Tag Info

2

You know that $2 \sin(\omega t) \cos(\omega t) = \sin(2\omega t)$ right? So it is clear that the instantaneous power can be written as $$P = -\dfrac{1}{2}\omega\dfrac{F_o/m}{{\omega_0}^2 -{\omega}^2} F_0 \sin{2\omega t}$$ As your textbook says, this is symmetric around zero - it spends as much time being positive as it does being negative. When you ...

2

I had missed this question. Most potentials physicists use to model the behavior of atoms and molecules, are symmetric in space. When the potential is unknown it is usual to guess at a first term in a Taylor expansion . For symmetric potentials the first term in a Taylor series expansion is the x**2, and that is why the use of the harmonic oscillator is ...

2

It turns out, as a general fact, that theories which contain only quadratic terms in position and momentum can be solved as a linear combinations of SHO (just look at the SHO Hamiltonian and you will find why). The quantum treatment of the harmonic oscillator enriches the interpretation adding the creation and annihilation operators which (as you can guess) ...

2

Absolutely. What Feynman means when he says the amplitude is infinite is indeed that the amplitude grows limitless as $t \to \infty$. When we talk about the amplitude of a harmonic oscillator, we typically refer to the oscillator in its steady state. However, the undamped system at resonance never reaches a steady state and grows to have an infinite ...

2

The equation $$\frac{d^2 x}{dt^2} + \omega^2 x = 0$$ is an example of a homogeneous second order linear differential equation, with a general solution of the form $$x = A\sin(\omega t) + B\cos(\omega t),$$ where $A$ and $B$ are constants to be determined from the boundary conditions of the problem. If you now put a constant on the right hand side of ths ...

2

Let me say basically the same thing as Rob, but in what seems to me a simpler way. If you start with the equation: $$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \omega^2x= c \tag{1}$$ We can rearrange it as: $$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \omega^2\left(x- \frac{c}{\omega^2}\right) = 0$$ Then you can define a new variable $z$ by:  z = x - ...

1

Einstein's Equivalence Principle (also derivable I believe from Newton's laws) states that being in an accelerating frame of reference is indistinguishable from being under a gravitational force. In particular, the mechanical laws in an accelerating frame of reference are the same as if a gravitational field of equivalent magnitude were added in the ...

1

If I understand the question. Assume you have a quantum oscillator, then if the system is an eigenvector of the hamiltonian, then a series of measurements will give a series of results reflecting eigenvalues of the system. The appearance of these eigenvalues must obey at a number of many measurements the probabilities of each eigenvalue to appear. It's ...

1

That system is called Primon gas (Wikipedia). The Riemann zeta function also pops up in another field theory context, see this table (nLab). Also, your use of $a$ and $a^\dagger$ appears flipped.

1

If you are asking, as suggested in the comments, if it is possible to define in a rigorous fashion $\ln (N+1)$, where $N$ is the self-adjoint number operator, then the answer is yes. The spectral theorem allows you to define the function $f(A)$ of a self-adjoint operator $A$, as long as $f$ is measurable wrt the spectral measure of $A$ (and this is the ...

1

The second Hamiltonian is different from the first! There is an extra term of -$\frac{\hbar\omega}{2}$ This terms comes from the fact that $im\omega(xp-px)=-\hbar m\omega$ So, obviously you have gotten an answer with a shifted ground state. But, I believe the answer for $E_n$ should $n\hbar\omega$, with $n=1,2,\dots$. Note that, $n=0$ is no longer the ...

Only top voted, non community-wiki answers of a minimum length are eligible