# Tag Info

-1

Have you had a 3rd or 4th year QM course yet? The simple answer is that the angular momentum is in the wave-function. For spherically symmetric (3-D) potentials the solution to the Schrodinger equation is in terms of spherical harmonics (at least the radial part of the solution.) For any of the solutions you can act on it with the angular momentum ...

0

Consider the kinetic energy operator $$\hat{K}=\frac{1}{2m}\hat{p}^{2}.$$ Then $[r^{-1}, \hat{K}]$ is ambiguous depending on the coordinate system. Moreover, in Cartesian coordinates, it is quite peculiar compared to its Poisson bracket... (At least, if I did my math correctly, which is possible considering how sloppy/quick it was done, as I am ...

2

I don't know how elementary you consider a simple position dependent mass, but due to ordering ambiguity in the kinetic term $\hat{p}^2/2m(\hat{r})$ such a system will have a quantum Hamiltonian different from the classical one. For example: Analytic results in the position-dependent mass Schrodinger problem Position-dependent effective masses in ...

0

You are not required to regard decay as quantum mechanical (QM) phenomenon. Regard it as a nuclear reaction, which occurs spontaneously. There are similar reactions in chemistry, like ozone decay. Modern QM is unable to calculate decay probabilities and half-lives, they are only measured experimentally. In principle it is possible to model nucleus in the ...

1

I don't know what "penetration power" is or why quantum tunneling needs to be invoked. Sr-90 decays entirely via beta emission with up to $0.546\ \mathrm{MeV}$ given to the electron, and its daughter isotope similarly decays with up to $2.28\ \mathrm{MeV}$ given to the electron. These energy ranges are right around the $1.71\ \mathrm{MeV}$ of P-32, whose ...

3

Yes, beta decay of Sr-90 produces a 546keV beta (note that Y-90, the daughter nucleus, is also a beta emitter, but at 2.284MeV). This energetic electron can then produce bremsstrahlung x-rays from interactions with electrons. For a given x-ray energy, lead will have some absorption coefficient - really this is no different than visible light interactions ...

0

The paper addresses the case of a circular, finite quantum well with different electron effective masses inside and outside the well. The equation you point to is the result of a derivation from the time-independent Schrodinger equation, effective mass approximation and BenDaniel-Duke boundary conditions. It is also, however, dependent on the statement ...

1

Yes, this is possible - but only for states with zero total angular momentum. To see why, the first step is seeing that if $\Delta\sigma_x^2=\Delta\sigma_y^2=0$ on state $\psi$, then $\psi$ is an eigenstate of both $\hat\sigma_x$ and $\hat\sigma_y$: $$... 0 When you measure the system's position you actually push it out of the ground state, so your "trick" of using the initial total energy to compute the momentum after the position measurement doesn't work. To be specific, when you measure the position, suppose you get x_0 as the result. Then the particle's wave function is now a reasonably sharp peak at ... 5 Good question. You're right, you can't really throw out one of the solutions. The author of this PDF is using a poorly explained shortcut, which goes like this: if you expand out the full complex solution, you get$$\begin{align} \Psi(\theta) &= c_1 e^{i\omega t} + c_2 e^{-i\omega t} \\ &= (a_1 + b_1 i)(\cos\omega t + i\sin\omega t) + (a_2 + b_2 ...

2

The problem here is how to quantize systems whose classical hamiltonian involves factors of the form (for example) $p^nx^m$, because these cannot be unambigously represented in a formalism where $p$ and $x$ do not commute. As such there are many alternatives (all of which are classically equivalent) but only one is quantum-mechanicaly relevant. In most ...

4

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 ...

0

From http://plato.stanford.edu/entries/bell-theorem/ : Each modulator is controlled by amplification from a very rapid generator, which randomly causes one of two rotations of the polarization of the traversing photon. http://www.phy.pku.edu.cn/~qiongyihe/content/download/2-13.pdf The settings a, a′, b and b′ are generally in practice chosen ...

1

Usually experimental Bell tests are performed using photon polarization. However, experiments have also been devised to test Bell inequalities using so-called "time-bin" entanglement. Essentially, the quantum information is encoded in the time-of-arrival of the photons. Also, there are Bell-type tests using continuous variables. These continuous variables ...

1

You are on the right track. Pushing one more step to the final answer may leave you disappointed: $\sigma_x \sigma_K$ can equal zero! To see this, I find it more helpful to think just in terms of $x$ and $K$ as linear operators satisfying certain commutation relations, rather than thinking explicitly in terms of integrals of wavefunctions. Specifically, we ...

0

Your intuition about inertia is essentially correct. The spin state of an electron does not change instantly. If the electron is in the spin-up state then the z-component of its angular momentum is $\frac{1}{2} \hbar$. If it is in the spin-down state then its angular momentum is $-\frac{1}{2} \hbar$. Classically, angular momentum cannot change instantly, ...

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.

