Dirac delta function of a function of several variables

[amjad-sh]

Homework Statement

In fact, I'm working on deriving the equations included in a theoretical condensed matter physics paper.
I reached the part where the normal charge current density is represented by $$J_y^N(z<0)=-\sum_{\mathbf{p},k}2i\dfrac{e}{m}p_yr_xsin(2k_z)\delta(\varepsilon_{p,k}-E_f)\triangle\mu_L^x$$ and we need to prove that it is equal to $\dfrac{-ek_f^2s}{(2\pi^2)}g(\nu,2k_fz)\triangle\mu_L^x$
It is not important to know what is normal charge current density, or other physical definitions I may mention Later in the post. Because I think knowing them will not serve in solving the problem.

Relevant Equations

$g(\nu,2k_fz)$ is a function.

We assume that $\hbar=1$ where $\hbar$ is the planck constant

$\varepsilon_{\mathbf{p},k}=\dfrac{k_x^2}{2m}+\dfrac{k_y^2}{2m}+\dfrac{k_z^2}{2m}$ is the energy corresponding to one particle and $k_x$,$k_y$ and $k_z$ are the momentum in x,y and z directions respectively.

$E_f=\dfrac{k_f^2}{2m}=$constant.

$\sum_{\mathbf p,k}$is the summation over all k's in the k-space.

$r_x=\dfrac{Ak_yk_z}{k_z+\sqrt{k_z^2-2mV})^2+c}$

$p_y=K_y$

$\delta(x)$ is the dirac delta function.

Form solid state physics, we know that the volume of k-space per allowed k-value is $\triangle{\mathbf{k}}=\dfrac{8\pi^3}{V}$
$\sum_{\mathbf{k}}F(\mathbf{k})=\dfrac{V}{(2\pi)^3}\sum_{\mathbf{k}}F(\mathbf{k})\triangle{\mathbf{k}}$
$\dfrac{1}{V}\sum_{\mathbf{k}}F(\mathbf{k})=\dfrac{1}{(2\pi)^3}\sum_{\mathbf{k}}F(\mathbf{k})\triangle{\mathbf{k}}$
When V$\rightarrow \infty \dfrac{1}{V}\sum_{\mathbf{k}}F(\mathbf{k}) \rightarrow \dfrac{1}{(2\pi)^3}\int F(\mathbf{k}) \, d\mathbf{k}$

Now $-\sum_{\mathbf{p},k}\dfrac{2ie}{m}p_yr_xsin(2k_z)\delta(\varepsilon_{p,k}-E_f)\triangle{\mu_L^x} \rightarrow -\dfrac{1}{(2\pi)^3}\int F(k_x,k_y,k_x)\delta(\varepsilon_{p,k}-E_f)dk_xdk_ydk_z$
Where $F(k_x,k_y,k_z)=\dfrac{2ie}{m}k_y\Bigg (\dfrac{2is\nu^2k_yk_z}{(k_z+\sqrt{k_z^2-2mv})^2+c}\Bigg )sin(2k_zz)$
$\delta(\varepsilon_{p,k}-E_k)=\delta(\dfrac{k_x^2}{2m}+\dfrac{k_y^2}{2m}+\dfrac{k_z^2}{2m}-\dfrac{k_f^2}{2m})$
then$$-\dfrac{1}{(2\pi)^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} F(k_x,k_y,k_x)\delta(\varepsilon_{p,k}-E_f)dk_xdk_ydk_z=$$
$$-\dfrac{1}{(2\pi)^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\dfrac{2ie}{m}k_y^2\Bigg (\dfrac{2is\nu^2k_z}{(k_z+\sqrt{k_z^2-2mv})^2+c}\Bigg )sin(2k_zz)\delta(\dfrac{k_x^2}{2m}+\dfrac{k_y^2}{2m}+\dfrac{k_z^2}{2m}-\dfrac{k_f^2}{2m})\, dk_xdk_ydk_z$$

What is stopping me of completing the derivation is that how we can use $\delta(\dfrac{k_x^2}{2m}+\dfrac{k_y^2}{2m}+\dfrac{k_z^2}{2m}-\dfrac{k_f^2}{2m})$ to solve the integral?
I searched a lot in google to find an identity regarding this type of Dirac delta function, but I couldn't find anything.

