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11

You are missing nothing. The Bohr model of the atom is false, and nowadays we replace the idea of the semi-classical "orbit" of Bohr with the fully quantum mechanical notion of orbitals or electron clouds, which give a probability distribution for the position of the electron around the nucleus, but do emphatically not imply that the electron is moving in ...


11

So I believe it's standard to place the operator inbetween the conjugate of the wavefunction and the wavefunction itself. For instance, $$\langle p\rangle = \int_{-\infty}^{\infty}\Psi * \frac{\hbar}{i}\frac{d}{dx}\Psi dx$$ Yes, that is correct, and Is it wrong to do this? $$\langle p\rangle = ...


7

The following passage has been extracted from Bohr's Nobel lecture: While in contradiction to the classical electromagnetic theory no radiation takes place from the atom in the stationary states themselves, a process of transition between stationary states can be accompanied by the emission of electromagnetic radiation, which will have the same ...


6

When position is measured, the uncertainty of the resulting delta spike's position is 0 This notion is the root of the problem. Quantum states which are actually eigenstates of the position operator are mathematically pathological and also completely unphysical. Some math tools Consider a one dimensional system. Suppose $\{|x\rangle \}$ is an ...


5

As far as I understand it, your claim is not per se about measurement but more like "if I prepare a particle in an eigenstate of either position or momentum, doesn't it mean that I have product of uncertainties $0.\infty$?" isn't it? One first problem with this claim as it is, is that the corresponding "wave functions" of these states are not ...


5

Consider the general case that we want to calculate $$ \langle p |F(r) |p'\rangle.$$ By inserting the resolution of the identity $\int d^3r\, |r\rangle\langle r|$ we find that we need to compute $$\tilde{F}(q = p-p') = \int d^3 r \, e^{i(p-p')r} F(r). \tag{1}$$ This integral will converge if $\int dr\, |F(r)|$ is finite. Such a function is said to be $L^1$. ...


5

Hints: OP's exercise is essentially a matter of checking an oscillatory Gaussian integral (3) over the initial position $q_1$. Let $\Delta t_M:=t_2-t_1>0$. To render the Gaussian integral convergent, insert Feynman's $i\epsilon$ prescription $\Delta t_M\to\Delta t_M-i\epsilon$. Or equivalently, Wick-rotate $\Delta t_E:=i\Delta t_M$, where ${\rm ...


5

Hint: Rather than using spherical coordinates, which are singular where the 3D Dirac delta function has support, work instead in Cartesian coordinates $\vec{r}=(x,y,z)$ and use the defining property $$ \iiint_{\mathbb{R}^3}\! d^3r ~f(\vec{r})~ \delta^3(\vec{r})~=~f(\vec{0}) $$ of the 3D Dirac delta function.


5

It doesn't matter what sign you choose. Notice that since $|A|^2 = \frac{2}{a}$, you could even pick $A = \sqrt{\frac{2}{a}} e^{i\phi}$, so $A$ doesn't have to be real. The reason is that a wavefunction is only defined up to a global phase. The reason is that we calculate probabilites with $|\psi|^2$ and mean values of operators with $\int \psi^* \hat{O} ...


4

Actually, all that is quite known from the foundational work by von Neumann and Birkhoff. In their formulation of QM they constructed the theory starting from the lattice of elementary properties (see my answer on quantum probabilities) for more details or "propositions" testable on a quantum system along the experimental common phenomenology of all quantum ...


4

There are a lot of misconceptions here so let's take it one step at a time. The entropy in classical mechanics is called the Gibbs entropy, $$S = - k_B \sum_i p_i \ln p_i,$$ where $p_i$ is the probability of some microstate $i$. This is essentially the same thing as Shannon entropy for physical systems. With this concept one can view knowing ...


4

The probability density of the ground state is time independent, so there is no motion in this sense. Yet the expectation value of the kinetic energy is non-zero, so there is movement in this sense. How are these notions of movement reconciled? First off, classically, if we had a particle in a $1/r$ potential and released it from rest, it would indeed bob ...


3

I think your confusion is cleared up quite simply: you're confusing the terms "orbital" and "electron shell". An orbital is characterized by the three quantum numbers $n,\ell,m_\ell$. This terminology makes sense, because these three numbers together completely determine the spatial component of the wave function. However, this leaves a freedom in the spin ...


3

The continuum states are different in several aspects. First, there are a countably infinite number of bound molecular states vs an uncountably infinite number of states in any finite range of the continuum. Thus, the ratio of total number of bound states over the number of states in even a small range at the beginning of the continuum is effectively ...


3

To explicitly verify this, one solves the problem for a box of finite depth $V_0$. If you additionally assume the wavefunction and its first derivative to be continuous across the potential step, the solution becomes a matter of Solve the Schrödinger equation in the distinct regions in- and outside of the box. Match $\phi$ and $\phi'$ at the potential ...


