Tag Info

New answers tagged

3

Mathematical reason is that the time evolution operator is unitary, which means that $U^\dagger = U^{-1}$. Therefore $\langle \psi(t) | \psi(t) \rangle = \langle \psi(0)| U^\dagger U | \psi(0) \rangle = \langle \psi(0) | \psi(0) \rangle$. We can see that it is unitary by considering the Schrodinger equation: $$ \newcommand{\ket}[1]{| #1 \rangle} ...


2

I have to admire the fact that in the face of an infestation like this you wonder about the physics instead of setting the bed on fire to kill the termites... Looking at the video it seems to me that the pellets are not "dropped" but "ejected". If that is the case, then if the termites exit hole is slightly angled, they will eject their pellet not straight ...


2

This may be the argument: you have $N$ particles, and for each one you can put it on the left side or on the right side. Each of these choices, for each particle, leads to a different microstate. There are $2^N$ possible choices you can make for how to distribute the $N$ particles between the left and right halves of the box (assuming the particles are ...


7

The ratio between avaible microstates is given by the following ratio: $$ \frac{n_{B}}{n_{A}}=\frac{\int_{V/2}d^{3N}q d^{3N}p}{\int_{V}d^{3N}q d^{3N}p}=\frac{\int_{V/2}d^{3N}q}{\int_{V}d^{3N}q}=\frac{\int_{V/2}d^{3}q_1 \int_{V/2}d^3 q_{2}\dots }{\int_{V}d^{3}q_1 \int_{V}d^3 q_{2}\dots}=\left( V/2 \right)^N/V^N=\frac{1}{2^N}$$ where $N$ is the number of ...


2

Termite pellets are the feces of a termite, the left overs of its eating wood and in no way of the size shown by the size of the bed. The pellets are from a colony of insects. Often, the only obvious signs of infestation are little mounds of fecal pellets building up underneath the infested wood or the appearance of “kick-out” holes in the surface of ...


1

There is an equation that helps a lot understanding this issue: Fermi's golden rule $$ W_{i\rightarrow f}=\frac{2\pi}{\hbar} \left|\left<f\right|H'\left| i\right> \right|^2 \rho $$ It describes the transition rates from one state to another. $\rho$ is the so called Density of States (DOS) of final states. This system has only two states: The initial ...


0

To obtain the result $\frac{\text d \rho }{\text d t}=0$ you need two facts: the first is that the hamiltonian flow preserves the volume of phase space. The second fact is the conservation of probability, that is, the probability that the system is found in a volume $U$ at time $t=0$ equals the probability of finding it within $\Phi _t U$ at time $t$, where ...


0

You can think about it as in the case of a simple fluid. You need to enforce conservation of probability (in the same way you could enforce mass conservation in fluid dynamics for instance or charge conservation in electrodynamics). As far as I understand it, probability conservation is an additional premiss to the conservation of phase space volume under ...


0

The phase-space density $\varrho(p,q)$ tells us how many dynamical possible trajectories (DPT) pass through a given unit volume of phase space. Therefore it is a measure for the probability of finding a system in state $(p,q)$ since if there are more DPTs, it is more likely to find a system in this state (if we assume that every DPT is equally probable, that ...


0

What worries me, if I add a constant term $F_a = F_a + F$. I clearly drive the system out of equilibrium and detailed balance should be broken. However such a constant term does not influence the equation: $\sigma^2_m \partial_m F_a =\sigma^2_a \partial_a F_m $



Top 50 recent answers are included