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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 ...


1

If we know there is a particle(s) in state $i_1$ why do we need $r_1$? Does the state $i_1$ not specify position? The fact that the particle is in a state $|\psi\rangle$ does not specify the particle's position - it specifies the wave-function $$ \langle x | \psi \rangle = \psi(x) $$ from which one can find $$|\psi(x)|^2$$ This is the probability ...


1

$\newcommand{\ket}[1]{\lvert #1 \rangle} \newcommand{\bra}[1]{\langle #1 \rvert}$The states of a quantum system are nothing else than the abstract vectors in the Hilbert space of states $\mathcal{H}$. For one particle, given a basis of position eigenkets $\ket{x}$ with $\hat{x}\ket{x_0} = x_0\ket{x_0}$ and a state $\ket{i}\in\mathcal{H}$, the wavefunction is ...


3

Before going further, I would suggest you to read Chapter 13 ("Spinors") of R.Wald's book "General Relativity". In that chapter, you will see that 2-spinors are simply vectors living in a two-dimensional complex vector space. The capital letters in the indices are simply the abstract index notation for these vectors (see Section 2.4 in Chapter 2 of the same ...


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 ...


1

Four component formalism is the "right" formalism, but it has negative energy eigenstates corresponding to the antiparticles. Most chemists and solid state physics are not interested in the antiparticles, and such negative energy solution causes trouble for conventional variational methods, where you might end up falling to negative infinity energy. It is ...


0

This is just a wild guess, but could it be the position vector of the element?


1

The symbol kW(e) or kW${}_e$ refers to the "kilowatt electrical". It is the part of the power that is actually used by the devices connected to the power station or the grid, effectively the average of $P = U\cdot I$. The "kilowatt electrical" should be contrasted with "kilowatt thermal" i.e. kW(th) which represents the power including the thermal losses. So ...



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