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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 ...
Your overall Hamiltonian for this system is actually $$H = H_{1} \otimes \mathbb{I} + \mathcal{I} \otimes H_2$$ where your overall Hilbert space is $L_{2}(\mathbb{R}_{3}) \otimes \mathbb{C}^2$, and $\mathcal{I}$ is the identity on $L_{2}(\mathbb{R}_{3})$, while $\mathbb{I}$ is the identity on $\mathbb{C}^2$. Let $|\mathbf{n} \rangle \in L_{2}(\mathbb{R}_{... 2 The action that goes in the path integral is a functional of the path$x(t)$. Namely, $$S[x(t)]=\int_0^T dt \frac{m}{2}\left(\dot{x}^2-\omega^2 x^2\right)$$ which can be discretized very simply $$S=\frac{m}{2}\sum_{i=1}^N\frac{(x_{i+1}-x_i)^2}{\epsilon}-\epsilon\omega^2 x_i^2$$ The$x$in this action is not necessarily a classical trajectory, but this is ... 2 You forget the direction of the damping force. In the linear equation the$-v$term always has the opposite sign to$v$, which is correct. But in your quadratic damping equation$-v^2$is always negative. If$v$is negative, that force is increasing the speed of the particle, not decreasing it, and the particle shot off to infinity as in your graph. ... 2$T=1.94$sec is certainly compatible with the big, but noisy peak at .5Hz. Why are you worrying about the tiny peak at about 2.6Hz? Your data looks very noisy so random but meaningless peaks are to be expected. I still don't understand why your FFT is so noisy given the smooth data of black points in the first plot. You say that you interpolate? Why do ... 1 The expression$a = -\omega^2 x$is not true in general, but in every SHM problem there exists one inertial coordinate system which makes it true. In your example, consider$ a = a_0 \sin(\omega t) $and$v(0)=0$as well as$x(0)=x_0$The general solution of the above is $$x(t) = x_0 + v_0 t - \frac{a_0}{\omega^2} \sin(\omega t)$$ which does not obey $$... 1 The number operator does not commute with the position operator. We have$$\hat{H}=\frac{\hat{P}^2}{2m}+\frac{1}{2}m\omega^2\hat{X}^2=\left(a^\dagger a+\frac{1}{2}\right)\hbar\omega=\left(\hat{N}+\frac{1}{2}\right)\hbar\omega$$Then,$$\hat{N}=\frac{\hat{H}}{\hbar\omega}-\frac{1}{2}$$where \hat{H} is the Hamiltonian for the harmonic oscillator. Using [\... 1 wa\sqrt{1-y^2} is the velocity of SHM while ƒ*Λ - velocity of on any wave 1 This isn't the definition of SHM I would go with, but it is equivalent so we'll run with it for now. The point is that the particle moving around the circle at uniform speed is purely fictitious, the real particle is its 'shadow' moving back and forth along the x axis. If you visualise this for a bit, you should see that when the particle is at the points ... 1 As far as the single Gaussian x_1 integration goes, OP is doing the correct thing (up to possible typos) in eq. (7). However since the time-increment$$\epsilon~\ll~ \omega^{-1}\tag{i}$$is supposed to be small (in order for Feynman's fudge factor 1/A to be valid), we have under the square root$$ \frac{\sin (\omega \epsilon)}{2\epsilon^2\omega\cos (\... 1 You have$[H_1, H_2] = 0$, so yes, it is possible to diagonalize$H = H_1 + H_2$using simultaneous eigenkets of$H_1$and$H_2$. And to the second question yes again, the two states are indeed orthogonal. You can use the reasoning that if one individual component is orthogonal then the whole state must be orthogonal or, since the two tensor products kets ... 1 My 2cents on it is studying harmonic systems was natural because any local or global potential minima can be approximated as a quadratic for modeling purposes. At least for basic quantum mechanics that would be my justification. QFT or particle physics someone else would have to answer the precise why it is powerful. I am aware it is the case, but it is well ... 1 I think instead of$x_i \le E_i$, we have$x_1 \le E_1$and$\omega x_2 \le E_2\$, from which the conclusion readily follows.