2

Since $\hat{p}$ is a Hermitian operator, one can always expand the wave function $|\psi\rangle$ as a linear combination of the eigenstates of $\hat{p}$, $$|\psi\rangle=\sum_{p}\psi(p)|p\rangle,$$ where the eigenstate $|p\rangle$ satisfies the equation $\hat{p}|p\rangle=p|p\rangle$. With this setup, we can first show $\langle\psi|\hat{p}|\psi\rangle=\langle ...


2

Spin of an electron is measured as a magnetic property. You should not visualize it as an electron "spinning" around its axis, which is what you seem to indicate if I'm not mistaken. Electrons are considered to be point particles. Also, the spin of an electron never changes instantaneously. For example, changes in the electron's spin in the Stern-Gerlach ...


2

The 2nd way $$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$ will produce a complex result in general (in the example above it is will simply be zero), not having a physical measurement analog. The operator operates on some vector (either $\Psi$ or $\bar{\Psi}$), whereas the $|\Psi|^2$ is a simple real number.


2

For the incoming state, you don't know which spin state the particle is in, so you should average over the possible states. But you can measure the spin of the outgoing state, so to get the total cross section you should add up the cross sections for each spin. More formally, an unpolarized incoming particle should be described as a density matrix, $$ ...


2

Typically, the Hilbert spaces one considers in quantum mechanics are $L^2$ spaces. The elements of these spaces are equivalence classes of functions which differ only on a null set of points, i.e. whose distance in terms of the $L^2$ norm is zero, $\|n-\tilde n\|_{L^2}=0$. That is, you are right, but it's the $L^2$ norm that matters, not the Sobolev norm. ...


2

Consider self-adjoint operators $A$ and $B$, on a Hilbert space $\mathscr{H}$. Roughly speaking (forgetting about domains of definition), it is possible only if your Hilbert space can be written as $\mathscr{H}=\mathscr{H}_1\otimes \mathscr{H}_2$; where $\mathscr{H}_1$ and $\mathscr{H}_2$ are Hilbert spaces such that: $A=A_1\otimes 1$ and $B=1\otimes B_2$, ...


2

Here is another version of the same proof. If $\langle \psi | A \psi \rangle \in \mathbb R$ for all $\psi \in \cal H$, then $\langle \psi | A \psi \rangle^* = \langle \psi | A \psi \rangle$ for all $\psi \in \cal H$. Since $\langle \psi | \phi \rangle^* = \langle \phi| \psi\rangle$ we have that $\langle \psi | A \psi \rangle = \langle A\psi | \psi ...


2

The potential energy function is the same for both. The energy level solutions are the same for both. The key difference is that in (most modern interpretations of) the Schrodinger model the electron of a one-electron atom, rather than traveling in fixed orbits about the nucleus, has a probablity distribution permitting the electron to be at almost all ...


2

1) Why don't we consider finite dimensional representations of this group? As you said, we ask (anti)unitarity, so it is impossible to find finite-dimensional representation. 2) Why associate the Lorentz group to fields? The essence of the answer is what Trimok already said in his comment: the "translational part" of the Poincarè group is already ...


2

If all the c's are different, then since $X_r(\xi)$ is the quotient when $\phi(\xi)$ is divided by $(\xi-c_r)$, $(\xi-c_r)$ cannot be a factor of $X_r(\xi)$ else two of the c's would equal $c_r$. $X_r(c_r)$ is, by the remainder theorem (I imagine this has some other name), the remainder when $X_r(\xi)$ is divided by $(\xi-c_r)$. Since $(\xi-c_r)$ is not a ...


2

A very explicit argument: $$\phi(\xi)=(\xi-c_1)\dots (\xi-c_{r-1})(\xi-c_r)(\xi-c_{r+1})\dots(\xi-c_n)$$ where $r\leq n$ and $c_k> c_l$ whenever $k>l$. We can order the $c_i$ like this because the $c_i$ are real. Now, $$X_r(\xi)=(\xi-c_1)\dots (\xi-c_{r-1})(\xi-c_{r+1})\dots(\xi-c_n) $$ Clearly, the set of zeros of $X_r(\xi)$ is $\{c_i\ \bigl|\ ...


2

Ok, to expand my comment into an answer: There are only two finite-dimensional division rings (that admit division) containing the real numbers as a finite subring: the complex numbers and the quaternions (application of Frobenius theorem). Also, a vector space (and Hilbert spaces are vector spaces) is usually defined over a field, that is a non-zero ...


2

When the first observer performs the measurement the result will be an eigenvalue of the corresponding observable being measured (e.g. if an electron's spin is measured the result will be either +1/2 or -1/2). Now the system is in the corresponding eigenstate of the observable which was measured (e.g., either spin-up or spin-down). If this is also the ...


1

Imagine an experiment where you are sending one particle of spin 1/2 into a "black box" and waiting for the result. You know you are sending in one particle at a time, but you do not know the spin and presume it's going to be fifty fifty - so you say in fifty percent of cases you will have the spin "up" (or better "right" for relativistic particles) and add ...


1

I will give a counter-example. Consider the potential below, $V$ is considered to be very large but not infinity. The ground state of the system should be approximate as the ground state of an infinite potential well. The uncertainty is greater than $\hbar/2$. While when the particle is confined in the parabolic potential well, the local "ground state" ...



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