# Tag Info

23

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 ...

8

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 ...

8

The ground state of the harmonic oscillator $|0\rangle$ obeys $$a|0\rangle = 0$$ which means that the action of $a$ can't be undone: once you act with it on a state, you set to zero the coefficient in front of $|0\rangle$ in the decomposition into occupation eigenstates. Any candidate inverse operator $a^{-1}$ acting on zero will give you zero again; you ...

6

The main problem is to determine what corresponds to zero mass of the harmonic oscillator. Remember that a fraction of the spring mass also participates in the motion. By introducing an intercept $\beta$, your friend takes into account that the true zero of the mass parameter $m$ may be shifted from what you think it is. So an affine model $T^2=Cm+\beta$ is ...

6

This is all about potential; it is common that a particle movement is described by a following ODE: $m\ddot{\vec{x}}=-\nabla V(\vec{x})$, where $V$ is some function; usually one is interested in minima of $V$ (they correspond to some stable equilibrium states). Now, however complex $V$ generally is, its minima locally looks pretty much like some quadratic ...

6

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 $$\tag{1}\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1}, \qquad\qquad \nu\in\mathbb{R},$$ $$\tag{2} \hat{N}~:=~\hat{a}^{\dagger}\hat{a},$$ $$\tag{3} ... 6 The Hilbert space {\cal H} of the one-dimensional harmonic oscillator in the position representation is the set L^{2}(\mathbb{R}) of square integrable functions \psi:\mathbb{R}\to\mathbb{C} on the real line. The Dirac delta distribution \delta(x-x_{0}) is not a function and it is not square integrable. See also this Phys.SE post. 5 Well, the reflection of a wave at the end happens always. One can picture this by imagining the succesive atoms being pushed off the equilibrium position as the wave propagates. Since the endpoint is fixed, it has nowhere to be pushed but the few atoms near it (I am considering idealized linear chain for simplicity) that have already being perturbed will, ... 5 Addressing just the physics part (go to stack overflow for the programming), and using the equation that you've been given:$$ x(t) = A \cos \left( \omega t + \delta \right) $$Let's look at the form of the solution. It is sinusoidal The curve will have a maximum value of A (because cosine has a maximum value of 1) When \delta is 0, \pm 2\pi, \pm ... 5 Yes, of course, they're the squeezed states. If your change involves a simple change of the frequency only, you may simply "rescale" all the energy eigenstates by x\to Cx and p\to p/C. This is achieved by the operation$$|n\rangle \to |n'\rangle = \exp[\frac {i\ln(C)}{2\hbar}(xp+px)] |n\rangle$$I inserted the Hermitian part of the operator xp ... 5 Yes, you are on the right track. The series you have there is called Dyson's series. First note that the n'th term looks like$$ 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) $$The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing ... 5 I think of 'quantize' as a verb that refers to converting the classical to the quantum picture. The Hamiltonian you wrote down is, after all, the classical one, which is not 'quantized'. Once you solve it by (1) writing down annihilation/creation operators, (2) finding that only particular wavefunctions are allowed, (3) numerically evaluated the energy ... 5 This question first posed to me by a friend of mine. For the subtleties involved, I love this question. :-) The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, \delta-functions are not valid \mathcal{L}^2(\mathbb{R}) functions, so we need to define a kosher function which gives the \delta-function as a ... 4 Your solution is correct (multiplication of 1D QHO solutions). Since the potential is radially symmetric - it commutes with with angular momentum operator (L^2 and Lz for instance). Hence you may build a solution of the form |nlm> where n states for the radial state description and lm - the angular. Is it better? Depends on the problem. It's just the other ... 4 Simple harmonic motion (SMH) describes the behavior of systems characterized by a equilibrium point and a restoring "force" (in some generalized sense) proportional to the displacement from the equilibrium. Example system A simple mechanical system with this behavior is a mass on a spring (which we will consider in one dimension for ease). There is some ... 4 A damped harmonic oscillator with a sinusoidal driving force is represented by the equation$$\ddot{x} + \gamma\dot{x} + \omega_0^2x = \frac{F_D \sin(\omega_D t)}{m}$$where \gamma = b/m (b is the damping coefficient, b=F/v) and \omega_0^2 = k/m is the resonant frequency of the oscillator. The particular solution to this equation can be determined ... 4 This is a way of giving systematic meaning to the radiation continuum in the context of a set of discrete states. You assume some set of boundary conditions on the EM fields where they hit the box {1}, derive a set of allowed modes in terms of the geometry of the box {2}, then allow the box to expand without limit. Thus you arrive at a continuum of allowed ... 4 Just to add to gigacyan's answer, the harmonic oscillator Hamiltonian may be written in terms of raising and lowering operators: \begin{eqnarray} \hat{H}\psi&=&-\frac{1}{2m}\frac{\partial^2\psi}{\partial x^2}+\frac{1}{2}m\omega^2x^2\psi\nonumber\\ &=&(a^{\dagger}a+\frac{1}{2})\omega\psi \end{eqnarray} where ... 4 1) There are many inequivalent quantum systems that have the same classical limit \hbar\to 0. 2) For instance, assume for simplicity that the quantum system is described by a single pair of creation and annilation operators,$$[\hat{a},\hat{a}^{\dagger}] ~=~\hbar {\bf 1}, \qquad\qquad ...

