96

Since no phenomenon is completely periodic (nothing keeps repeating from minus infinity to infinity), you could say that sine waves never occur in nature. Still, they are a good approximation in many cases and that is usually enough to consider something physical. Or are they a mathematical construct that helps us interpret nature? I would even go ...


55

The harmonic oscillator is important because it's an approximate solution to nearly every system with a minimum of potential energy. The reasoning comes from Taylor expansion. Consider a system with potential energy given by $U(x)$. You can approximate $U$ at $x=x_0$ by $$ U(x) = U(x_0) + (x-x_0) \left.\frac{dU}{dx}\right|_{x_0} + \frac{(x-x_0)^2}{2!} \...


46

This is really more of a supplement to jinawee's answer, but you might want to consider what, if anything, makes your question different from the following analogous questions: Are there lines in nature, or are they a mathematical construct that helps us understand more complex phenomena? Are there points in nature, or are they a mathematical construct that ...


45

The pendulum is being driven by the magnet: the fixed magnet in the clock is actually the pole of an electromagnet which the clock is using to drive the pendulum: the clock is putting energy into the pendulum via the electromagnet. Almost certainly the clock 'listens' for the pendulum by watching the induced current in the electromagnet, and then gives it a ...


33

To begin, note that there is more than one incarnation of "the" harmonic oscillator in physics, so before investigating its significance, it's probably beneficial to clarify what it is. What is the harmonic oscillator? There are at least two fundamental incarnations of "the" harmonic oscillator in physics: the classical harmonic oscillator and the quantum ...


32

The reason for having two prongs is that they oscillate in antiphase. That is, instead of both moving to the left, then both moving to the right, and so on, they oscillate "in and out" - they move towards each other then move away from each other, then towards, etc. That means that the bit you hold doesn't vibrate at all, even though the prongs do. You ...


32

A simple pendulum does not strictly show simple harmonic motion unless you allow some approximations and uncertainties. It approximately behaves as a harmonic oscillator for small amplitudes. An object is said to be executing simple harmonic motion (no damping; not a forced oscillation) if and only if it satisfies the following condition: $$\frac{d^2 \phi}{...


29

This follows from the uniqueness theorem for solutions of ordinary differential equations, which states that for a homogeneous linear ordinary differential equation of order $n$, there are at most $n$ linearly independent solutions. The upshot of that is that if you have a second-order ODE (like, say, the one for the harmonic oscillator) and you can ...


28

It is actually true, in an almost trivial way. The Ehrenfest theorem states that, \begin{equation} \frac{d}{dt}\langle x\rangle=\langle p\rangle,\quad \frac{d}{dt}\langle p\rangle =- \langle V'(x)\rangle \end{equation} However for all eigenfunctions for the Harmonic oscillator $\langle x\rangle=0$ (and therefore $\langle V'(x)\rangle=0$) and $\langle p\...


27

The actual restoring force in a simple pendulum is not proportional to the angle, but to the sine of the angle (i.e. angular acceleration is equal to $-\frac{g\sin(\theta)}{l}$, not $-\frac{g~\theta}{l}$ ). The actual solution to the differential equation for the pendulum is $$\theta (t)= 2\ \mathrm{am}\left(\frac{\sqrt{2 g+l c_1} \left(t+c_2\right)}{2 \...


26

This question was first posed to me by a friend of mine; for the subtleties involved, I love this question. :-) The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, $\delta$-functions are not valid $\mathcal{L}^2(\mathbb{R})$ functions, so we need to define a kosher function which gives the $\delta$-function as ...


25

The harmonic oscillator is common It appears in many everyday examples: Pendulums, springs, electronics (such as the RLC circuit), standing waves on a string, etc. It's trivial to set up demonstrations of these phenomena, and we see them constantly. The harmonic oscillator is intuitive We can picture the forces on systems such as pendulum or a plucked ...


24

The Hilbert space ${\cal H}$ of the one-dimensional harmonic oscillator in the position representation is the set $L^2(\mathbb{R})={\cal L}^2(\mathbb{R})/{\cal N}$ (of equivalence classes) of square integrable functions $\psi:\mathbb{R}\to\mathbb{C}$ on the real line. The equivalence relation is modulo measurable functions that vanish a.e. The Dirac delta ...


22

To answer the question directly - No. Your clock is not violating the Laws of Thermodynamics. If (as you suspect) the pendulum is purely decorative, then, as tfb states, the fixed magnet is an electromagnet that is being driven by an oscillating current. This provides an oscillating field to kick the pendulum in phase with its swing; same way a dad keeps ...


21

Yes, you are on the right track. The series you have there is called Dyson's series. First note that the $n$'th term looks like $$ U_n = (-\frac{i}{\hbar})^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n) $$ The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing ...


