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8

A classical turning point is a point at which the system's total energy $E$ equals the potential energy $V.$ Past this point, i.e. for $E<V$ the potential is greater than the total energy, such cases we denote as classically forbidden regions, because from a purely classical point of view, the system has 0 chance of being in a state where its potential ...


7

$\rho_\psi$, the density matrix, is not an observable/operator evolving in the sense of the Heisenberg equation of motion $$ \mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t} A = - [H,A]$$ since it is defined, as you correctly write, as a projector on states, hence it is time-dependent in the Schrödinger picture (since there the states it projects on are ...


6

My answer has a lot of overlap with the earlier answer by NoEigenvalue, but grew too long for a comment. The orbitals are segregated by the amount of angular momentum that they carry. No angular momentum, $\ell=0$, are called $s$-orbitals, and are spherically symmetric. The first two examples in your figure are $s$-orbitals with $n=1$ and $n=2$. Angular ...


4

The Schrödinger equation is only correct in the non-relativistic limit $v << c$, for particles without spin. The correct equation for spinless (=spin $0$) particles is the Klein-Gordon equation, which reduces in the non-relativistic limit to the Schrödinger equation. If we want to talk about spin $\frac{1}{2}$, the correct, relativistic equation is ...


3

Mathematical reason is that the time evolution operator is unitary, which means that $U^\dagger = U^{-1}$. Therefore $\langle \psi(t) | \psi(t) \rangle = \langle \psi(0)| U^\dagger U | \psi(0) \rangle = \langle \psi(0) | \psi(0) \rangle$. We can see that it is unitary by considering the Schrodinger equation: $$ \newcommand{\ket}[1]{| #1 \rangle} ...


3

Just wanted to add that it's not totally true that the drawn orbitals are "the regions where an electron can be found". But my answer grew and grew... Let's take a neutral boron atom where it has filled 1s and 2s shells and one electron in the 2p shell. Suppose it is floating in space, far away from messy interactions with other things. So, you wonder, ...


3

The orbitals are only spherically symmetric for $S$-states, for which the angular momentum number $l$ is zero. So in you picture the first two orbitals are the $s=1$ and the $s=2$ state, while the other three pictures correspond to the $n=2,l=1$ with the three different possibilities of $m_l$. There are several ways to see that the wavefunctions with $l=0$ ...


3

I know of no such publication. However, this issue may simply be on one hand too trivial and on the other hand too far removed from practical relevance. Let's derive what you are after to see: A creation ladder operator $\hat{a}^\dagger$ for arbitrary states would have to be of the form $$\sum_{n=0}^\infty c_n \left| n+1 \right\rangle \left\langle n ...


2

No, the (elementary solution for the position representation of the) wavefunction of a free particle, $\psi(x) = \mathrm{e}^{\mathrm{i}px}$ is not an "explicit function of both" position and momentum. It is a function of position - the momentum of the plane wave is fixed, and the momentum space wave function of this is its Fourier transform $\tilde{\psi}(k) ...


2

Indeed, in non-relativistic quantum mechanics, the equation of evolution of the quantum state is given by Schrödinger's equation and measurement of a state of particle is itself a physical process and thus, should and is indeed be governed by the Schrödinger's equation. Indeed, people like to predict probabilities using Born's rule, and sometimes they do ...


2

Actually, you can use duality: the normal states of quantum mechanics are objects of the (unique) predual of the von Neumann algebra of quantum observables. Using a concrete example: if the algebra of observables are the bounded operators on a Hilbert space, the predual are the trace class operators. Of them, the normal states are the ones positive, ...


2

Your question apparently stems from a lack of understanding of the different pictures in quantum mechanics, that are Schrödinger picture, Heisenberg picture and Interaction picture. In the Schrödinger picture, states are time-evolving, while observables are time-independent. The density matrix is another (more general) way of writing the state vector; its ...


2

I suspect the preprint is wrong at least in this phrase from the abstract: "We illustrate a simple derivation of the Schrodinger equation, which requires only knowledge of the electromagnetic wave equation and the basics of Einstein's special theory of relativity." They use the following in their derivation: "Recall from Einstein and Compton that the energy ...


