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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 ...


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

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 ...


0

Ψ is supposedly a probability density amplitude. ΨΨ* is the probability density which in theory can be measured. For example in electron diffraction through a crystal a statistical measure of the electrons in, divided into the electrons out in a small region divided by the volume would allow ΨΨ* to be approximately measured. Phase information is lost when ...


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

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 ...


0

The $|00\rangle$ and $|1,M_S\rangle$ represent the spin singlet and triplet states. The overall wavefunction must contain both the 'space' part and the 'spin' part. We can schematically express this as follows: $$\psi \sim \psi(\mathbf{r}_1, \mathbf{r}_2) |s\rangle$$ Now, Pauli's exclusion principle demands the antisymmetry of the overall wavefunction. For ...


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} = ...


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.


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} + ...


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) ...


0

You're right, the Hamiltonian doesn't change. The Hamiltonian is that of a free particle, and doesn't depend on $L$. It is the boundary condition, which in turn determines the energies, that changes when the wall expands. A wall that expands to 2x its original width will give energy eigenvalues reduced by a factor of $2^2=4$.


0

Since the wavefunction depends on r, which is the spherical coordinate representing the distance from the origin, we use spherical coordinates to perform the integration because it is most convenient. And yes, this is a triple integral, $\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta\int_0^{\infty}r^2\Psi^*\Psi dr$. The wave function doesn't depend on the ...


0

Integrate the complex square of the wavefunction $\Psi^*\Psi=\frac{A^*A}{r^2}$ over all space and set the result to 1. Since the wave function is given in spherical coordinates, it would be easiest to integrate in spherical coordinates using the volume element $dV=r^2\sin\theta drd\theta d\phi$ where $0\leq r\leq \infty$, $0\leq \theta\leq \pi$ and $0\leq ...


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 ...


0

In general, when you make any measurement on a system, the wave function collapses to the eigenstate of the measured observable with the eigenvalue corresponding to the measured value. So for a position measurement, it is the delta function in space; for energy measurements, it is one of the eigenstates of the Hamiltonian (in this case, sinusoidal ...


0

it is the dirac delta function with the spike at L/2. if you make a a measurement of energy some time after - the function is still a delta function in spacetime with a definite energy.


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 $\sqrt{\frac{2}{3}}$ ...


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, ...


0

$k^2$ is easy and I assume you know how to expand that into its real and imaginary parts. As for $\ln k^2$, use the fact that (the principal branch of) $\ln(r e^{i\theta}) = \ln(r) + i\theta$. So, if you can write $k^2$ in the form $r e^{i\theta}$, you're done. Now, writing a complex number $a + i b$ in its polar form $r e^{i\theta}$ isn't hard: $r = ...


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.


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 ...


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

It is a real wave associated with electron. And we adopt its mathematical description using wave function in configuration space


0

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.


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 ...


-2

The big questions with this experiment is something else. It is about of recording of where The particule goes. On The Left ore right. Or Both. When is recording The particule is moving as a particule , no interference. But When it is no recording by sensors it has a wave behaviour. Moreover other experiments confirm That The light is Smart. It know That it ...


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 ...


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

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

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}) = ...


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

To be sure, the TISE solutions (in Cartesian coordinates) for a free particle are not normalizable and thus are not physical states. Quoting Griffiths "there is no such thing as a free particle with a definite energy" However, these free particle solutions are a basis for normalizable states including the wave packet. That is to say, by Fourier ...


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: $$ ...



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