# Tag Info

0

Starting from the last equation and taking into account the braket one gets: $$e^{-|\alpha|^2}\frac{\partial}{\partial\alpha}\sum_{n=0}^{\infty}\frac{\alpha^{n+1}}{\sqrt{(n+1)!}}|n+1\rangle\sum_{m=0}^{\infty}\frac{\alpha^{*{m}}}{\sqrt{(m)!}}\langle m|$$ Relabeling the index according to $l=n+1$: $$e^{-|\alpha|^2}\frac{\partial}{\partial\alpha}\sum_{l=1}^{\... 0 The missing term in the summation containing the vacuum state is independent of \alpha, isn't it? In other words,$$\frac{\partial}{\partial \alpha} \sum_{n=0}^\infty \frac{\alpha^{n+1}}{\sqrt{(n+1)!}}|n+1\rangle = \frac{\partial}{\partial \alpha} \left( \sum_{n=0}^\infty \frac{\alpha^{n}}{\sqrt{n!}}|n\rangle - |0\rangle \right) = \frac{\partial}{\partial \...

1

As is shown in Griffiths (equations 2.54 or 2.57 in the third edition), the Hamiltonian can be written as $$\hat{H}=\hbar\omega\left(\hat{a}_+\hat{a}_-+\frac{1}{2}\right)\text{ or }\hat{H}=\hbar\omega\left(\hat{a}_-\hat{a}_+-\frac{1}{2}\right).$$ The energy corresponding to the $n$th state is $E_n=\hbar\omega\left(n+\frac{1}{2}\right)$ by equation 2.62. You ...

-1

Assuming no friction and conservation of energy, the maximum height on the other side would be 100 km. You need to find the time to fall 100 km, double it and add to the time from your formula (which assumes constant density and no friction). If you allow for the change in g over a height of 100 km, the fall time is given by a very complex expression. I ...

3

$\hat{a} f(\hat{n})= f(\hat{n}+1)\hat{a}$, so that $$\left[\hat{a}, f(\hat{n}) \right]= \Bigl (f(1+\hat{n})- f(\hat{n})\Bigr )~ \hat{a} \\ = \hat{a} \Bigl (f(\hat{n})- f(\hat{n}-1)\Bigr ) .$$ Note its action on $|0\rangle$, $|1\rangle$, ...

0

I'll be appealing to you physical intuition, so that I don't have dive deep into functional analysis. Wavefunctions are mathematical objects which are defined in a Hilbert space, which is square integrable. Now, the operators you see in QM are defined on this Hilbert space. Roughly, there are kind of linear functions which map from one Hilbert space to ...

0

$a_-\psi_0$ results in the zero-vector. Call this vector $\vert\hbox{0 vector}\rangle$. Then in any computation $$\langle \psi_n|\hat T \vert\hbox{0 vector}\rangle =0$$ for any operator. In particular the length of $\vert \hbox{0 vector}\rangle$ is $0$ and in this sense it cannot be normalized.

2

The $\delta$ distribution is defined by the result of integrating it against some collection of test functions. Specifically \begin{align} \int dt\,\delta(t - t') f(t) = f(t'). \end{align} If we replace $\delta(t)$ with $g(t) = e^{-i \omega |t|}\delta(t)$, \begin{align} \int dt\,g(t-t') f(t) &= \int dt\, \delta(t-t') e^{-i\omega|t-t'|}f(t)\\ &=e^{-i\...

0

For an intuitive explanation, imagine pushing a child on a swing. Once the swing starts moving, you wait to push it again until it reaches the high point of its backwards motion. If you try pushing the swing when it is already swinging forward, it is not going to be as effective. If you try pushing the swing while its swinging backwards, you're pushing ...

0

I don't believe that the phenomenon is caused by math although it is hard to explain without using some math. The reason is that if a system is pushed slightly out of equilibrium, there is a force that opposes thus. This force is proportional to the deviation, for example the displacement of a string or a pendulum. This leads to the equation of motion $$m \... 0 Things actually don't vibrate at just one frequency. That's a simplification. It happens to be a good simplification for a heavy mass on a spring that's oscillating a small distance. Your title question cannot be answered without math because we have to define "natural frequency." With math, we can describe that using variables and equations. ... 0 The frequency depends on the spring stiffness, k, and the mass, m, as the math shows. For a more intuitive explanation, suppose we watch as the mass moves through the middle (equilibrium) position and the spring starts to stretch. There are two competing factors: the stiffness of the spring, which wants to pull the mass back to the equilibrium position, and ... 1 Why the mass spring system prefers to oscillate at natural frequency, why does it rather oscillate at other frequencies? Because the math tells us so! For a bob of mass m connected to a Hookean spring with constant k the Newtonian equation of motion is:$$ma=-kx$$Or with a=\ddot{x}, we get:$$\ddot{x}+\frac{k}{m}x=0$$Or with \omega^2=\frac{k}{m}$$...

