Recent content by matteo137

I found the mean to be $$\langle n\rangle=\vert\alpha\vert^2 \tanh(\alpha^2)$ and $\langle n^2\rangle=\vert\alpha\vert^2 \left( \alpha^2\sech(\alpha^2)^2 + \tanh(\alpha^2) \right)$$. Do you know if there is any reference where I can check if this is correct?

Consider two harmonic oscillators, described by annihilation operators a and b, both initially in the vacuum state. Let us imagine that there is a coupling mechanism governed by the Hamiltonian H=P_A P_B, where P_i is the momentum operator for the oscillator i. For example P_A =...

I think it is usually the average weighted by the isotope abundance.

I would call it point-particle theory...

Thank you!

The sidebands carry information because they ARE the information. Think about the Fourier decomposition of the signal (hoping you are familiar with it). Let's say your "information" is a 1 KHz signal, so the spectrum of this signal is a dirac-delta at that frequency. When you modulate a carrier...

According to quantum mechanics it is not possible to have a region of space without any photon. If you want a region of space without photons you can detect, probably you can have it between two metal plates, separated by a length which is smaller than half of the smallest measurable wavelength...

Often it is not numerically stable to blindly solve equations with coefficients in such a large range. What you have to do first, is to understand which terms you can drop/neglect, for example because they are many orders of magnitude smaller than the others.

B is not an up-conversion of the signal you want to transmit. In other words, let's say you want to transmit a 100 kHz signal, you do not want to amplify it an send into the antenna... what you want to do is to up-convert it in frequency, and then send an electromagnetic wave of frequency e.g. >...

Why can't you drop the terms multiplied by ##10^{-35}## and ##10^{-36}##? I tried doing it, and it works.

First, I think your starting equation is wrong: inside square brackets you have a scalar, and then you take a dot product? Try with ##\nabla^2P = - \rho \nabla \cdot (V \cdot \nabla)V##

There is an i too much in the term in the middle The term in the middle is again wrong... Basically the last two equations you wrote before the Griffiths insert are wrong. Once you have them right, you know that the Hamiltonian is H=\dfrac{p^2}{2 m} + \dfrac{1}{2} m \omega^2 x^2

1. No. If a state is accessible by the system it does not mean that the probability to find the system in that state is the same as finding it in an other one. 2. A gas of ideal particles, what are the micro states? they are the positions and the velocities of every single particle. If you...

##\Psi(x,t)## is a time-depemdent wave function, ##\psi(x)## is a time-independent wave function / stationary state. Since the Schroedinger equation is linear, if you have a solution of the type ##\psi(x) = \phi_1(x) + \phi_2(x)##, you can study the two ##\phi_i(x)## separately, and then...

My first guess would be to first amplify the photodiode signal using an opamp and then send the amplified signal to a low pass filter (first/second order?).