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29

You have your "prove" in the wrong place. The way to prove that ground-state electrons in hydrogen atoms don't emit radiation is the following: Construct a sample of ground-state neutral hydrogen atoms. Place this sample near a detector which is sensitive to the sort of EM radiation you expect. Die of old age waiting for a signal, because ground-state ...


17

This is definitely not a dumb question. If we work in a (linear) Hilbert space, then our inner product $\langle \cdot,\cdot \rangle$ induces the usual natural flat metric (given by $d(\psi,\phi) = || \psi - \phi ||$). However, often we take the viewpoint that our states are elements of projective Hilbert space $\mathbb CP^n$. Then it is more natural to ...


15

The existence of hydrogen atoms is enough to demonstrate that the electrons don't emit radiation. If they did, that energy would have to come from somewhere. The only place it could come from would be a reduction of orbital radius until the electron finally reaches the nucleus. If you accept that electrodynamics applies, then you have to accept that atoms ...


9

Because of its wave nature, the electron in its ground state is actually smeared symmetrically about the proton (ignoring spin-spin effects), and spherically symmetric charge distributions do not radiate (there's no special direction). Accelerated charges do not always radiate em radiation. See also How to find the magnetic field due to a revolving electron ...


8

I believe some of the answer in the links are correct, others are less obvious and might even be confusing. I am not gonna repeat the arguments there, but to stress the following idea. You cannot demonstrate that using classical electrodynamics. The theory as is does not apply to quantum objects and thus it was modified. The equations are the same, they are ...


8

The decay of potassium-40 to argon-40 is either a $\beta^+$ decay in which what is emitted is not an electron but a positron $$ {}^{40}{\rm K} \to {}^{40}{\rm Ar} + e^+ + \nu_e $$ or, more frequently (if we have whole atoms), an electron capture that you mentioned in which no charged leptons are emitted at the end! About 11% of the potassium-10 decays ...


8

There doesn't exist any procedure to uniquely associate a Hermitian operator $L$ to a function of the phase space $f(x,p)$. Quantum mechanics is a theory that exists independently of classical physics. Quantum mechanics is not just a cherry on a classical pie that needs the classical theory to exist at every moment. If we want to define a quantum theory, we ...


5

When you have a bunch of interrelated phenomena in physics, trying to figure out which one is the "reason" for the other ones is often just a recipe for confusion. Different people will start from different postulates, so they will disagree on which results are trivial and which aren't, but hopefully everyone agrees on what's true. In a first course on ...


5

Electron capture Electron capture (K-electron capture, also K-capture, or L-electron capture, L-capture) is a process in which the proton-rich nucleus of an electrically neutral atom absorbs an inner atomic electron, usually from the K or L electron shell. This process thereby changes a nuclear proton to a neutron and simultaneously causes the emission ...


5

Let me preface by saying that "coupling" is a favorite physicist word that is perhaps best described linguistically than rigorously; it's deployed in a few different situations. In general, we say that a coupling exists in quantum mechanics if the evolution of one part of the system depends on another quantity, which could be either classical or quantum. I'...


5

In addition to the answers already given, which answer the question pretty-well, I'll say that, historically, this exact question was the one which puzzled Niels Bohr enough to inspire him to advance his famous theoretical-explanation for the several observed frequencies of the radiations emitted from hydrogen-atoms ... in general, the fact that electrons in ...


4

I'll address your issues with definition (1): $E$ is a function of $\vec p$ because $\lvert \lambda_{\vec p}\rangle\sim\lvert \lambda_0\rangle$ where by $\sim$ I mean that they are related by a Lorentz boost. That is, to "construct" these states, you actually first sort out all the states $\lvert \lambda_0\rangle,\lambda_0\in \Lambda$ ($\Lambda$ now denotes ...


4

Imagine something oscillating in space and time, for example a plane wave propagating across the axis $x$. This propagation is expressed via the so-called phase $$ \phi(x,t)=\omega \cdot t - k\cdot x = \dfrac{2\pi}{T}\cdot t -\dfrac{2\pi}{\lambda}\cdot x \tag{01} $$ and the magnitude of the plane wave as $$ E(x,t)=A\cos\phi(x,t) \tag{02} $$ As the ...


4

Considering only the spin, ignoring translational DOF, the Hamiltonian is $$H = -\mu ~\mathbf{B} \cdot \mathbf{S}$$ If $\mathbf{B}$ is directed along $z$, it's easy to see that $S_z$ states are energy eigentstates and thus are stationary. Applying time evolution and taking the expectation value shows that if the spin is not oriented along $z$ initially, ...


3

When in doubt go back to the masters. From Dirac's Principles of QM When we make an Observation we measure some dynamical variable. It is obvious physically that the result of such a measurement must always be a real number, so we should expect that any dynamical. variable that we can measure must be a real dynamical variable. One might think ...