0

If the wall disappears after the touch impact, the ball will bounce back,even if the impact is infisinitessimal in time. You are forgetting conservation of momentum, which is calculated by what is taken up at impact by the wall , and the bounced ball. No possible oscillations . Thinkof it. As far as the ball is concerned, the impact is at one point in time,...

3

I think that the ball will start oscillating at its position. I don't think it oscillates—rather, its size oscillates. At the moment of maximum potential energy in the ball, its velocity is zero. If the wall disappears at this instant, the ball just falls vertically downwards with its size oscillating in a damped manner (due to continuous expansion and ...

1

Let the pseudoforce be $f$. Then the equation of motion in the non inertial frame is $$m \ddot{x} = -kx + f = -k\left(x - \frac{f}{k}\right)$$ Change variables to $X =x - \frac{f}{k}$, then the equation of motion is $$m\ddot{X} = -kX$$ which is the equation for shm with solution $$X = A \cos (\omega t + \phi)$$ where $\omega = \sqrt{k/m}$ and $A$ is the ...

0

I suppose that you are in euclidean space and so placing yourself in a non-inertial frame corresponds to apply a global transformation, in the sense that every point of space has associated this acceleration vector that corresponds to the opposite of your acceleration seen from an inertial frame of reference. Consider first the case in which you have an ...

0

I don't know what the author of the question thinks, but no one answered his question. He asks why the electromagnetic wave oscillates, the answers explain that these oscillations are induced by the oscillations of other particles, but then, why does the first source which transmitted oscillations oscillate? I would add this: why must all the particles also ...

1

It follows from the spring's force law: $F = kx$, or $a = (k/m)x$. The important point is acceleration is proportional to distance from the center. You are aware that one solution to the equation is $x = A_1sin(t/T)$ So why is $x = A_2sin(t/T)$ also a solution? Physicists will answer because both solve $F = kx$. This is correct, but it doesn't give any ...

2

An analytical explanation with classical mechanics may be useful to some \begin{array}{l} T = \frac{1}{2}m{{\dot x}^2}{\rm{ (kinetic)}}\\ V = \frac{1}{2}k{x^2}{\rm{ (potential)}} \end{array} Lagrangian approach \begin{array}{l} L = T - V\\ \frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial \dot x}}} \right) - \frac{{\partial L}}{{\partial x}} = 0\\ \...

3

I mean, if you compress a spring more like push it more in, then it would have more distance to travel and hence need more time Intuitively you can think like this : If you compress a spring greater , and then release it the force on the block by spring is greater. Hence it's acceleration(avg) is faster due to which it can cover more distance in the same ...

0

Roughly speaking, the reason the time period of a spring-block system is independent of initial displacement is that when the displacement is high, even though the block has to travel greater distance to reach the equilibrium position, the force becomes larger with more displacement (and hence the acceleration) in such a way that the block is able to ...

0

The short answer is: because of conservation of energy and momentum. If energy and momentum are conserved, nothing should get in the way of the movement, no matter the mass. So the only two things that actually matters are the properties of the spring (the$k$) and how much mass is there to oscillate. Similar stuff happens in the pendulum.

0

Your formula will give the velocity at any height on the swing down, and that can be used in the Doppler equation. Recording the sound is not a problem, but measuring the (variable) frequency may be a challenge. If you can freeze (or video) the display on an oscilloscope, you might measure the period of each cycle of the sound wave.

