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The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a $\Delta x$ amount with $F_{spring}(0) = k\Delta x$. So the spring force on point A is a little smaller than in point -A, since $ F_{spring}(A) = -k(A-\Delta x)$ and $ F_{spring}(-A) = k(A+\Delta x)$ so it compensates the "extra" force. You have ...


2

In accordance with Hooke's law the force is linear with distance. Incorporating gravity only means that the equillibirum position of the spring has changed, the "zero" around which it oscillates. The gravitational pull is already compensated by the spring. Thus the magnitude of the force is euqal at $-A$ and $+A$. Edit: When the gravitational pull on the ...


2

The terminology "continuous variable system" is non-standard, but likely refers to the fact that any canonical quantization of a classical Hamiltonian system (i.e. a system described by a continuous phase space) must have an infinite-dimensional Hilbert space since the canonical commutation relation $$ [x,p] = \mathrm{i}\mathbf{1}$$ cannot be realized on ...


1

It's always possible to expand the potential in Taylor series around any local minima (in this example $U(x) $ has local minima at $x_0$ , thus $U'(x_0)=0 $ ) $$ U(x) \approx U(x_0)+\frac{1}{2}U''(x_0)(x-x_0)^2 $$ Setting $ U(x_0)=0 $ and $ x_0=0$ (for simplicity, the result don't depend on this) and equating to familiar simple harmonic oscillator ...


1

No, it has not discrete spectrum (on $L^2(\mathbb{R}^d)$). In fact $a+a^*$ is proportional to the position operator (or the momentum one, depends on your definition of $a$ and $a^*$; by the usual one the position operator $x$ is proportional to the real part $a+a^*$ and the momentum $p$ to the imaginary part $\frac{1}{i}(a-a^*)$). Both position and momentum ...



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