# Tag Info

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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, $$\frac{d}{dt}\langle x\rangle=\langle p\rangle,\quad \frac{d}{dt}\langle p\rangle =- \langle V'(x)\rangle$$ However for all eigenfunctions for the Harmonic oscillator $\langle x\rangle=0$ (and therefore $\langle V'(x)\rangle=0$) and $\langle p\... 20 The issue with measuring at the ends is that the pendulum “dwells” at the end point as it turns around, so that there is a greater spread of time for which it looks “at the extreme point” than for which it looks “at the midpoint”. This spread of time introduces error, whether you are triggering a stopwatch by hand or having some kind of sensor make the ... 13 It's not so bad if you just set all the constants to one,$m = \omega = \hbar = 1$. You can always recover these later by dimensional analysis. Classically, we have the Hamiltonian $$H = \frac{p^2 + x^2}{2}.$$ Even without knowing quantum mechanics, you can think of phase space as the complex plane, with$x$as the real axis and$p$as the imaginary axis. ... 13 First, let's start by completing the square as suggested. The equation will now be of the form $$\frac{d^2\Psi}{dz^2}+(\nu+\frac{1}{2}-\frac{1}{4}z^2)\Psi=0$$ with boundary conditions$\Psi(z=L)=\Psi(z=\infty)=0$and $$z=\sqrt{\frac{2m\omega}{\hbar}}x+L \\ \nu=\frac{E}{\hbar\omega}+\frac{L^2}{4}-\frac{1}{2} \\ L=-\sqrt{\frac{2mg^2}{\hbar\omega^3}}$$ This ... 13 To measure the time period it is useful to use a fiducial (reference) mark which in this case could be a vertical line drawn on a piece of card and placed “behind” the pendulum bob/string. It is assumed that the time for a number of complete oscillations will be measured to enable one to find a more accurate value of the period than just measuring the ... 7 Observe that, from the spectral decomposition of$e^{-\beta H}we have that $$e^{-\beta H} \psi - \sum_{n=0}^N e^{-\beta (n+1/2)}|n\rangle \langle n|\psi\rangle = \sum_{n=0}^{+\infty} e^{-\beta (n+1/2)}|n\rangle \langle n|\psi\rangle-\sum_{n=0}^N e^{-\beta (n+1/2)}|n\rangle \langle n|\psi\rangle= \sum_{n=N+1}^\infty e^{-\beta (n+1/2)}|n\rangle \langle n|\... 7 These are all good and correct answers, but I will answer from a different perspective. Any linear-differential equation of degree n has n linearly independent solutions, ie. these n solutions span a vector space, with sets of solutions forming a basis. For simple harmonic motion, the differential equation is:$$m(\dfrac{d^2x}{dt^2})+kx = 0As ... 7 "Angular velocity" can be used inter-changably with "angular frequency", but you want to distinguish clearly between those and "cyclic frequency" which is the thing usual just termed "frequency". The angular quantities are measured in radians per second, while the cyclic frequency is "cycles per second" AKA hertz (Hz). The "angle" here is not obvious the ... 7 You have almost the entire answer in your question. But first: I hope you can get past your intuition that triangle waves are more natural. That is an example of one of the biggest stumbling blocks to progress in physics education: the preconceived notion. There are three important things that contribute. \, F=ma\quad\ \ — Newton's Second Law. ... 7 The reason is that it is (supposedly) marginally easier to judge when the pendulum sweeps past a point when it is moving quickly than when it is moving slowly. In think the point is debatable if you are judging it by eye and manually triggering a stop watch, as your own reaction time comes into play. What's much more important is to measure the elapsed time ... 5 The difference is one is wrong (probably a typo). The two formulas are: \begin{align} E&=\hbar\omega\left(n+\frac{1}{2}\right)\\ &=hf\left(n+\frac{1}{2}\right). \end{align} The relationship between them is that \hbar=\frac{h}{2\pi} and \omega=2\pi f. We use two different forms because angular frequency, \omega, appears naturally in a lot of ... 5 The most classical-like states of harmonic oscillator are Coherent states of harmonic oscillator, not \delta(x). The reason is that coherent state has balanced uncertainty of coordinate and momentum:\Delta x =\sqrt{\frac{\hbar}{2m\omega}}; \Delta p = \sqrt{\frac{m\hbar\omega}{2}},$$and in the classical limit (\hbar\rightarrow 0) \Delta x \approx 0,... 5 Your attempts to take a classical limit don't really make sense. By the nature of classical mechanics, outside of statistical mechanics, there is no such thing as a "position distribution" of a classical system. A classical system always has definite position and momentum, so its position and momentum "distributions" would be \delta(x) and \delta(p), ... 5 I have also studied this relation. May be things get clearer in my explanation: For a general pendulum, it is obvious that if the l is substituted to a large value the time period corresponding also increases. For a pendulum of very large length(comparable to radius of earth), we have to make a change in derivation. An assumption which is valid only for ... 5 No, your calculation is not correct - you're using a non-canonical transformation in ways which assume that it is canonical. More specifically, it is correct to say that if your hamiltonian maps into the form$$ H = \frac{1}{2\mu} P^2 + \frac12 \mu\tilde\omega^2 X^2, $$where X and P are operators such that [X,P]=i\hbar, then the eigenvalues of H ... 5 Let us solve the differential equation to see what we get. (spoiler: it is not what you originally posted). You have$$\ddot{x}+2\gamma\dot x+\omega^2x=0$$The characteristic equation of the DE is,$$\lambda^2+2\gamma\lambda+\omega^2=0$$And solving for \lambda you get:$$\lambda_{1,2}=\frac{-2\gamma\pm \sqrt{4\gamma^2-4\omega^2}}{2}=-\gamma\pm\sqrt{\... 5 Both of your statements are true for SHM and can be seen as true for approximations as well. Each statement is essentially saying the same thing, becauseF=-\text dU/\text dx$, and we can choose$U(0)=0$. Whether or not you use approximations to make these statements applicable to your system doesn't change that. In other words, if you have approximated$U\...

