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Any "harmonic oscillator", seen as the second quantization operator $$d\Gamma(1)=\int a^*(k)a(k)dk$$ of the symmetric Fock space $\Gamma_s(\mathscr{H})$ over a (separable) Hilbert space $\mathscr{H}$, has the natural numbers $\mathbb{N}$ as spectrum (i.e. evenly spaced spectrum). In addition, if e.g. $\mathscr{H}=\mathbb{C}$, the operator $aa^*$ has ...

3

You are right that $$\frac{T_M}{T_E} = \sqrt\frac{g_E}{g_M}>1,$$ which means that a pendulum period on the Moon is longer than the period of the same pendulum on the Earth. The rest, in my opinion, is just not precise enough. I think it is very confusing to use the early/late terminology when discussing two modes of operation of the same clock. I suggest ...

2

Let's start from $$H = \hbar \omega \left(f^\dagger f - \frac{1}{2}\right),$$ with $\{f, f^\dagger\}=1$, $\{f, f\} = 0$ and define fermionic position and momentum coordinates by $$\psi_1 = \sqrt{\frac{\hbar}{2}} \left(f + f^\dagger\right) \\ \psi_2 = i\sqrt{\frac{\hbar}{2}} \left(f - f^\dagger\right)$$ with the following anticommutation relations: $$... 2 The units of \hbar are in fact J.s/rad. (thanks AV23) this is because \hbar = \frac{h}{2\pi} the units of h are J.s and the units of \pi are rad. Thus we have J.s/rad. (thanks Noiralef) Thus the ladder operators are in fact unitless. On reflection this is the only logical possibility as they move between different eigenstates - which must all be in ... 1 You are at a point where you'll need v_1 and v_2. Observe from the original transformation that:$$v_2 = v_1 - v\implies V = \frac{(m_1+m_2)v_1 -m_2v}{m_1+m_2}v_1 = V + \frac{m_2v}{m_1+m_2}$$We also get, by a similar procedure,$$v_2 = V - \frac{m_1v}{m_1+m_2}$$We have expressed v_1 and v_2 in terms of the new variables, V and v. ... 1 Well, consider a Lax pair for the Harmonic Oscillator; $$L = \begin{pmatrix} p & \omega q \\ \omega q & -p \\ \end{pmatrix}, \quad M=\frac{\omega}{2} \begin{pmatrix} 0 & -1 \\ 1& 0 \\ \end{pmatrix}$$ Since the Hamiltonian is$$H(q, p)=\frac{1}{2}(p^{2}+\omega^{2}q^{2})$$It is eay to check that the Lax ... 1 At low Reynolds number, as in a creeping flow, one can ignore the advective acceleration terms in the Navier-Stokes equation. If we also assume a steady state, the equation becomes $$0 = -\nabla p + \mu\nabla^2\vec{v},$$ where p is the hydrostatic pressure, \mu is the viscosity of the fluid and \vec{v} is the flow velocity. ... 1 Assuming that X=X^\dagger, P=P^\dagger and [X,P] = i\hbar, let me try$$f = \sqrt{\frac{m\omega}{2\hbar}}\left( \alpha X + \frac{\beta}{m\ \omega } P \right) $$where \alpha and \beta are complex numbers of modulus one. From this follows that$$ \hbar \omega \left( f^\dagger f - \frac{1}{2} \right) = \frac{P^2}{2m}+ \frac{1}{2} m \omega^2 X^2 + ...

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