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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 ...


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 $d-1$ ...


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 ...


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 ...


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 ...


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 ...


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 ...


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

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 ...


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.


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 ...


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 ...


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 ...


1

Essentially, separation of variables in the time-independent Schroedinger equation amounts to diagonalizing the Hamiltonian. One can see this easily by considering the case where the Hilbert space is finite-dimensional, and the Hamiltonian is a Hermitian matrix. In case of a partially continuous spectrum one gets the same, except that the sum must be ...



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