5

Absolutely. The Hermite polynomials $$H_n(x) = (-1)^n e^{x^2} \partial_x^n e^{-x^2} = \left(2x - \partial_x\right)^n \cdot 1$$ are orthogonalized by $$\int_{-\infty}^\infty H_m(x) H_n(x)\, e^{-x^2} \,dx = \sqrt{\pi}\, 2^n n! ~ \delta_{nm} ~,$$ whereas the (nonpolynomial) Hermite functions $$\psi_n(x) = \left (2^n n! \sqrt{\pi} \right )^{-\frac12} e^{... 4 The state \psi(x) = \delta(x) is a perfectly valid state for the harmonic oscillator to occupy. (With caveats, though: it is not normalizable, so it's not a physically-accessible state. Still, it's a perfectly reasonable thing for the mathematical formalism to handle.) As you note, it has a position uncertainty equal to zero, as well as a vanishing ... 4 One way of deriving it is using Taylor series (although to be fully rigorous, this requires further justification for restricting to analytic functions). We have that f(x) = \sum a_n x^t, so f''(t)=\sum (n+1)(n+2)a_{n+2}t^n. If f''(t)=-\frac k m f(t), then (n+1)(n+2)a_{n+2} = -\frac k m a_n, so a_{n+2} = \frac {-k a_n}{(n+1)(n+2)m}. So a_0 ... 4 \def\PD#1#2{{\partial#1 \over \partial#2}} I would suggest a different approach. If l is very large the circumference of radius l becomes a straight line, tangent to a circumference centred in Earth's centre C and radius R. If x is bob's displacement along this line, bob's distance from C is D=\sqrt{R^2+x^2} and gravitational potential energy is ... 4 Well, sinusoids per se are not that common in nature at all. Even a tiny bit of nonlinearity essentialy corrupts the pure sine behavior of the idealized oscillator (see the van der Pol and the Duffing oscillators for some popular weakly nonlinear extensions). Based on what you have already stated, maybe a bit better assertion would be that the sinusoids are ... 4 Hint:$$ x^2+y^2+xy = \frac34 (x+y)^2 + \frac14 (x-y)^2. This means that if you transform to the new variables \begin{align} \xi & = \frac1{\sqrt{2}}(x+y) \\ \eta & = \frac1{\sqrt{2}}(x-y), \end{align} and with a similar transformation from p_x,p_y to p_\xi,p_\eta to make sure that [\xi,p_\xi] = i = [\eta,p_\eta] and [\xi,p_\eta] = 0 = [\... 4 No. There are two points of view on this. First, it's a definition of what a simple harmonic oscillator/ideal spring is. Second, it's an empirical formula that is approximately correct in many circumstances. It is literally the same thing as searching for the minimum of a function by setting the derivative of the function equal to zero, and then ... 4 In the second line of Eq. (6), you have to be a bit careful about when the Green's function is zero. The expression should be \frac{A}{m^2}\int^{\textrm{min}(t_1,t_2)} dt'\; G(t_1, t') G(t_2, t')  where $\textrm{min}(t_1,t_2)$ is the smaller of $t_1$ and $t_2$. Eq. (7) doesn't look, right; in this kind of calculations you always want to take the ...

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There is no prior difference about naming the concept. IT depends on how you call things, this is just one more. A quantum free particle = a wavefunction that obeys the Schrödinger's equation with no potential energy. A particle enclosed in an infinite-well = a wavefunction that obeys the Schrödinger's equation with such infinite-well potential. A quantum ...

3

The problem with your method is that it implicitly assumes the errors are (anti)correlated. Well, what does that mean? It means you assume that when the length fluctuates high (low), then the period measurement must fluctuate low (high). Since a larger (smaller) length and a shorter (longer) time both increase (decrease) the value of $g$, the fluctuations ...

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