3

In the Heisenberg picture, one simply has $$ A(t) = \exp(-Ht/i\hbar) A(0) \exp(+Ht/i\hbar) $$ The Hamiltonian $$ H = \frac{E_1+E_2}{2}\cdot {\bf 1} + \frac{E_1-E_2}{2} \cdot \sigma_z $$ while $$ A(0) = a\sigma_x $$ The term in $H$ proportinal to ${\bf 1}$ cancels in $A(t)$ so we have $$ A(t) = a\cdot \exp(-(E_1-E_2)t\sigma_z/2i\hbar) \sigma_x \exp(+(E_1-E_2)...


3

Both $$\sum_i |i\rangle \langle i | $$ and $$\sum_j |j\rangle \langle j | $$ are summations over basis vectors. The indices $i,j$ run over the same values – values of indices that identify the basis vectors in the same basis (set of vectors) – but the particular values of the indices $i,j$ are independent. Can you calculate how much is the expression below?...


2

Let us make things clear. Protons and electrons are quantum mechanical entities and there is little meaning to project classical electrical attractive behavior to the micro framework of quantum mechanics, nor classical electric field calculations . Classically, a negative charge attracted to a positive charge will experience acceleration, and accelerating ...


2

For context, any general quantum operation $\Phi$ on a bipartite system $AB$ with finite dimensional Hilbert space $H_{A} \otimes H_{B}$ has a Kraus representation $\Phi(\rho) = \sum_j K_j \rho K_j^{\dagger}$ where the "Kraus operators" $K_j$ are linear operators on $H_{A} \otimes H_{B}$ such that $\sum_j K_j^{\dagger} K_j = I_{AB}$. The Kraus ...


2

The word "coherent" is used in Physics in a rather sloppy way. Your first state is a linear combination of harmonic oscillator eigenvectors that turns into a gaussian in momentum/position representations. In a more general background, a coherent state is just a state where coherences (off-diagonal terms in the density matrix) are non-zero, which means the ...


2

No, magnetic lines of force don't flow. They have a direction, which shows the direction of the magnetic field but there is nothing flowing. If you were to place a small magnetic dipole at the location of the magnetic field line its north pole would feel a force in the direction of the line of force. The phrase "line of force" was introduced by Michael ...


2

Direction of deflection of electrons in magnetic field It is not the full picture you are describing. A moving electron in a magnetic field gets deflected according the rule $ \vec F = q \vec v \times \vec B $. This vector product has a direction and and this is what we observe in natur: all electrons get deflected in the same direction. Otherwise no ...


2

Hermitian operators (or more correctly in the infinite dimensional case, self-adjoint operators) are not used because measurements must use real numbers, but rather because we almost always decide to use real numbers. As the OP mentions at one point, you might choose to use complex numbers to label a two-dimensional screen, and in that case you'll be able ...


2

Basically it is whatever you need to multiply a distance by to find a phase difference (in radians). For a traveling wave, the wave number is the amount of phase difference per unit length. For a physical sine wave, it is the ratio between the maximal slope of the wave surface and the amplitude. In other words, it measures how dramatic the local ...


2

In addition to the answer by hsinghal it is worth point out some historical notational quirks. An expression such as $^2P_{3/2}$ is called a Term Symbol. The superscript is the multiplicity of the electron spins, i.e. 2$S$+1 for total spin S. The capital letter, P in this example is the total orbital angular momentum and has letters and values of S=0, P=1, D=...


2

Any quantised system has a ground state and excited states, and in any quantised system relaxing into the ground state requires energy to be shed in some fashion. In an isolated system like a hydrogen atom the energy is normally emitted as photons. However add other hydrogen atoms and this opens new routes for energy to be lost. For example an excited ...


2

The Heisenberg Uncertainty Principle has two distinct aspects: One is the identification of matter as a wave and, in particular, the relationship between a particle's momentum $p$ and its wavelength $\lambda$ through de Broglie's relationship $p=h/\lambda$. This is the crucial bit of physical input. The second one is purely mathematical, and it's the ...


2

Hilbert spaces are vectorspaces by definition. If you interpret a vector space as a manifold (which you can do) then it's a flat manifold.


2

As commentators have indicated Hilbert space is a vector space. A manifold is a space with an atlas-chart construction with maps on overlapping regions that define connection coefficients and ultimately curvature. It is certainly possible to think of a finite dimensional complex vector space that is a locally flat region in an otherwise curved space. This ...


1

In a comment elsewhere you write that you're interested in understanding how quantum-mechanical theory describes the radiation that a hydrogen atom does and does not emit. In your question you ask about another answer that suggests some significance to the electron having zero total momentum; I think that's a feature of the coordinate system choice rather ...



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