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? $$\langle p\rangle = ... 4 The probability density of the ground state is time independent, so there is no motion in this sense. Yet the expectation value of the kinetic energy is non-zero, so there is movement in this sense. How are these notions of movement reconciled? First off, classically, if we had a particle in a 1/r potential and released it from rest, it would indeed bob ... 2 Spin of an electron is measured as a magnetic property. You should not visualize it as an electron "spinning" around its axis, which is what you seem to indicate if I'm not mistaken. Electrons are considered to be point particles. Also, the spin of an electron never changes instantaneously. For example, changes in the electron's spin in the Stern-Gerlach ... 2 Since \hat{p} is a Hermitian operator, one can always expand the wave function |\psi\rangle as a linear combination of the eigenstates of \hat{p},$$|\psi\rangle=\sum_{p}\psi(p)|p\rangle,$$where the eigenstate |p\rangle satisfies the equation \hat{p}|p\rangle=p|p\rangle. With this setup, we can first show \langle\psi|\hat{p}|\psi\rangle=\langle ... 0 If I understand correctly the question (not sure I do...), The question is basically a linear algebra one. Consider an operator \hat{A}, that has a eigenvalue a, the eigenfunctions/eigenvectors of \hat{A} are denoted by |a\rangle such that:$$ \hat{A}|a\rangle=a|a\rangle $$Now consider a composite of \hat{A}\circ\hat{A}, that operates on ... 0 Both Rabi oscillations and the revivals are quantum mechanical effects. However, they consequences of the quantization of two different systems. Rabi oscillations can be explained and derived with a semi-classical theory in which the atomic system has quantized energy levels but the incident light fields are classical. Rabi oscillations do not require ... 1 Both L^p-spaces and Sobolev spaces are actually defined via equivalence classes (this is at least one of many equivalent ways). By definition, two functions that do not differ in norm are representants of the same function as already explained. Now, note that the Sobolev norm is just the sum of the L^p norms for weak derivatives (to whichever order you ... 5 It doesn't matter what sign you choose. Notice that since |A|^2 = \frac{2}{a}, you could even pick A = \sqrt{\frac{2}{a}} e^{i\phi}, so A doesn't have to be real. The reason is that a wavefunction is only defined up to a global phase. The reason is that we calculate probabilites with |\psi|^2 and mean values of operators with \int \psi^* \hat{O} ... 1 However then I say: Why do you make things so complicated? Suppose you want to calculate \exp(A). Why don't you define$$\exp(A)~:=~1+A+1/2 A^2 + \ldots $$and require convergence with respect to the operator norm. An example: Consider the vectorspace spanned by the monomials 1,x,x^2,\ldots and let A=d/dx. Then you can perfectly define ... 1 I'm trying to give a less technical answer. It's not rigorous but should give you the idea how spin and the regular rotation related. Maxwell's equations say in order to have magnetic field, you need a ring current. This can be achieved by giving angular momentum to charged particles. This can be orbital or simply because the particle is spinning. This was ... 2 Typically, the Hilbert spaces one considers in quantum mechanics are L^2 spaces. The elements of these spaces are equivalence classes of functions which differ only on a null set of points, i.e. whose distance in terms of the L^2 norm is zero, \|n-\tilde n\|_{L^2}=0. That is, you are right, but it's the L^2 norm that matters, not the Sobolev norm. ... 1 I want to post my attempt at the solution based on Qmechanic's hint (thanks!): Rewriting in cartesian coordinates$$ \psi_{100} \propto \exp[-\sqrt{x^2+y^2+z^2}/a]  <\psi_{100}|H|\psi_{100}> \propto \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dy \int_{-\infty}^{+\infty} dz \exp\left[-2\sqrt{x^2+y^2+z^2}/a\right] \alpha ...

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.

0

If in the Stern-Gerlach experiment we prepare an ensemble of particles in a superposition of spin up and spin down with respect to the z-axis and you could predict in every instance if we would obtain either spin up or spin down with respect to this axis but still reproduce the statistics corresponding to that superposition, there would at first sight be no ...

0

If we suppose that $\Psi$ is confined we must have $\Psi=0$ in the external region, so in the border of internal region we must have $\Psi=0$ and $\frac{d\Psi}{dx}=0$. But with our flat $U$, stationary K-G equation say that $\Psi$ is sinusoidal or exponential, so this requisites cannot be satisfied sumultaneously. Conclusion: the confinement of $\Psi$ is ...

0

As an example, http://en.wikipedia.org/wiki/Particle_decay, you can regard $\Delta t$ as the lifetime of the particle decayed.

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 ...

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 ...

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 ...

1

This is simply a consequence of basic properties of probabilities. Let us take an event $A$ made of independent events $A_i$ (so that $p(A) = \sum\limits_i p(A_i)$), and an event $B$ made of independent events $B_j$ (so that $p(B) = \sum\limits_j p(B_j)$) Then you have : $p(A/B) = \dfrac {\sum\limits_{i,j} p(A_i/B_j) \,p(B_j)}{\sum\limits_{j}\,p(B_j)}$

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" ...

0

$\sigma_{\vec{\imath}}$ is the rotation matrix about the vector $\vec{\imath}$: $$\sigma_{\vec{\imath}} = \imath_x\sigma_{x}+\imath_y\sigma_{y}+\imath_z\sigma_{z}$$ After expanding it out completely and after some shuffling around, the solution is as follows: $$e^{-i\theta/2 \sigma_{\vec{\imath}}^A} \otimes e^{-i\theta/2 \sigma_{\vec{\imath}}^B} ... 6 Consider the general case that we want to calculate$$ \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}$$This integral will converge if \int dr\, |F(r)| is finite. Such a function is said to be L^1. ... 0 (I will write \ell instead of l, because it confuses me less frequently.) Assuming that \delta definition is a typo, and it is supposed to be$$ \delta=(\ell+1)[2(\ell-1)j_{\ell-1}(kR)-kRj_\ell(kR)] Your matrix's determinant is nonzero... \begin{align} \alpha\delta-\beta\gamma &= \ell \left[R j_{\ell}(kR) k - 2 j_{\ell+1}(kR) \left(\ell + ... 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 ... 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 ... 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 ... 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, ...

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 ...

0

The only crucial point is the degeneracy of eigenspaces. Consider the finite dimensional Hilbert space $\cal H$ (the extension to the infinite dimensional case is more difficult also because a part of continuous spectrum may appear) and a pair of commuting Hermitian operators $A$ and $B$ on that space such that the following requirement is satisfied. R.: ...

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 ...

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 ...

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 $B=1\otimes B_2$, ...

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