4

I did a bit of discussion on this subject in this thread on Music.SE. The fundamental doesn't necessarily have the strongest amplitude. As said by Alfred Centauri, it depends on the initial configuration: ideally, the string returns to exactly that configuration after each $\tfrac1{\nu_1}$, and the amplitude of each harmonic in frequency space is ...

4

My approach would be: first determine the time evolution of $\hat{x}(t)$ and $\hat{p}(t)$. For $\hat{x}$ you have $$\frac{d}{dt}\hat{x}_H(t) = i[H_H,\hat{x}_H(t)] = \frac{i}{2m} [\hat{p}_H(t)^2,\hat{x}_H(t)] = \frac{\hat{p_H(t)}}{m}$$ and for $p$ you have (assuming $0\leq t \leq T$)  \frac{d}{dt}\hat{p}_H(t) = i[H_H(t),\hat{p}_H(t)] = -m\omega_0^2 ...

4

Your problem essentially amounts to multiplying two sums of numbers. I would also say this seems like more of a homework problem than a research level question, but since I'm new here and feel like answering my first question, I will help you out. Let $A = (a_1 + a_2 + \dots)$ and $B = (b_1 + b_2 + \dots)$. So the product is $AB = a_1 (b_1 + b_2 + \dots) ... 4 Quantization is a process of constructing a quantum (field) theory out of a classical (field) theory. In your case you have a classical Hamiltonian (which governs the dynamics of your classical system, in your case the harmonic oscillator), the underlying classical theory is Hamiltonian mechanics. If you say to quantize$H$one means to find the ... 3 I just purchased Feynman's Thesis, which provides some insight on how Feynman saw the world, and provides some context here. One of the key issues Feynman was trying to reconcile in his Lagrangian approach was how to describe quantum mechanics without rely on a field defined by harmonic oscillators, from page 5: "In particular, the problem of the ... 3 The phase constant is needed only if you have a specific initial condition, e.g. if I told you where$x$was at time$t = 0$, you could solve for$\varphi$. Otherwise you can just choose whatever you want for it: Note that it is the same in all functions. Choosing some value for$\varphi$is analogous to you manually setting the time origin to something ... 3 I guess this thread shows everyone has their own tastes when it comes to this topic! First of all, it's called harmonic motion because sine and cosine are the elementary harmonic functions. Recall that in general a harmonic function is a solution of Laplace's equation (which shows up everywhere in physics), and in we initially study ... 3 with the given$m$and$k$you indeed cannot calculate the damping coefficient$c$. Remember that you just have a model where you put some constants in and you can only derive other constants which somehow depend on them. The question concerning an actual measurement was answered in Investigating the dampening of a spring. Greets 3 and welcome to the site. I suppose that you have to do this as some kind of homework so I would not try to invent new things and just look what mathematical description(s) of such a system exists. The governing equations will have some constants in it and it would be already a nice thing to look if the given framework can explain the dynamics and if you can ... 3 You should already be familiar with damping. It simply refers to the fact that if you set a spring going, it eventually stops. The wikipedia article should cover most of what you want to know. Any particular spring may be damped for all sorts of reasons. Any way the spring can lose energy contributes to damping, so it could be lost internally to heat due ... 3$\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}$is the second time derivative of angular displacement.$\frac{\mathrm{d}\theta}{\mathrm{d}t}$would be first time derivative. In order to understand this displacement, let's compare it with linear displacement$x\frac{\mathrm{d}x}{\mathrm{d}t}$is speed, while$\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\$ is ...

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