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4

First of all I guess that what you wrote is the Hamiltonian and not the Lagrangian of the system and $\dot{x}$ stays for $p_x$ and $\dot{y}$ stays for $p_y$. You can decouple the problem redefining $$(X,Y)^t = R(x,y)^t$$ for a suitable $R\in O(2)$ diagonalizing the symmetric matrix in the potential part of your Hamiltonian. This way you see the final ...


3

The potential energy of the pendulum is $U(θ)=mgl(1-\cos θ)$. For small angles, $U(θ)≈mglθ^2\!/2$ and you get a harmonic oscillator. So $θ$ (or equivalently $lθ$) may be taken as the oscillating variable. However, since we are considering small angles, we may as well use $x=l\sin θ≈lθ$. Addendum: You may wonder which approximation is better. Let's look at ...


2

To find the instantaneous energy eigenstates, you need to treat $t$ as a parameter and solve the problem for a time independent Hamiltonian depending on the extra parameter $t$. The best way to do that is to complete the square and write the Hamiltonian as: $$H = \hbar \omega (A^{\dagger} A + \frac{1}{2}) - \frac{ f(t)^2}{\hbar \omega}$$ where $$A = a - ...


2

The annihilation operator is a linear operator. A linear operator can be applied to ANY state. And yes, it returns zero when applied to the ground state. You can really take this as a definition. The definition of the momentum operator $\hat{p}$ is the operator such that $\langle x|\hat{p}|\psi\rangle=-i\hbar \psi'(x)$. One could write this as $\langle ...


2

Echoing knzhou's comment, the expression $$ \langle 0|\mathcal{T}x(t_i) x(t_f) | 0 \rangle = \frac{1}{2\omega} e^{-i\omega|t_i-t_f|} \tag{1} $$ only depends on $t_i,t_f$, while $$ \langle x_f,t_f|x_i,t_i\rangle \tag{2} $$ is a function of $t_i,t_f$, but also of $x_i,x_f$, that is, it depends on four variables, and therefore is more general than $(1)$. This ...


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Yes, it is possible to match the QM with a QFT in 1+0 dimensions. However, the Fock vacuum $|0\rangle$ (which is annihilated an annihilation operator $a|0\rangle=0$) is naturally related to the coherent states $$\hat{a} |z\rangle~=~ z|z\rangle \tag{1}$$ rather than position eigenstates $$\hat{q} |q\rangle~=~q|q\rangle.\tag{2}$$ [Of course, it is possible ...


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Do u really think that velocity is constant?i think in this equation nothing is constant.if tension increases per unit mass decreases and it may make change in velocity or not if the ratio remains the same.further tension depends on some variables such as intermolecular force,elasticity etc


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The tension of the string is a constant, if there is no vibration on the string. A wave is produced on the string when you give an unbalanced force on the string which varies the original tension of the string. The velocity of the wave now depends on the value of the tension. The given equation is valid only for small amplitude vibrations. The tension is ...


2

When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it. If we take a segment small enough that we can consider it as a straight line, then the ...


2

The answer is given by Prahar in his comments : \begin{equation} {\bf T} \cdot {\bf T}^\dagger= {\rm T}_{1}{\rm T}^\dagger_{1}+{\rm T}_{2}{\rm T}^\dagger_{2}+{\rm T}_{3}{\rm T}^\dagger_{3} \tag{01} \end{equation} For $k=1,2,3$ \begin{equation} {\rm T}_{k}{\rm T}^\dagger_{k}=\dfrac{1}{2\hbar}\left(\sqrt{m\omega}\ {\rm R}_{k}-\dfrac{i}{\sqrt{m\omega}} {\rm ...


2

When you ignore higher orders in the force, assuming it is linear in the position, you are assuming the potential energy is quadratic at most (quadratic, parabolic or harmonic approximation). Then you loose global information about the potential. We are not able to say for example whether the movement of the particle is bounded or unbounded as we increase ...


2

Your wording is a little off. SMH doesn't care about the size of displacement. SHM is the lowest order approximation to a general oscillation. A general oscillation does care about the size of the displacement, and when you ignore the higher order effects you will miss various phenomena. Examples of things you will not be able to describe are harmonic ...


2

I first thought, that you have a $\frac 0 0$ or $\frac\infty\infty$ in both cases, but it's wrong. In (a) you get $R = l/\omega$ and in (b) too you can just naively insert $\infty$ for $\omega$ and get $R = \frac{(l\omega)^2}{\omega^4}$, and thus an $\frac 1{\infty^2}$ as the result. But actually I think the physical meaning is more interesting and I'm not ...


1

Harmonic motion is sinusoidal: $x(t)=x_0\sin(ωt)=x_0\sin(2πt/τ)$. The argument of the sine function is called phase, here an increasing funtion of time: $x(t)=x_0\sin(φ(t))$ with $φ(t)=ωt=2πt/τ$. Hence, a phase difference $δφ$ corresponds to a time difference $δt=δφ/ω=δφ\,τ/2π$. With $δφ=π/2$ you find $δt=τ/4$. Remember: $1\ \text{period} ↔ 2π ↔ τ$


1

Let me continue on from BeastRaban's exposition: In 1866, Mehler figured out how to carry out the sum for the $u_n(x)$ eigenfunctions of the quantum harmonic oscillator, Hermite polynomials, adding them all to an elegant and compact eponymous Mehler kernel, or equivalently. The Green's function (propagator) for the quantum harmonic oscillator is then: $$ ...


1

For example, if the first particle is moving on a spring and its position sets the electric potential that controls the second, electrically charged, particle. This way you'd have the potential energy of the coupling in the form of the product of coordinates.


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A wave propagates through the medium due to the interacting particles in the medium. The medium is nothing but a certain region of space containing some interacting particles. If there are only identical particles, the the medium is homogeneous. Otherwise it is heterogeneous. A wave in such a medium is nothing but a disturbance in a particle's energy in that ...


1

The equation of motion for a simple pendulum is actually not the same as for a simple harmonic oscillator. In fact, the pendulum motion can is described by the differential equation $$\frac{d^2\theta}{dt^2}+\frac{g}{l} \sin{\theta}=0$$ It is only in the small angle approximation (where $\sin{\theta} \approx \theta$) that this equates to a simple harmonic ...



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