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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 ...


8

This is an important question. You are correct that the energy expectation values do not depend on this phase. However, consider the spatial probability density $|\psi|^{2}$. If we have an arbitrary superposition of states $\psi = c_{1} \phi_{1} + c_{2} \phi_{2}$, then this becomes $|\psi|^{2} = |c_{1}|^{2}|\phi_{1}^{2} + |c_{2}|^{2} |\phi_{2}|^{2} + ...


4

You are correct in one thing: if an atom in an isotropic medium spontaneously emits a photon, it can do so in any direction at all, and the overall emission will be evenly spread over the unit sphere. However, lasers work using stimulated emission, which is slightly different: if an atom is excited, you can induce it to emit its energy by shining an initial ...


4

The equation you phrase as $$|l,m\rangle=\int_\text{all space}\psi_{lm}(r,\theta,\phi)\,\left|r,\theta,\phi\right\rangle r^2\,\mathrm dr\,\mathrm d\Omega$$ is, and must be, wrong. The reason is that $|l,m⟩$ inhabits the orbital part of Hilbert space, $\mathcal H_\Omega$, and the right-hand side is a vector in the full Hilbert space $\mathcal H$, which is the ...


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 ...


3

Also you can modify the wavefunction with a global phase $\psi(x)\rightarrow e^{i\phi}\psi(x)$ without affecting any expectation values because the phase factor will cancel when taking inner products, so this global phase doesn't contain any information. Only relative phases are meaningful in quantum mechanics.


3

What does measurable mean ? It means that one can do an experiment and get a value for a+ib , the complex number. A complex number to be measurable one should be able to measure a value at the same time for a and b and put a point on the complex plane. This means two independent variables, a and b can be measured and a point defined. In quantum mechanics ...


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

By normalization condition you get$$\int_0^{2\pi}\frac13+c^*c+\frac1{\sqrt3}c^*e^{i \phi} + \frac1{\sqrt3}ce^{-i \phi}=2\pi$$ Now we know that $e^{i\theta}=\cos{\theta}+i\sin{\theta}$ thus its integration over a period of $2\pi$ is 0. Thus our equation reduces to $$cc^*=\frac{2}{3}$$ Thus any complex number who's magnitude or modulus is $\sqrt{\frac{2}{3}}$ ...


2

First I want to point out that most of these questions do not bring up issues specific to Bohmian mechanics. That's not a criticism, I'm just pointing out that these notations and concepts are already employed in standard quantum mechanics, or even in classical mechanics. I am going to answer this a little casually, and then make my answer "community ...


2

In it's simplest form, de Broglie's hypothesis is meant to be applied to fundamental, indivisible particles, like an electron (an electron is fundamental and indivisible to within our current experimental precision at least). In that case it doesn't make semse to talk about half an electron, or to divide the mass of the electron among its parts. There is 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

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

How come a free particle (take a free electron) has a representation as a plane wave and not as something with spherical symmetry? Because we specified a single outward momentum vector $\vec{k}$, with a direction to it. This breaks spherical symmetry, but it's not required. We can do spherical waves too, and we often do in scattering theory: $$ ...


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

The mean value of the position is given by $\langle \psi|\mathbf{X}|\psi\rangle$. Now insert the completeness relation for position states to get $$\langle \psi|\mathbf{X}|\psi\rangle=\int_{\mathbb{R}^3} \mathbf{x} |\psi(\mathbf{x})|^2\,\mathrm{d}V$$ From the OP, we see that the modulus squared of the wave function is ...


1

For a particle of mass $m$ with a simple Hamiltonian in position-space $\mathcal{H} = -\frac{\hbar^2}{2m}\nabla^2+V(\vec{x})$, if you write a general wavefunction as $$\Psi(t;\vec{x}) = \sqrt{\rho}e^{iS/\hbar}\text{,}$$ where $S$ and $\rho\geq 0$ are real, then the phase information $S$ directly corresponds to the probability current $$\mathbf{J} = ...


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 ...


1

Further to Anna V's answer, in the case of an electron there is an important physical meaning to the "lack" measurability of the phase of the electron's wavefunction. This is because the electron is coupled to the electromagnetic field. And, if one models this by the Minimal Coupling between the electron and the electromagnetic field, one gets the ...



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