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The quantity you should consider first is the Bohr radius, this tells you an idea of the relevant atomic scales, $$ a_0 = 5.29\times 10^{-11} ~{\rm m} $$ For hydrogen (the most abundant element), in its ground state, the probability of finding an electron beyond a distance $r$ from the center looks something like (for $r \gg a_0$) $$ P(r) \approx e^{-2r/...


62

Take a gaussian (or any function that decays sufficiently quickly), chop it up every unit, and turn all the pieces sideways.


50

Real particles are never completely localised in space (except possibly in the limit case of a completely undefined momentum), due to the uncertainty principle. Rather, they are necessarily in a superposition of a continuum of position and momentum eigenstates. Pauli's Exclusion Principle asserts that they cannot be in the same exact quantum state, but a ...


44

An animation is worth a million words:


44

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 ...


38

What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results. Take into account that the models are based on hypothesis, which are easily violated. For example, spherical symmetry, which allows finding the solution in the ...


33

The other answer shows nicely how one may interpret the Pauli exclusion principle for actual wavefunctions. However, I want to address the underlying confusion here, encapsulated in the statement If time and space are continuous then identical quantum states are impossible to begin with. in the question. This assertion is just plainly false. A quantum ...


31

Complex numbers are used in all of mathematics, and therefore by extension they are used in other fields that require math; not just physics, but also engineering and other fields. Trying to assign a "physical interpretation" to a complex number would be like assigning a physical interpretation to a real number, such as the number 5. A complex number is ...


28

The fact is that there are two kind of things: 1) the wave function and 2) the physical observables. The evolution of the wave function is dictated by the Schrödinger equation and is deterministic meaning that if you know the wave function at some time, then you know it at any time just using Schrödinger equation $$ i \hbar \frac{d}{dt} \left \lvert \Psi (...


28

It is actually true, in an almost trivial way. The Ehrenfest theorem states that, \begin{equation} \frac{d}{dt}\langle x\rangle=\langle p\rangle,\quad \frac{d}{dt}\langle p\rangle =- \langle V'(x)\rangle \end{equation} However for all eigenfunctions for the Harmonic oscillator $\langle x\rangle=0$ (and therefore $\langle V'(x)\rangle=0$) and $\langle p\...


27

Let $$ \psi(x) = \begin{cases} 1 & \exists\, n \in \mathbb N: x \in [n, n+\frac 1 {n^2}]\\ 0 & \text{otherwise.} \end{cases} = \sum_{n \in \mathbb N} \mathbf 1_{[n,n+\frac 1 {n^2}]}(x) , $$ where $\mathbf 1_A$ is the characteristic function of the set $A$. Then $$ \int_{-\infty}^\infty |\psi(x)|^2 dx = \sum_{n=1}^\infty \frac 1 {n^2} < \infty, $$ ...


27

Complex numbers are used in "macro" physics. They are used in analysis of electrical circuits (especially when AC is involved) and in fluid dynamics. Solution of differential equations is simplified if complex numbers are used, as is Fourier analysis. Any scenario that involves periodic or cyclic functions can be modeled using complex numbers.


26

The Photon more or less has it right, the probability distributions of the orbitals (even in the state of being bound in an atom) extend infinitely far away radially, albeit with increasingly vanishing probability. I think the confusion comes from the fact there are certain "regions", planes and other 2D surfaces really, quite close to the atom, where ...


25

This year-old question popped up unexpectedly when I signed in, and it's an interesting one. So I guess it's OK just to add an intuition-level "addendum answer" to the excellent and far more complete responses provided long ago. Your kernel question seems to be this: "Why is the wave function complex?" My intentionally informal answer is this: Because ...


25

If you have a copy of Griffiths, he has a nice discussion of this in the delta function potential section. In summary, if the energy is less than the potential at $-\infty$ and $+\infty$, then it is a bound state, and the spectrum will be discrete: $$ \Psi\left(x,t\right) = \sum_n c_n \Psi_n\left(x,t\right). $$ Otherwise (if the energy is greater than the ...


25

Warning: long post ahead. To simply watch an animation, scroll to the bottom ;) Requirements on the wavepacket First let's do some estimations, using the exact analytic expression for evolution of a Gaussian wave packet in free space. Hopefully, it'll not be too wrong in the case when Coulomb potential is on. The expression looks like this: $$\...


24

All bound states can typically be chosen to have real-valued wavefunctions. The reason for this is that their wavefunction obeys a real differential equation, $$ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf r)+V(\mathbf r)\psi(\mathbf r)=E\psi(\mathbf r)$$ and therefore for any solution you can construct a second solution by taking the complex conjugate $\psi(\...


