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Consider, two fields interact with each other and the interaction term of the action is given. Now the Lagrangian density is fourier transformed and the interaction term of the action is expressed as an integral over the momentum space. How is the integrand related to the form factor?

Consider a Lagrangian: \begin{equation} \mathcal{L} = \mathcal{L}(q_1\, \dots\, q_n, \dot{q}_1\, \dots\, \dot{q}_n,t) \end{equation} From this Lagrangian, we get a set of ##n## equations: \begin{equation} \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}_i} - \frac{\partial...

This will transform the PDE into a wave equation. But this exercise asks to solve this problem not using this coordinate transformation. Thanks for your suggestion anyway.

Homework Statement Find out the Green's function, ##G(\vec{r}, \vec{r}')##, for the following partial differential equation: $$\left(-2\frac{\partial ^2}{\partial t \partial x} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right) F(\vec{r}) = g(\vec{r})$$ Here ##\vec{r}...

I was looking for a rigorous derivation.

Homework Statement Suppose two successive coordinate rotations through angles ##\Phi_1## and ##\Phi_2## are carried out, equivalent to a single rotation through an angle ##\Phi##. Show that ##\Phi_1##, ##\Phi_2## and ##\Phi## can be considered as the sides of a spherical triangle with the angle...

If ##\frac{d\theta}{dt} = 0## and ##\omega## is nonzero, will the piston move? What do you think? Drawing diagrams may help.

Photon has spin 1 and Higgs boson has spin 0. (Source: Wikipedia) You may find this thread on spin 0 particle helpful.

Why didn't you consider ##\omega## in your solution?

You forgot to upload the figure.

Let $$\Psi(x,t) = A(t) \psi(x)$$ Applying Schrodinger's Time dependent equation: $$\begin{equation} i\hbar\frac{\partial}{\partial t}\left(A(t)\psi(x)\right) = H\left(A(t)\psi(x)\right) \end{equation}$$ Let ##\psi(x)## is an eigenfunction of ##H## with eigenvalue ##E##. So, we get...

Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?

How do we experimentally apply the operator ## \exp{\left(-i\phi\frac{ S_z}{\hbar}\right)}## on a quantum mechanical system? (Here ##S_z## is the spin angular momentum operator along the z-axis) For example, on a beam of electrons?

Yes. I just wanted to show that the energy eigenkets are also eigenkets to the operator ##i\hbar \frac{\partial}{\partial t}##.

Can't ##H:=-\frac{\hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V(x) ## act on ##\Psi (x,t)## as well?

What is the four-vector related to electric and magnetic dipole moment?

Suppose, the energy of a particle is measured, say ##E_1##. So now the state vector of the particle is the energy eigenket ##|E_1>##. Then the position of the particle is measured, say ##x##. As the Hamiltonian operator and the position operator are non-commutative, the state vector is changed...

I thought writing more equations means spending more time on physics.

How many equations does a physicist write in his/her lifetime on average? Is there any approximate statistics on this? Also how much is this correlated to his/her contribution to physics?

Consider the particle in a box problem. The number of energy eigenbasis is 'countable' infinity. But the number of position eigenbasis is 'uncountable' infinity. x can take any value from the interval [0,L] Whichever basis I choose, shouldn't the dimensionality of the vector space be the same?

Look, functions like ##f(z)=z^2##, gives you the same value for a particular ##z##, no matter you write ##z=||z||e^{i\theta}## or ##z=||z||e^{i(\theta+2\pi)}##. The problem arises when you deal with functions like ##g(z) = log (z)## or ##g(z) = z^{1/2}##. In those cases, you notice...

Suppose a second rank tensor ##T_{ij}## is given. Can we always express it as the tensor product of two vectors, i.e., ##T_{ij}=A_{i}B_{j}## ? If so, then I have a few more questions: 1. Are those two vectors ##A_i## and ##B_j## unique? 2. How to find out ##A_i## and ##B_j## 3. As ##A_i## and...

Thanks for your help. But still could not get it.

Why ##\mu_1, \mu_2## must be the same as ##\mu_1^*, \mu_2^*## ? What I thought is : If ##\mu_1\mu_2 = \mu_1^*\mu_2^*## and ##\mu_1+\mu_2 = \mu_1^*+\mu_2^*##, then ##\mu_1, \mu_2## are the same as ##\mu_1^*, \mu_2^*## It can be shown by taking the complex conjugate of (27.5) that $$\mu_1\mu_2 =...

I did not understand why (3) is the Fourier transform of -k instead of k. Look, I used the defination, $$ \int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)\\ $$ Taking complex conjugate on both sides, $$...

I used the defination, $$\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)$$

[##f^*## represents complex conjugate of ##f##. ] [##\widetilde{f}(k)## represents fourier transform of the function ##f(x)##.] $$\begin{align} \int_{-\infty}^{\infty}f^*(x)e^{ikx}\,dx&=\int_{-\infty}^{\infty}f^*(x)\left(e^{-ikx}\right)^*\,dx\\...

Using fourier transformation, we have, Comparing with the equation, $$f(t)=\int_{-\infty}^{\infty}\delta(t-u)f(u)\,du$$ we have, Thus, $$\begin{align} \delta(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\,d\omega\\ &=\frac{1}{2\pi}\lim_{\Omega\rightarrow...

We know, $$\delta(x) = \begin{cases} \infty & \text{if } x = 0 \\ 0 & \text{if } x \neq 0 \end{cases}$$ And, also, $$\int_{-\infty}^{\infty}\delta(x)\,dx=1$$ Using Fourier Transformation, it can be shown that, $$\delta(x)=\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x}$$ Let's...