 

[Cryo]

I think similar functions occur when looking at covariant electrodynamics (e.g. Jackson). In general , assume $f$ is a 'nice' function, and so is $h$. Let function $h$ go to zero at $x_0$, and let $\frac{dh}{dx}_{x=x0}=h'_0\neq 0$

$\int dx f\left(x\right) \delta\left(h\left(x\right)\right)=f\left(x_0\right) \int^{\epsilon}_{-\epsilon} \delta\left(h'_0 x\right)dx=f\left(x_0\right) \int^{h'_0\left(\epsilon\right)}_{h'_0\left(-\epsilon\right)} \delta\left(y\right)\frac{dy}{h'_0}=\frac{f\left(x_0\right)}{h'_0}$

Now let the first derivative vanish too, but keep the second one, i.e. $h\left(x\approx x_0\right)=\frac{h''_0}{2}\left(x-x_0\right)^2$. This time the integral will only converge if $f\left(x\approx x_0\right)=g\left(x\right)\left(x-x_0\right)$

$\int dx f\left(x\right) \delta\left(h\left(x\right)\right)=\int^{\epsilon+x_0}_{-\epsilon+x_0} dx f\left(x\right) \delta\left(\frac{h''_0}{2}\left(x-x_0\right)^2\right)=\int^{\epsilon+x_0}_{-\epsilon+x_0} dx g\left(x\right)\left(x-x_0\right) \delta\left(\frac{h''_0}{2}\left(x-x_0\right)^2\right)$

Tidy up and change integration variable

$\int dx f\left(x\right) \delta\left(h\left(x\right)\right)=g\left(x_0\right)\,2\int^{\epsilon}_ 0 x\,dx\delta\left(\frac{h''_0}{2}x^2\right)=g\left(x_0\right)\,2\int^{\frac{h''_0}{2}\epsilon^2}_ 0 \frac{dy}{h''_0}\delta\left(y\right)=g\left(x_0\right)\int^{\frac{h''_0}{2}\epsilon^2}_ {-\frac{h''_0}{2}\epsilon^2} \frac{dy}{h''_0}\delta\left(y\right)=\frac{g\left(x_0\right)}{h''_0}$

$\int dx f\left(x\right) \delta\left(h\left(x\right)\right)=\lim_{x->x_0}\frac{f\left(x\right)}{h''_0 \cdot \left(x-x_0\right)}$

This is a generic approach. In your case I would seek to go to spherical coordinates for $k$ and evalate the delta-function in the integral over k-radius.

 

[amjad-sh]

Thanks for your reply!

This is a generic approach. In your case I would seek to go to spherical coordinates for kkk and evalate the delta-function in the integral over k-radius.

I went to spherical coordinates, but I got stuck with a a little bit nasty integral.
I will show you how I proceeded.
1-First I want to note that I wrote $r_x$ wrongly.
$r_x=\dfrac{2is\nu^2k_yk_z}{(k_z+\sqrt{k_z^2-2mv})^2+(s\nu^2)^2(k_x^2+k_y^2)}$
2- Now in spherical coordinates $\delta(\dfrac{1}{2m}(k^2-k_f^2))=2m\delta(k^2-k_f^2)=2m(\dfrac{1}{2k_f})\Big[\delta(k-k_f)+\delta(k+k_f)\Big]=\dfrac{m}{k_f}\Big[\delta(k-k_f)+\delta(k+k_f)\Big]$
Where I have used the relations $\delta(ax)=\dfrac{1}{a}\delta(x)$ and $\delta(x^2-a^2)=\dfrac{1}{2a}\Big[\delta(x-a)+\delta(x+a)\Big]$ and$k^2=k_x^2+k_y^2+k_z^2$

3- $$j_y^N(z<0)=-\dfrac{1}{(2\pi)^3}\int_0^{+\infty}\int_0^{\pi}\int_{0}^{2\pi}k^2sin(\theta)\dfrac{2ie}{m}k_y^2\Bigg(\dfrac{2is\nu^2k_z}{(k_z+\sqrt{k_z^2-2mV})^2+(\nu^2s)^2(k_x^2+k_y^2)}\Bigg)sin(2k_zz)\triangle\mu_L^x\dfrac{m}{k_f}\delta(k-k_f)\, d\theta \, d\phi \, dk$$
Now by using the change of variable $k_x=ksin(\theta)cos(\phi)$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad k_y=ksin(\theta)sin(\phi)$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad k_z=kcos(\theta)$
$$j_y^N(z<0)=-\dfrac{1}{(2\pi)^3}\int_0^{+\infty}\int_0^{\pi}\int_{0}^{2\pi}k^2sin(\theta)\dfrac{2ie}{m}\dfrac{k^2sin^2(\theta)sin^2(\phi)2is\nu^2kcos(\theta)}{(kcos(\theta)+\sqrt{k^2cos^2(\theta)-2mV})^2+(\nu^2s)^2k^2sin^2(\theta)}sin(2kcos(\theta)z)\triangle\mu_L^x\dfrac{m}{k_f}\delta(k-k_f)\, d\theta \, d\phi \, dk$$

