Can we use this approximation for k_D<<k_F?

[Petar Mali]

For $$k_D<<k_F$$

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

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

$$|\frac{\hbar^2k_F^2}{2m}-\frac{\hbar^2k^2}{2m}|=|\frac{\hbar^2k_F\delta k}{m}+\frac{\hbar^2 \delta k^2}{2m}|$$

since $\delta k<k_D\ll k_F$ we can neglect the last term (quadratic in $\delta k$ and get

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

 

[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

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

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

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

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!