1
$\begingroup$

Ashcroft Mermin chapter 9 number 2

Hey, I'm just sutck on the problem 2 of the Ashcroft Mermin. I did prove the density of state but I'm having a hard time finding $k_\text{min}$ and $k_\text{max}$ (for the subquestions).

The question is to calculate $g(E)$. We start with :

$$E = \frac{\hbar^2 k^2_\bot}{2m}+h_\pm (k_\parallel)$$

where

$$h_\pm (k_\parallel) = \frac{\hbar^2}{2m} \left\lbrack k^2_\parallel + \frac{1}{2}(G^2 - 2 k_\parallel G) \right\rbrack \pm \left\lbrace \left\lbrack \frac{\hbar^2}{2m}\frac{1}{2} (G^2 - 2 k_\parallel G) \right\rbrack^2 +|W_{\mathbf{G}}|^2\right\rbrace^{1/2}$$

Using:

$$g_n (E) = \int \frac{d \mathbf{k} }{4 \pi^3 } \delta(E- E_n (\mathbf{k}))$$

and doing the integral in polar coordonates, I've obtained:

$$g(E) = \int^{2 \pi}_0 d \theta \int^{k_{\parallel}^{max}}_{k_{\parallel}^{min}} d k_\parallel \int^{\infty}_{-\infty} d k_\bot k_\bot \delta(E- E_n (\mathbf{k}))$$

With some properties of the Delta functions (notably $\delta(a c ) = \delta (a)/|c|$ and $\delta(f(x)) = \delta (x - x_i )/ |\partial f(x) / \partial x|_{x_i}$) I end up with

$$g(E) = \frac{1}{4 \pi^2} \left(\frac{2m}{\hbar^2} \right) (k_\parallel^{\text{max}}-k_\parallel^{\text{min}})$$

The subquestion is: Show, for the lower band that: $k_\parallel^\text{min} = - \sqrt{\frac{2 m E}{\hbar^2}} + O(W_G^2)$

for $E>0$ and $k_\parallel^\text{max} =G/2$ , if the constant energy surface go through the Bragg plan ($E_{G/2}-|W_G| < E < E_{G/2}+|W_G|$). I suspect it has something to do with the integral limits (for the lower bands, they are from $-\infty$ and $G/2$, for the higher band they are $G/2$ to $\infty$), but I just can't figure out how to complete the math properly. Then we have to prove that for the higher band, we must find that: $$g_+ (E) = \frac{1}{4 \pi^2} \left(\frac{2m}{\hbar^2} \right) (k_\parallel^{\text{max}}-G/2)$$ Which should be easily done once you give the physic argument that the minima of the upper band is at $k_\parallel = G/2$, but I'm not sure it's a complete answer/justification for it.

Thank you :)

$\endgroup$
1
  • $\begingroup$ Thank you very much for the corrections $\endgroup$
    – Jacques
    Feb 23 at 21:10

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.