$$j_y^N(z<0)=-\dfrac{1}{(2\pi)^3}\int_0^{+\infty}\int_0^{\pi}\int_{0}^{2\pi}\dfrac{k^5\dfrac{2ie}{m}sin^3(\theta)sin^2(\phi)2is\nu^2cos(\theta)sin(2kzcos(\theta))\triangle\mu_L^x}{k^2cos^2(\theta)+k^2cos^2(\theta)-2mV+2kcos(\theta)\sqrt{k^2cos^2(\theta)-2mV}+(\nu^2s)^2k^2sin^2(\theta)}\dfrac{m}{k_f}\delta(k-k_f)\, d\theta \, d\phi \, dk$$

4- Now by using the integral $\int_0^{2\pi} sin^2(\phi) \,d\phi=\pi$$$j_y^N(z<0)=-\dfrac{1}{(2\pi)^3}\Big(\dfrac{-4e\pi}{k_f}\Big)\int_0^{+\infty}\int_0^{\pi}\dfrac{\nu^2k^5sin^3(\theta)scos(\theta)sin(2kzcos(\theta))\triangle\mu_L^x}{k^2cos^2(\theta)+k^2cos^2(\theta)-2mV+2kcos(\theta)\sqrt{k^2cos^2(\theta)-2mV}+(\nu^2s)^2k^2sin^2(\theta)}\delta(k-k_f)\, d\theta \, dk$$

5- By using $\int_0^{+\infty}f(k)\delta(k-k_f)dk=f(k_f)$

$$j_y^N(z<0)=\dfrac{1}{(2\pi)^3}\Big(\dfrac{4e\pi}{k_f}\Big)\int_0^{\pi}\dfrac{\nu^2k_f^5sin^3(\theta)scos(\theta)sin(2k_fzcos(\theta))\triangle\mu_L^x}{k_f^2cos^2(\theta)+k_f^2cos^2(\theta)-2mV+2k_fcos(\theta)\sqrt{k_f^2cos^2(\theta)-2mV}+(\nu^2s)^2k_f^2sin^2(\theta)}\, d\theta $$

$$j_y^N(z<0)=\dfrac{1}{2(2\pi)^3}(4e\pi)\int_0^{\pi}\dfrac{\nu^2k_f^4sin^3(\theta)scos(\theta)sin(2k_fzcos(\theta))\triangle\mu_L^x}{k_f^2\Bigg(cos^2(\theta)-\dfrac{mV}{k_f^2}+cos(\theta)\sqrt{cos^2(\theta)-\dfrac{2mV}{k_f^2}}+\dfrac{(\nu^2s)^2sin^2(\theta)}{2}\Bigg)}\, d\theta $$

6- Let $u=cos(\theta)$ then $du=-sin(\theta)d\theta$

$$j_y^N(z<0)=\dfrac{-k_f^2es}{4(\pi)^2}\int_1^{-1}\dfrac{\nu^2(1-u^2)usin(2k_fzu)\triangle\mu_L^x}{u^2-\dfrac{mV}{k_f^2}+u\sqrt{u^2-\dfrac{2mV}{k_f^2}}+\dfrac{(\nu^2s)^2(1-u^2)}{2}}\, du $$

So we can say now that :
$j_y^N(z<0)=\dfrac{-k_f^2es}{(2\pi)^2} g(\nu,2k_fz)\triangle\mu_L^x$
where$$ g(\nu,2k_fz)=
\int_1^{-1}\dfrac{\nu^2(1-u^2)usin(2k_fzu)}{u^2-\dfrac{mV}{k_f^2}+u\sqrt{u^2-\dfrac{2mV}{k_f^2}}+\dfrac{(\nu^2s)^2(1-u^2)}{2}}\, du $$

Does my approach look right, did I do something wrong?
and do you think this integral can be solved analytically? or I need a computer?

 

[Cryo]

Sorry for delay. The approach seems right to me, although I do not have time to check the maths thoroughly. The final integral does look nasty, but it is easy to compute numerically (finite domain), so I usually do not bother in such cases.

If I did want to bother with it, I would first try to plot the integrand, estimate the rough magnitude of different terms, and see if there are any approximations I could make. Maybe some of the nastyness never comes into play in your problem.

I would also try looking at the integrand on the complex plane and deforming the path of integration to simplify the evaluation. But, I am not sure this would help.