# Tag Info

0

I cannot be sure, but I suspect that you can get analytical solutions of the Pauli equation by taking a non-relativistic limit of analytical solutions of the Dirac equation. The latter can be found in many books, say Bagrov, Vladislav G. / Gitman, Dmitry, The Dirac Equation and its Solutions (http://www.degruyter.com/view/product/177851) (you can find a ...

4

In the quantum mechanical description of any physical system, including a quantum field or a collection of interacting quantum fields, there is always one state vector – one collection of numbers (probability amplitudes) that generalizes what is referred to as the "wave function" in quantum mechanics of particles. In quantum field theory, a better name is a ...

0

A progressive wave is a function of space and time where the dependence on space and time can be modeled by a function of one variable, composed with an “argument” function, which combines space and time and describes the geometry of the wave. Wavefronts are the loci where the “argument” function is constant. Let’s give a few examples. Plane wave If ...

4

The basis is still $\{|\boldsymbol r\rangle\}$. The abstract Schrödinger equation is $$i\frac{\mathrm d}{\mathrm dt}|\psi\rangle=H|\psi\rangle$$ where $|\psi\rangle$ is a set of four kets, (with a slight abuse of notation) $$|\psi\rangle=\begin{pmatrix}|\psi_1\rangle\\|\psi_2\rangle\\|\psi_3\rangle\\|\psi_4\rangle\end{pmatrix}$$ Time is still a ...

1

Momentum and position are conjugate variables in classical mechanucs, which means they satisfy the Poisson bracket relationship. When quantum mechanics was invented the Poison bracket relation was replaced by the operator commutation relationship which results in the relation under consideration.

2

These relations are found in every book on QM, but the usual notation is $$X|x\rangle=x|x\rangle$$ and $$P|p\rangle=p|p\rangle$$ To go from these equations to the ones you've written, you just have to project them into the position basis $|x'\rangle$ (and use $\langle x'|x\rangle=\delta(x-x')$ and $\langle x'|p\rangle\sim\exp[ipx]$). Edit Important: ...

-1

Ab initio the momentum operators can be constructed using de Broglie Plane waves In one dimension, using the plane wave solution of the Schrodinger equation,the wave function Psi = exp. i (kx -wt) , if one takes the partial derivative w.r. to x of the wave function delta/delta x (Psi) = ik. Psi and using de-Broglie relation p = hbar . ...

-1

Momentum is the generator of spacial translations, even in classical physics. Anyway, you can find a derivation here or in Sakurai's book Modern Quantum Mechanics. They are more or less the same and go like this: The translation operator is the operator $T( a)$ such that $$T( a) \mid x \rangle = \mid x+a\rangle$$ From the definition it follows that the ...

0

That $\hat{P} = - i \hbar \partial_x$ generates translations comes from a straight-forward computation: if $\psi$ is continuously differentiable, and $\Psi$ as well as its derivative are square integrable, then you can prove that \begin{align} i \frac{\mathrm{d}}{\mathrm{d} y} \bigl ( \psi(x - y) \bigr ) \big \vert_{y = 0} = - i \partial_x \psi(x) ...

4

Historically, you probably want to start with the de Broglie relations (i.e. $p = \hbar k$), which are just a wild guess. This immediately pops out the form of $p$ as an operator if the wavefunction is a plane wave. Mathematically, $p$ should be defined as the generator of translations (or equivalently the conserved quantity corresponding to translational ...

0

If a particle is a wave function describing a probability amplitude distributed through space, what happens when two wave functions meet? Well, I just try to think allowed and one can not portray a 'wave function' picture unless a physical system is at hand to use the picturization or description. Let us think about a hole in the reactor and neutron ...

1

Wavefunctions combine trough tensor products, which is not the addition that one would expect naively. The reason for this is that a wavefunction contains the description of all possible futures of the system at once, so if there are multiple subsystems, then the wavefuntion of the entire system has to describe all possible futures for each part ...

0

To get A normalise the wavefunction. To get probabilities note you can expand any wavefunction in terms of energy eigenfunctions. Work out the co-efficients for each eigenfunction in the expansion by multiplying each side of the expansion with eigenfunction with quantum number n then noting that eigenfunctions with different n are orthogonal. the ...