21

The problem here isn't so much calculus as the assumptions that are made about the system. The solution can only be as accurate as the assumptions made, no matter how accurate the solutions of the equations are. For a pendulum where the distance between the centre of gravity and the anchor point is $l$ and the mass is $m$, the equation of motion is: $$\...


20

The higher you are, the greater the maximum velocity and maximum potential energy. Consider one pendulum lifted higher than a second both released at the same time. When the higher pendulum reaches the starting point of the second, it already has a velocity greater than 0. This higher velocity allows the higher pendulum to complete its swing in the same ...


20

The issue with measuring at the ends is that the pendulum “dwells” at the end point as it turns around, so that there is a greater spread of time for which it looks “at the extreme point” than for which it looks “at the midpoint”. This spread of time introduces error, whether you are triggering a stopwatch by hand or having some kind of sensor make the ...


18

Ladder operators are usually constructed to form a Lie algebra (we want them to have specific conmutation relations). The mathematical basis is weight theory. The important thing of Lie algebras is that they are a vector space and their elements, which are called generators obbey this conmutation rule: $$[X_i,X_j]=f_{ijk}X_k$$ Where we have used the ...


17

I) It depends on how abstract OP wants it to be. Say that we discard any reference to 1D geometry, and position and momentum operators $\hat{q}$ and $\hat{p}$. Say that we only know that $$\tag{1}\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1}, \qquad\qquad \nu\in\mathbb{R},$$ $$\tag{2} \hat{N}~:=~\hat{a}^{\dagger}\hat{a}, $$ $$\tag{3} [\hat{a},\hat{a}^...


16

One needs to be careful about what one mean by the "size" of a vector space. A theorem of functional analysis tells us that any two Hilbert bases for a Hilbert space must have the same cardinality. This allows us to define the Hilbert dimension of a Hilbert space as the cardinality of any Hilbert basis. The Hilbert space for the one-dimensional harmonic ...


16

Usually, the (sinusoidal) driven harmonic oscillator is damped, and the first two parts of your solution (which depend on the initial conditions, while the third term does not) are transient, i.e. not relevant after a short time. That the solution $$ x(t) = \frac{F_0\sin(\omega t)}{m(\omega_0^2 - \omega^2)}$$ cannot be the "full" solution to the equation of ...


16

The coherent state $\vert \alpha\rangle$ is just a vacuum state $\vert 0\rangle$ translated in $x$ and $p$ space so $\alpha=x_0+ip_0$. Thus the vacuum state is a coherent state that has not been displaced, i.e. $x_0=p_0=0$. In fact, a nice way to see this is in the Wigner function formalism. The vacuum state is just a Gaussian sitting at the centre of $(x,...


15

The reason that creation and destruction operators don't commute is that, on top of 'moving a state up and down energy levels', they multiply it by a number in the process, and this number depends on where you are in the ladder. More specifically, $$\begin{cases} \hat{a}|n\rangle&=\sqrt{n}|n-1\rangle,\text{ while}\\ \hat{a}^\dagger|n\rangle&=\sqrt{n+...


15

$y(t)=A \sin(\omega t+d)=A \cos(\omega t+d-\left(\frac\pi2\right))$ The two forms are the same except for the phase term, so which one you use depends on when you started your stopwatch. If you started off at max displacement at t=0, then you use the cosine form, if you start out with max velocity at t=0, then you use the sine form. If you are deducing ...


15

The main idea is that you can take a complicated interacting or coupled system and write its solution as a sum of non interacting or free modes. Even in classical mechanics, if you have a linear chain of $N$ oscillators, one can show that the general solution is a sum of $N$ normal modes each of them being a simple harmonic oscillator. In Quantum Field ...


14

You can get a proportionality using one of the most basic techniques in science: dimensional analysis. What you do is look at the dimensions of the various quantities in the problem and try to see how they can fit together. Here you want the period, which is a time. I'll write $[T]$ to indicate that it has dimensions of time. What could it depend on? If we'...


14

The previous answers are all correct, but I thought I'd give a more conceptual explanation for why the delta-function basis is the "wrong" basis in which to expand when counting degrees of freedom. Since the situation is much, much more complicated in QFT, for simplicity I'll only consider first-quantized wavefunctions for a system with a fixed, finite ...


14

Short version $\newcommand{\ket}[1]{\lvert #1 \rangle}\newcommand{\Ket}[1]{\left| #1 \right>} % $Because you can use beamsplitters to split a coherent sate into a tensorial product of many independent low photon number coherent states. Longer version If you send $\ket{\alpha}$ on a beamsplitter of transmission coefficient $t$ and reflection coefficient ...


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