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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 ...


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 ...


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 ...


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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 ...


0

Apart from (physically motivated) asymptotic prescribed behaviour at infinity $|x|\to\infty$, we don't actually impose/demand/require any conditions on the wave function $\psi$ beyond the TDSE. Continuity and (possibly higher) smooth conditions are instead derived from a standard bootstrap argument, see e.g. my Phys.SE answer here for details.


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Because in ODE you only need $n$ conditions according the $n$-order of ODE, but these conditions could be on the same point, value of function and derivative, or two points, value of function first and last point of interval.


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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} ...


0

So if I understand your question correctly, you are saying that every solution to a mathematically formulated physical problem should necessarily correspond a physical reality. Let me give you a very elementary example. Suppose I give you the following problem 'Person X and person Y are brothers. Y is younger to X by 6 years and the product of their ages is ...


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) ...


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

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 ...


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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

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, ...


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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.


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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 ...


0

MATHEMATICALLY, if a function $\phi$ is a solution of the SE: $$ H\phi=E\phi $$ then $c\phi$ (c is a constant) is also a solution of it due to its linearity. PHYSICALLY, however, we cannot choose any mathematical solution of the above SE equation to be the physical solution. Remember that we are working with Physics, not Mathematics. In Quantum Mechanics, ...


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 ...


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, ...


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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 ...


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 $l=0$ ...


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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 ...


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It happens because of quantum tunneling. In the first scenario, we cannot possibly find the particle beyond the barrier. in the second, however, the particle can quantum tunnel over.


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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

The problem can be solved by introducing the relations \begin{equation} n_2-n_1=1,\:\:\:n_3-n_2=1 \end{equation} Since the resonance occurs for the energies $E_1$,$E_2$, and $E_3$ in a sequence the argument for the sine $n \pi$ also occur in a sequence in this case. This leads to one of possible solvable equations for the well potential \begin{equation} 2 ...


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 ...


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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, ...


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$\rho_\psi$, 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 ...


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 ...


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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 ...


0

Firstly, the assumption that you could build a working Maxwell Daemon (a three state machine with Szilard-engine actuation hardware) to extract work from the system already gainsays the assumption of a steady state universe. The problems of mixing classical and quantum statistical mechanics aside, the tunnelling here is in principle no different from ...


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The flaw in this argument is your assumption that only the hotter molecules would tunnel.


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}) = ...


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 ...


0

The matrix exponential for any Pauli matrix using the general formula [see general formula][1]. This procedure gives, $$\exp[-iHt] = I_{2\times2}\cos \nu_{F}t - i {\bf\sigma}\cdot\left(q-By\hat x\right) \sin \nu_{F}t$$ I tryed to solve using this method but seems not convincing. Any comments would be appreciated.


0

The answer is yes: Obviously it is possible to solve radially symmetric Schroedinger equations, including the specialized treatment (radial component only) that you ask about. However, you will need a well-defined problem to meaningfully begin something as rigorous as an analytical (or numerical) treatment. The differential equation you have given for the ...


0

Usually one requires, that the propability density $\varrho = |\Psi(x)|^2$ is integrable over the whole range (suchs that it can be normalzied). In this sense $\Psi(x) = \exp(kx)$ is just as divergent as $\Psi(x) = \exp(ikx)$, since the limit $$\lim_{\epsilon \rightarrow \infty } \int_{-\epsilon}^\epsilon |\Psi(x)|^2 \ dx$$ does not exists in both cases. ...



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