_x,p_y$ to $p_\xi,p_\eta$ to make sure that $[\xi,p_\xi] = i = [\eta,p_\eta]$ and $[\xi,p_\eta] = 0 = [\... 4 No. Take, for instance, the fully depolarizing channel, where$A_a=\{I,X,Y,Z\}$. Since$\mathcal E(\rho)=\tfrac12 I$, your condition$(1)$holds for all$H$. On the other hand, there is no operator which commutes with all$A_a$. (Let me take the opportunity to advertise my list of canonical counterexamples for quantum channels ;) ). 3 Photons are elementary particles, part of the SM, they are traveling at speed c in vacuum, when measured locally. The world line (or worldline) of an object is the path that object traces in 4-dimensional spacetime. It is an important concept in modern physics, and particularly theoretical physics. https://en.wikipedia.org/wiki/World_line Now ... 3 There is no particular reason to expect that energy and angular momentum should have a one-to-one coupling. If we take the case of classical two-body orbits as a rough guide, we see that each (bound) orbital energy is associated with a upper limit on angular momentum but that the acutal angular momentum can range from that limit all the way down to zero. ... 3 I believe that you can do all of physics apart from QM without using complex numbers: complex numbers are a convenience (generally because$e^{ix} = \cos x + i \sin x$), but they are only a convenience. However if you want to do QM you either end up using complex numbers or creating mathematical objects which have all the properties of complex numbers: you ... 3 One could say that all numbers are imaginary in that sense, they are abstract concepts that we force on a relation into nature just to allow us to have some descriptive power of our surroundings. The fact that natural numbers seem more "natural" or easy to relate to has nothing to do with what they really are; abstract ideas without physical meaning until we,... 3 The term "imaginary" for referring to the so-called "imaginary numbers" (which, by the way, are only a one-dimensional sliver of the full complex numbers that are actually used here) is a historical artifact that has caused way more confusion than it should, now. Let me say one thing that is crucially important here: There is no ontological difference ... 3 Writing the requirement explicitly $$\mathcal{E}(U\rho U^\dagger)=U\mathcal{E}(\rho)U^\dagger$$ in terms of Kraus operators $$\sum_a A_aU\rho U^\dagger A_a^\dagger=\sum_a U A_a\rho A_a^\dagger U^\dagger$$ Hence we want the channel with Kraus operators$ A'_a=A_aU $and$A_a''=UA_ato be equal. We know that two channels are equal if and only if their ... 2 Your expansion is not quite the most general and should more generally read $$\Psi(x,t)=\sum_m c_m e^{-i E_m t/\hbar} \psi_m(x)$$ so that \begin{align} \frac{d\rho}{dt}=0\quad\Rightarrow 0= \sum_{mn} c_m c_n^* (E_m-E_n) e^{-i(E_m-E_n)t/\hbar} \psi_n(x)^*\psi_m(x)\tag{1} \end{align} (up to a factor ofi\hbar$). For simplicity order the eigenvalues so ... 2 There are two aspects to this. Firstly, as long as the interaction energies are well below the energies necessary to create new particles the Schrodinger equation is usually an excellent approximation. For the majority of atoms and molecules it gives an excellent description of the electronic structure. (There is also the pragmatic consideration that bound ... 2 It's important to understand that we don't directly observe quarks and gluons in scattering experiments. What we observe is an unruly shower of well known and understood particles hitting our detectors. To try and understand what happens in the collision we use a mathematical model called the Standard Model. That is, we have equations that describe how ... 2 Are you referring to hydrogen? In other atoms different$l$values with the same principal quantum number$n$will have different energies. It is only for hydrogen with its single electron that all states with the same$n$have the same energy. In this sepcial case there is an extra, hidden,$O(4)$symmetry possesed by the Kepler$V(r)=k/r$potential. ... 2 How can the annihilation of an electron and a positron create a quark-antiquark pair or a muon-anti muon pair? .... But the total rest energy of the electron and positron(1.102 Mev) is less than the the total energy required to produce either the quark-antiquark pair or the muon-anti muon pair(211.4 Mev) Certainly if the annihilation happens at rest, i.... 2 It is possible to simulate quantum computers with classical computers. However, the time (and typically memory) needed for such a simulation will scale exponentially with the time a quantum computer needs. Thus, if a quantum computer needs$10$times the computation time for a more complicated case, the simulation would need$2^{10}$times more time, and so ... 2 Quantum mechanics as a theory was invented in order to explain why classical mechanics and classical electrodynamics did not work in certain situations: small scales, spectra of atoms , black body radiation. Classical mechanics and classical electrodynamics cannot explain or predict the behavior of atoms (their interactions) , neither that of ... 2 This is a partial answer. There is a class of proteins called enzymes. Enzymes facilitate chemical reactions. Many enzymes facilitate in the following way: a particular small region of the enzyme has a high affinity for a particular molecule. That is, when that molecule comes in contact with that active region it tends to get stuck there. The fit is good, ... 2 Insofar as your question regarding whether or not it is a statement about "actual" particles, as @knzhou suggests, we don't really have access to "actual" reality in general in any kind of science (or perhaps we do - this really more depends on your philosophical viewpoint). You see, a scientific theory is perhaps best understood as what I'd call a "useful ... 2 Your question is a good one, though it's somewhat unclear. I think what you're asking, in essence, is whether one needs quantum mechanics to accurately model the activity of proteins, or whether classical physics is sufficient. If one takes the proteins as already existing, the answer, in the vast majority of cases, is that classical mechanics is sufficient. ... 2$d\tau$is the element of volume$dxdydz$, which in spherical coordinates is$r^2dr\sin{\theta}d\theta d\phi$. When integrating a function that doesn’t depend on the two angles, the integration over those angles gives$4\pi$. 2 The existence of a minimum energy follows from the wave-like aspects of matter. The allowed values of energy are those corresponding to stationary wave states, so the trite answer to your question is that where you have a set of allowed energy values one of them has to be a minimum. To get some physical insight into why the minimum level ends up where it ... 2 Absolutely! The Hermite polynomials $$H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2} = \left(2x - \frac{d}{dx} \right)^n \cdot 1$$ are orthogonalized by $$\int_{-\infty}^\infty H_m(x) H_n(x)\, e^{-x^2} \,dx = \sqrt{\pi}\, 2^n n! ~ \delta_{nm} ~,$$ whereas the Hermite functions$\$ \psi_n(x) = \left (2^n n! \sqrt{\pi} \right )^{-\frac12} e^{-\frac{x^2}{...

2

To simulate means to mimic or model the characteristics or appearance of something. Cloning means to make an exact replica- another actual instance of the subject being cloned. The two actions are utterly different in essence. So yes, you could straightforwardly simulate the two slits experiment, and no that would not violate the no-cloning theorem.

1

The double slit experiment, regardless of the size of the particles (electrons, neutrons, molecules) does not prove that those particles exist in two places at once, as claimed by the SciAm article. The difficulty of understanding this experiment in classical physics is caused by the use of an unsuitable classical model, rigid body Newtonian mechanics with ...

1

There are no forces acting between photons, so they do not interact with each other directly (eg attract, repel, scatter, etc). If one photon interacts with a charged particle to change its energy in some way, it is possible for that particle to interact with another photon as a consequence of its interaction with the first. There can therefore be a chain ...

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