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Your equation in the Liouville form is elementary for numerical integration, it is structurally just a linear advection equation with spatially varying coefficients. The transformed equation with the kernel F is not useful at all for numerical solution, don't bother with it. All we have here is a 2D advection equation (I use y instead of p): $\partial_{t} ... 3 If there is no external force with explicit time dependence, then the harmonic oscillator contains no explicit time dependence. Then the system has time translation symmetry, i.e. the result can only depend on the difference$T =t_b-t_a$, not on$t_a$and$t_b$individually. 2 If an object is accelerating upwards at a rate of$a$m/s$^2$, then the gravitational force felt by this object is effectively, $$g_{eff}=g+a$$ where$g\sim9.8$m/s$^2$is the canonical gravitational acceleration we all know and love. In your particular case, the common equation of motion for a pendulum, $$\frac{d^2\theta}{dt^2}=mg\sin\theta$$ replaced ... 2 The multi-dimensional analog of simple harmonic motion is an object subject only to a harmonic potential,$U = \frac 1 2 k ||\vec r - \vec r_0||^2$, where$k$is a positive constant (oftentimes called the spring constant),$\vec r$is the object's position, and$\vec r_0$is the position of the center of the potential. By choosing the origin to be the center ... 1 In the first part of the linked document are not giving a formal derivation of the uncertainty principle. It is giving a particular example to show the general idea. The gaussian wavefunction is chosen because it is a particularly simple and happens to exactly satisfy the lower bound$\sigma_x\sigma_p = \frac{1}{2}\hbar$. For the harmonic oscillator being ... 1 Your reasoning is correct up to the point where you compare the accelerations of the mass and the spring. But the effect on the period is actually inverted. Intuitively, when the spring acceleration is higher (lower), this actually means that the spring is more (less) deformed and pulls back with a force correspondingly stronger (weaker) compared to case 1, ... 1 Well it depends on the context of your question. If you're being introduced to General Relativity, then you're just going to assume, in the spirit of the equivalence principle, that gravity and the acceleration cannot be told apart from the pendulum's standpoint, so the acceleration is obviously$a+g$. If you need to do it from first principles in a ... 1 It is true in both cases and an even more general statement can be made. For a potential of the form$V(r)=\alpha r^n$, the expectation value of the kinetic an potential energy is related by $$2 \langle T \rangle = n \langle V \rangle.$$ For the case of the harmonic oscillator,$n=2$. See virial theorem for details. 1 The fact that you can eliminate the$\sin$term tells you that its coefficient$[-\omega^2 B_{2}-2\beta\omega B_{1}+\omega_{0}^{2}B_{2}]$must be zero. This gives you another relation between$B_{1}$and$B_{2}$, and with two equations and two unknowns, you can solve cleanly for$B_{1}$and$B_{2}$. 1 You can think of the continuos formalism as being the limiting case of the discrete-momentum one: if the momentum is taken as a discrete variable (which amounts to constraining the particles to be in some finite volume$V) the fourier expansion of the (real, scalar) field is: $$\tag{1} \varphi(x) = \sum_{\textbf{k}} \frac{1}{\sqrt{2V \omega_{\textbf{k}}}} ... 1 Just look at the equation of motion. Suppose you hang a spring from the ceiling, and that it hangs a distance y_0 from the ceiling in equilibrium (we orient our axis so that positive y points downward). Then, the equation of motion is$$ m\ddot{y}=-k(y-y_0)+mg, $$and so$$ \tfrac{\mathrm{d}^2}{\mathrm{d}t^2}(y-y_0)+\tfrac{k}{m}(y-y_0)=g. $$The square ... 1$$ \sum _nf(n)\int \mathrm{d}p\, |\left< p|n\right> |^2=\sum _nf(n)\int \mathrm{d}p\int \mathrm{d}q\, \left< n|p\right> \left< p|n\right> =2\pi \sum _nf(n)\left< n|n\right> =2\pi \cdot \sum _nf(n), $$where I used the fact that$$ \int \mathrm{d}p\, \left| p\right> \left< p\right| =2\pi, $$which follows from$$ \left< ... 1 We know that: \begin{align} x(t) &= A \cos(\omega t - \phi), && x(0)= A\cos(-\phi)=0.3,\\ \dot x(t)&=-A\omega\sin(\omega t - \phi), && \dot x(0)=-A\omega\sin(-\phi)= 0.07,\\ \ddot x(t)&=-A\omega^2\cos(\omega t - \phi), && \ddot x(0)=-A\omega^2\cos(-\phi)=-0.33. \end{align} We know that\phi=12.5º$or$\phi=-12.5º$, lets ... 1 By definition of the momentum basis,$p$is diagonal in that basis as per $$\langle k \rvert p \lvert k' \rangle = \delta(k - k')$$ since the momentum basis states are defined by$p \lvert k \rangle = k \lvert k \rangle$. 1 I don't see why you wouldn't get any imaginary terms in the matrix elements. Your suggestion of aproach is right and in this case the eigenstates$|n\rangle$of the harmonic oscilator are represented by Hermite Functions in position representation, which happens to be Fourier transform's eigenfunctions as well. As the unitary map that changes form the ... 1 The first solution is not correct since it implies a strange connection$\dot x (0)=-\beta x(0)$. Check the general solution for its region of validity. 1$a^\dagger a\! \mid \! n \rangle = n\mid \! n \rangle$. For vacuum you can write$a^\dagger a\! \mid \! 0 \rangle = 0\mid \! 0 \rangle$, if you like. No need to write 0 solely. 1 Hints: OP's eq. (1) is the equation for a constant of motion$\frac{df}{dt}=\{f,H\}_{PB}+\frac{\partial f}{\partial t}=0$of a harmonic oscillator$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2$. Let us assume for simplicity that$m\omega=1$, and leave it to the reader to generalize to arbitrary$m$and$\omega$. Complexify$z=x+ip\in\mathbb{C}\$. Then the ...