Ashcroft and Mermin's problem 13.4.

[MathematicalPhysicist]

Homework Statement

I need some help on how to solve this problem.

I am attaching the problem:
[![enter image description here][1]][1]
[![enter image description here][2]][2]

For item (a) in the problem, I think I need to use eq. (13.21)
in the following pic:
[![enter image description here][3]][3]

I think I need to Fourier transform equation (13.21) but don't see how exactly to get the form in equation (13.77).

What I mean is that:
$$g(r,k,t)=\int g^0(k)e^{i\omega t}d\omega + \int \int_{-\infty}^t d\omega dt' e^{-(t-t')/\tau(\epsilon(k))}(-\partial f /\partial \epsilon) \times v(k(t'))[-eE(t')-\nabla \mu(t')] e^{i\omega t}$$
I think I need to use here the identity: $g^0(k)=f(k)-f_{coll}$, but to tell you the truth I don't know how to proceed from there?

For (b), the induced current is $j = -nev(k), so if q\cdot j(q,\omega) = -ev(k)\cdot q \Re(i\partial n_{eq}(\mu)/\partial \mu \delta \mu (q,\omega)e^{i(q\cdot r -\omega t)}) = -\partial \rho /\partial t = -e\omega\Re(\partial n_{eq}/\partial \mu \delta \mu(q,\omega)ie^{i(q\cdot r -\omega t)})$; so by equating terms I get the condition $v(k)\cdot q =\omega$, but I didn't use equation (13.77) for item (b), so perhaps I am wrong here?

For (c), just plug in (13.77) the fact there's no induced charge i.e that $\rho=0$ which means that $\delta \mu(q,\omega)=0$.
But how to show that a sufficient condition for (13.79) to be valid is that the electric field be perpendicular to a plane of mirror symmetry in which the wave vector q lies?

Thanks for your help.

[1]: https://i.stack.imgur.com/YeJyP.png
[2]: https://i.stack.imgur.com/KzopQ.png
[3]: https://i.stack.imgur.com/JgzOv.png

Homework Equations

The Attempt at a Solution

 

[mark726]

Hello,

Thank you for posting your problem and providing the relevant equations and figures. I will do my best to help you solve this problem.

For item (a), you are correct in thinking that you need to use eq. (13.21) in order to solve the problem. To Fourier transform this equation, you can use the identity:

$$\int_{-\infty}^{\infty} e^{-i\omega t}f(t)dt = F(\omega)$$

where F(ω) is the Fourier transform of f(t). Applying this identity to eq. (13.21), we get:

$$g(r,k,t) = \int_{-\infty}^{\infty} g^0(k)e^{i\omega t}d\omega + \int_{-\infty}^{\infty} \int_{-\infty}^t d\omega dt' e^{-(t-t')/\tau(\epsilon(k))}(-\partial f /\partial \epsilon) \times v(k(t'))[-eE(t')-\nabla \mu(t')] e^{i\omega t}$$

Next, we can use the identity you mentioned, g^0(k) = f(k) - f_{coll}, to rewrite the first term in the equation as:

$$\int_{-\infty}^{\infty} g^0(k)e^{i\omega t}d\omega = \int_{-\infty}^{\infty} [f(k) - f_{coll}]e^{i\omega t}d\omega = \int_{-\infty}^{\infty} f(k)e^{i\omega t}d\omega - \int_{-\infty}^{\infty} f_{coll}e^{i\omega t}d\omega = F(k) - F_{coll}$$

where F(k) is the Fourier transform of f(k) and F_{coll} is the Fourier transform of f_{coll}. Similarly, we can rewrite the second term as:

$$\int_{-\infty}^{\infty} \int_{-\infty}^t d\omega dt' e^{-(t-t')/\tau(\epsilon(k))}(-\partial f /\partial \epsilon) \times v(k(t'))[-eE(t')-\nabla \mu(t')] e^{