3

The uncertainty of quantum mechanics does not refer to the uncertainty in the mind of a scientist (human or otherwise). The uncertainty is inherent in the system under observation. Say we are trying to measure the position and velocity of an electron. The uncertainty principle does not limit the quality of measurement we can make of the electron. The ...

1

Any interaction conveys "information" from one particle to another about its state. That doesn't require anything as involved as consciousness: if a photon interacts with an electron, and that electron is knocked out, the remaining electrons will rearrange themselves to lower their energy. What sequence of events (and therefore, what photon emissions) will ...

5

A nice overview of the problem is given in arXiv:1205.3740. I'll summarise the most important points here. Let $d$ be the number of space dimensions. Then the Laplace operator is given by $$\Delta=\frac{\partial^2}{\partial r^2}+\frac{d-1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_S\tag{1}$$ where $\Delta_S$ is the Laplace operator on the $d-1$ ...

0

Photon C is in some quantum state near Alice. A and B are entangled photons. Pairs of entangled particles can be described by a wave function that is a superposition of four possible states, each with a prob. of 1/4. Photon B is sent to Bob far away, e.g. via laser bean. Now Alice has A and has C. A wave function describes three particles, C, A and B, again ...

2

First, the geometric extent of the (quantum mechanical) wave packet doesn't mean that the particle (atom in your case) has become diluted all over the volume like if it were fog. Instead, the right interpretation of the wave packet is in terms of probabilities. For example, assume that there is 1 atom in a box. The initial wave packet is nearly localized ...

2

Yes, the gaussian wavepacket can get narrower as the time passes indeed. It's a matter of phases. You know that a gaussian wavepacket is the superposition of plane waves, each one having a precise wavevector. So it really depends on how you "prepare" this superposition, i.e. on how you set the phase of each chromatic component. If at $t=0$ all the plane ...

0

There are explanations for why the probability of a measurement of a particular observable is the square amplitude of the relevant eigenstates. The square amplitude is the only quantity that fits the calculus or of probability, and the rules of decision theory, and quantum physics: http://arxiv.org/abs/quant-ph/9906015 http://arxiv.org/abs/0906.2718. The ...

1

Differentiating $$\sigma=\sqrt{a^2+\left({\hbar t \over 2 m a}\right)^2}$$ you get - $$\frac{\partial \sigma}{\partial t} = {\hbar^2t \over 4 a^2 m^2\sqrt{a^2+\left({\hbar t \over 2 m a}\right)^2}}$$ In the limit of $t \rightarrow\infty$ you get $$\left(\frac{\partial \sigma}{\partial t}\right) _{t\rightarrow \infty} = {\hbar \over 2 m a}$$

1

I know that the usual interpretation of the wavefunction in QM is that it´s a probability distribution of measurable quantities. It is a postulate that is necessary to choose the subset of mathematical sets that can be useful to modeling nature. The postulate was chosen because observations were fitted by the hypothesis. Not a deterministic ...

10

This is indeed possible only in some situations, e.g. when the continuous spectrum is absent (it may also consist of a single point, see Valter Moretti's comment below). A sufficient condition for that to be true is that either the Hamiltonian is compact or it has compact resolvent. Sadly, very few interesting Hamiltonians satisfy that property (an example ...

-1

Quantum mechanics is a theory that can only predict probability distributions. It cannot predict trajectories. It is ruled by differential equations which have as solutions the wavefunctions, and the complex conjugate square of the wavefunction gives the probability of a specific, photon, electron, to be at (x,y,z,t) given the boundary conditions of the ...

1

According to quantum physics, when certain different polarizers are placed over the slits in the double-slit experiment (for instance, one vertical and one horizontal polarizer, or one circular clockwise and one circular counter-clockwise), thus "marking" each photon with which-way information, the photon indeed passes through only one slit, resulting in no ...

0

A qualitative understanding can be gained from the animation provided by 'TheGhostOfPerdition', each small peak on the dark blue wave (envelope of the wave packet) has an associated $k_{0}$ for which $\phi(k)$ peaks at $k = k_0$. So you can work out the velocity $v_g : = \frac{d \omega}{d k}|_{k = k_0}$ of the dark blue (envelope of wave packet) using the ...

Top 50 recent answers are included