2

It essentially follows from a bunch of relabelings of the dummy indices. In the second line of eqn (1), you have the sum $$\sum_{i j} \sum_{\kappa \alpha, \nu \beta} \phi_{\alpha \beta}\left(\boldsymbol{R}_{i}+\boldsymbol{d}_{\kappa}-\boldsymbol{R}_{j}-\boldsymbol{d}_{\nu}\right) u_{i \kappa, \alpha} u_{i \kappa, \beta}.$$ First relabel the indices $j\to ... 3 For a one dimensional harmonic oscillator,the energy eigenfunctions are given by.. $$\psi_n = \Biggr(\frac{m\omega}{\hbar\pi }\Biggr)^{1/4}\frac{1} {\sqrt {2^n n!}} \exp \Biggr(\frac{-\xi^2}{2}\Biggr)H_n(x)$$ $$\langle m \vert x^6 \vert n \rangle = \Biggr(\frac{m\omega}{\hbar\pi }\Biggr)^{1/2}\int_{-\infty}^{\infty}\frac{1}{\sqrt {2^m m!}} \exp \Biggr(\... -1 Oscillation happens around one fixed point while vibration happens around two fixed points. 1 I think this is it: It seems like we can treat X as the Fourier transform of p to explain the factors. 1 The idea of a photons dates back to 1905, when Einstein discussed the photo-electric effect. According to Einstein a photon is a quantised light package, and therefore an elementary excitation of the electromagnetic field. However, it is not immediately clear how we could formulate this conceptional idea mathematically. The mathematical description of a ... 0 The Hamiltonian of the quantum harmonic oscillator is given by$$ H = (N+\frac{1}{2}) \hbar \omega$$N corresponds to the number operator and its eigen values corresponds to number of photons in the cavity. The author has ignored the zero point energy which is quite common thing to do. Hence the Hamiltonian becomes$$H = N \hbar \omega $$Now consider an ... 0 Actually, this is the rigorous definition of a photon in quantum field theory: the excitation of a mode (or, in the langauge of the cited book, the excitation of a harmonic oscillator associated with the mode). The problem with this definition is that it is at odds with more intuitive notions of a photon as a particle. These are not without their merits: e.g.... 2 The other answers have begun by assuming a force linear in displacement. Let us take a step back and see why harmonic oscillators are popular in physics. Consider perturbing a system slightly off a stable equilibrium. This means that it is perturbed from a minima (by a small amount \epsilon). Taylor expanding the potential energy (as ideally, kinetic ... 1 Let's consider the one-dimensional case for the sake of simplicity. Further, assume the oscillations are simple-harmonic and that the force is conservative. Then by definition for such oscillations,$$F_{\text{rest}} = -m \omega^2 x$$where F_{\text{rest}} is the restoring force. But we know that F_{\text{rest}} can be written as the negative-gradient of ... 1 In a harmonic oscillator,$$F=-kx$$where k is a constant and x is the displacement from the mean position. Thus,$$\frac{\mathrm dF}{\mathrm dx}=-k\tag{1}$$about the equilibrium point. Now,$$F=-\frac{\mathrm dV}{\mathrm dx}\tag{2}$$where V is the ootential energy at any point. Now substituting equation (2) in equation (1), we get$$-\frac{\... 2 I could not find the reference I mentioned where they perform perturbation theory starting from both the quantum harmonic oscillator limit and the quantum rotor limit and connect their solutions and energies with the exact solutions of the quantum pendulum. Here I will just briefly scheme the basics of this procedure. The Hamiltonian for a quantum pendulum ... -5 My best guess would be$\dfrac{h\nu}{2}$. Note that it requires quantum gravity to check my answer . 1 Generally solutions of quantum mechanics equations are necessary when the phase space of the problem is of dimensions of the order of magnitude of the Planck constant. The mathematical transition from the microworld of particles, molecules and lattices described by solutions of quantum mechanical equations to the classical dimensions is the density matrix ... 0$e^{at}$only represents an oscillation when$a$is imaginary. If$a$is real, it is just an exponential curve. Consider what happens when$b = \beta = 0$. Obviously there is no damping at all. The solution is then $$x(t)=C_1e^{iω_0t} + C_2e^{−iω_0t}$$ which is undamped simple harmonic motion. For an underdamped system, the damping is the$e^{-\beta t}$term,... 0 If a second order linear homogeneous ODE has a Characteristic Equation with complex roots, then the general form of the solution is: $$y(t)=c_1e^{\lambda t}\cos \mu t+c_2 e^{\lambda t}\sin \mu t$$ For roots: $$r_{1,2}=\lambda \pm \mu i$$ Reference and derivation. So that form is different from the one you described. 1 The damping constant$\beta$and the angular frequency (eigenmode) of the oscillator$\omega_0$are both positive (not negative). However, the damping constant is smaller than the eigenfrequency,$0\le \beta \le \omega_0\$. Here an example: \begin{align} \beta &= 189 Hz \\ \omega_0 &=200.1 Hz \end{align} Although both are positive the damping constant ...

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