Calculate the expectation value of $L_{z}=-ih \left(x\frac{d}{dy}-y\frac{d}{dx} \right)$ for the state : $\phi(x,y,z) = a^{5/2} e^{-a(|x|+|y|+|z|) } (x+iy),$ where $a>0$. So, I need to calculate the integral: $ \langle L_{z} \rangle = \int_{-\infty}^{\infty}\overline{\phi}L_{z} \phi \hspace{1mm} dx \hspace{1mm} dy \hspace{1mm} dz = \int a^{5/2}e^{-a(|x|+|y|+|z|)}(x\frac{d}{dy}-y\frac{d}{dx})a^{5/2}e^{-a(|x|+|y|+|z|)}(x+iy) \hspace{1mm} dx \hspace{1mm}dy \hspace{1mm} dz .$
The possible cases are:
- $x>0, y>0, z>0$;
- $x<0, y<0, z<0$;
- $x>0, y<0, z<0$;
- $x>0, y>0, z<0$;
- $x<0, y>0, z<0$;
- $x<0, y<0, z>0$;
- $x>0, y<0, z>0$
which gives the six integrals:
$\displaystyle \int a^{5} e^{-2a(x+y+z)}(x^{2}+y^{2}\;dx\;dy\;dz$
$\displaystyle \int a^5 e^{-2a(-x-y-z)} (x^2+y^2)\;dx\;dy\;dz$
$\displaystyle \int a^5 e^{-2a(x-y-z)} (x^2+y^2)\;dx\;dy\;dz$
$\displaystyle \int a^5 e^{-2a(x+y-z)} (x^2+y^2)\;dx\;dy\;dz$
$\displaystyle \int a^5 e^{-2a(-x+y-z)} (x^2+y^2)\;dx\;dy\;dz$
$\displaystyle \int a^5 e^{-2a(-x-y+z)} (x^2+y^2)\;dx\;dy\;dz$
$\displaystyle \int a^5 e^{-2a(x-y+z)} (x^2+y^2)\;dx\;dy\;dz$
Does anybody see how to reduce this to a integral of one variable? Please do tell.
Edit: there is a horrible mistake, I forgot to multiply within the integral by $L_{z}$, Srivatsans solutions for the "wrong" integral stays true, though.
I come to:
$\displaystyle{ \int \int \int (x-iy)a^{5/2}e^{-a(|x|+|y|+|z|)}(-i\hbar(x\frac{d}{dy}-y\frac{d}{x})e^{-a(|x|+|y|+|z|)}a^{5/2}(x+iy)}dxdydz$
$z\frac{d}{dx}(e^{-a(|x|+|y|+|z|}(x+iy))$, according to wolframalpha this gives an a' in the derivative… how can this be?