24

The way you phrase your question violates quantum mechanics: saying "there must be a portion of all atoms on Earth whose electron lies outside the Milky Way" is not a statement that makes sense within Quantum Mechanics. What you can ask, and what others have answered, are variations of the question of how probable it is to find a bound electron at galactic ...


24

The Hamiltonian for the He atom is: $$H = -\frac{\hbar^2}{2m_e}(\nabla_1^2 + \nabla_2^2) - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 r_{12}}$$ where the electrons are denoted 1 and 2, and $r_i$ is the distance to the nucleus at the origin and $r_{12}$ the distance between the electrons. Since the ...


23

The precise theorem is the following, cf. e.g. Ref. 1. Theorem 1: Given a non-positive (=attractive) potential $V\leq 0$ with negative spatial integral $$ v~:=~\int_{\mathbb{R}^n}\! d^n r~V({\bf r}) ~<~0 ,\tag{1} $$ then there exists a bound state$^1$ with energy $E<0$ for the Hamiltonian $$ H~=~K+V, \qquad K~=~ -\frac{\hbar^2}{2m}{\bf \...


23

Why are states rays? (Answer to OP's 1. and 2.) One of the fundamental tenets of quantum mechanics is that states of a physical system correspond (not necessarily uniquely - this is what projective spaces in QM are all about!) to vectors in a Hilbert space $\mathcal{H}$, and that the Born rule gives the probability for a system in state $\lvert \psi \...


23

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 ...


22

In my opinion, manipulations involving $\hat p$ and position bras and kets are most easily done by considering the action of $\hat p$ on the position bras, which is simply $$ \boxed{ \vphantom{\begin{array}{}make\\the box\\taller\end{array}} \quad\,\,\, \langle x|\hat p=-i\hbar\frac{\text d}{\text dx}\langle x|. \quad\,\,\,} \tag 1 $$ You can get ...


22

The interpretation of the Dirac equation states depend on what representation you choose for your $\gamma^\mu$-matrices or your $\alpha_i$ and $\beta$-matrices depending on what you prefer. Both are linked via $\gamma^\mu=(\beta,\beta\vec{\alpha})$. Choosing your representation will (more or less) fix your basis in which you consider the solutions to your ...


22

I) The solution to the time-dependent Schrödinger equation (TDSE) is $$ \Psi(t_2) ~=~ U(t_2,t_1) \Psi(t_1),\tag{A}$$ where the (anti)time-ordered exponentiated Hamiltonian $$\begin{align} U(t_2,t_1)~&=~\left\{\begin{array}{rcl} T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right] &\text{for}& t_1 ~<~t_2 \cr\cr AT\exp\left[-\frac{i}{\...


22

Nobody is "doing the normalization". Normalization is not even necessary. We often normalize for convenience, since that means that the Born rule for $\lvert \psi \rangle$ being the state $\lvert \phi \rangle$ reads $$ P(\psi,\phi) = \lvert \langle\psi\vert\phi\rangle \rvert ^2$$ which is certainly easier to recall/write than $$ P(\psi,\phi) = \frac{\...


22

IF you know that your Hamiltonian is of the form $$ \hat H=\frac{-\hbar^2}{2m}\nabla^2+V(\mathbf r) \tag 1 $$ for a single massive, spinless particle, then yes, you can reconstruct the potential and from it the Hamiltonian, up to a few constants, given any eigenstate. To be more specific, the ground state $\Psi_0(\mathbf r)$ obeys $$ \hat H\Psi_0(\mathbf r) =...


22

In chapter 3, he says, if I understood this right, that electrons can only exist in specific quanta - that is they can only be in certain regions, and will perform a quantum leap over regions that are not stable energy levels. Electrons (that are confined by a potential well, such as when they are part of an atom) are indeed limited to certain quantized ...


21

A wave function is a complex-valued function $f$ defined on ${\mathbb R}^1$ (if your electron is confined to a line) or on ${\mathbb R}^2$ (if your electron is confined to a plane) or ${\mathbb R}^3$ (if your electron ranges over three-space), and satisfying $$\int |f|^2=1$$ (where the integral is defined over the entire line or plane or 3-space). Every ...


20

There is nothing to prove; this just involves making definitions as follows: Let an element $|\psi\rangle$ of the Hilbert space $\mathcal H$ of a particle moving in three dimensions be given. Let $|\mathbf x\rangle$ denote a simultaneous eigenstate of the position operators $X,Y,Z$ corresponding to eigenvalues $x,y,z$ where $\mathbf x = (x,y,z)$. Then for ...


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