Can we use this approximation for k_D<<k_F?

[Petar Mali]

For k_D&lt;&lt;k_F

|\frac{\hbar^2k^2_F}{2m}-\frac{\hbar^2k^2}{2m}|\approx \frac{\hbar^2k_F}{m}|k_F-k|

Where k goes from k-k_D to k+k_D

k_F - Fermi wave vector
k_D - Debay wave vector

 

[jensa]

I suppose you mean k goes between k_F+k_D and k_F-k_D?

Yes it is valid. Let's denote k=k_F+\delta k, then

<br /> |\frac{\hbar^2k_F^2}{2m}-\frac{\hbar^2k^2}{2m}|=|\frac{\hbar^2k_F\delta k}{m}+\frac{\hbar^2 \delta k^2}{2m}|<br />

since \delta k&lt;k_D\ll k_F we can neglect the last term (quadratic in \delta k and get

<br /> |\frac{\hbar^2k_F\delta k}{m}|=\frac{\hbar^2k_F}{m}|k_F-k|<br /> [/itex]

 

[Petar Mali]

Thanks a lot! :) Yes from k_F-k_D to k_F+k_D.

I think that you have just a little mistake

You must write like

<br /> <br /> |\frac{\hbar^2k_F^2}{2m}-\frac{\hbar^2k^2}{2m}|=|-\frac{\hbar^2k_F\delta k}{m}-\frac{\hbar^2 \delta k^2}{2m}|<br /> <br />

To get C|k_F-k| or in case you wrote you will get

C|k-k_F|

You helped me a lot!

 

[jensa]

Thanks a lot! :) Yes from k_F-k_D to k_F+k_D.

I think that you have just a little mistake

You must write like

<br /> <br /> |\frac{\hbar^2k_F^2}{2m}-\frac{\hbar^2k^2}{2m}|=|-\frac{\hbar^2k_F\delta k}{m}-\frac{\hbar^2 \delta k^2}{2m}|<br /> <br />

To get C|k_F-k| or in case you wrote you will get

C|k-k_F|

I assumed that |\ldots | meant taking the absolute value. If this is so then overall signs do not matter.

Anyway, You're welcome.

 

[Petar Mali]

Yes! My mistake!