1

Solving $$e^{ikL}-e^{-ikL}=0$$ we write: $$e^{ikL}=e^{-ikL}$$ then, dividing both sides of the equation by $e^{-ikL}$ we find that $$\frac{e^{ikL}}{e^{-ikL}}=e^{2ikL}=1$$ From Euler's Formula, $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ the solution is that k is quantized: $k =\frac{n\pi}{L}$ for positive integer n. Thus $$\psi (x)= A(e^{ikx}-e^{-ikx}) = ...


1

You can search for eigenvalues using the bisection method. Priliminaries: To get the eigenvalues from Numerov method you will need to know the wavefunction at the boundaries. Generally this would mean that you need to set the potential to infinity at the boundaries hence putting the wavefunction to zero at those points. For your potential, modify it as ...


1

For an infinite well (actually infinitely high barriers) the probability to find an electron in the barrier vanishes. Therefore the wavefunction in the barrier has to be 0. For barriers with a finite height, the commonly used, but actually wrong boundary conditions require the wavefunction and the first derivative to be continuous. This relies on the wrong ...


1

Comments to the question (v2): The fact that the TISE is invariant under a symmetry group $G$ (in this case the Lie group $G=SO(3)$ of 3D rotations) does not imply that the orbital/wave-function solutions $\psi$ must be $G$-invariant as well. (Think e.g. on spontaneous symmetry breaking where the governing laws of a physical system are invariant under a ...


1

Modern electronic devices like quantum well lasers, resonant tunneling diodes, quantum cascade lasers and detectors heavily rely on the spatial and energetic position of such bound states. This defines their transport and optical properties. On a separate notice: any well, no matter how shallow or narrow, has at least one bound state.


1

There are two important points to keep in mind when working through this problem. (1) Since the Hamiltonian for the system changes suddenly, the wavefunction just after the change is the same as the wavefunction just before the change. (2) Then energy eigenstates after the change are different from the energy eigenstates after the change. It follows that, ...


1

The issue is that the electromagnetic field and its gauge transformations are treated classically here - they are not operators of the quantum theory, but "tacked on" because we want to describe how a quantum object interacts with the electromagnetic field without treating the EM field itself as a quantum object. "Gauge-invariance" in this half-quantized ...


1

Does this mean that the Hamiltonian doesn't describe a true physical quantity like in classical mechanics? Even in classical mechanics, Hamiltonian for one particle in external field EM is function $$ H(\mathbf r,\mathbf p) = \frac{(\mathbf p - \frac{q}{c}\mathbf A(\mathbf r, t))^2}{2m} + q\phi(\mathbf r,t) $$ where $\mathbf A,\phi$ are any of the ...


1

The three directions $x$, $y$ and $z$ are separable for the particle-in-a-box problem - the behaviour in each is independent of that in the others. Thus, each direction when separately considered only gives the contribution to the energy due to the limits of the box, or equivalently, the 'part' of the wavefunction, in that direction. In the case of $l_z ...


1

Comments to the question (v2): Normally we assume that $-\infty< x_1 < x_2<\infty$ are the turning points in the 1D potential well. This means $$\forall x\in ]x_1,x_2[:~~ E ~>~ V(x).$$ Hence OP's third last equation $$0 ~<~ \int_{x_1}^{x_2}{dx\ \sqrt{2m(E-V(x))}} ~=~ -\frac{\pi}{2}\hbar~<~0$$ can never be fulfilled. The correct WKB ...


1

The Dirac notation is simply an alternative to vector notation. Certainly there are PDEs describing the quantum state of a lone particle with spin and they are: The Pauli equation (see Wiki page of this name) was historically the first, and here the quantum state is two $\mathcal{L}^2(\mathbb{R}^3, \mathbb{R})$ functions of space and time. The two ...


1

The Dirac equation is mentioned in other answers as PDE describing spin. As you ask "what would Schrodingers equation (or some kind of generalization that allows for you to include spin) look like?", the following may be relevant. Yes, the Dirac equation adequately describes spin. However, it is actually a system of four partial differential equations for ...


1

Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance $\Psi(x)=\sum_nc_n\psi_n(x)$ is a general quantum state for a particle in a box, where $\psi_n(x)$ are the energy ...


1

It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation. For example, particle in a box has discrete eigenvalues, bounded ...



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